basic school mathematics

Number Systems 1

Module 1Algebra

Since the dawn of civilization, man tried to have a count of his belongings–goods, stones,animals, trees, etc. Whenever the animals were taken out of the enclosure for grazing, for eachanimal taken out, a scratch was put on the ground/stone. Thus, the number of animals takenout for grazing were equal to the number of scratches. (This is analogous to tally marks usednow a days). On return, for every one animal returning to the enclosure a scratch was erased.In this way, even without knowing the counting, they used to save their belongings. Slowlythe civilizations advanced and first came the counting numbers (natural numbers). You willbe happy and proud to know that the present system of numeration, including numerals 0, 1,2, 3, ..., 9 and place value system were the discoveries of our ancient Indians to the world.From India, these numbers reached the reign of Arabian king Al-mansur whose wise manAl-khowarizmi, translated the works of Indian scholars and mathematicians. From Arabia, thenumerals reached the western world. Therefore, these are called Hindu-Arabic Numerals.

You know that Algebra is generalised form of Arithmetic in which variables are used fornumbers. Aryabhatt (476 AD) and Brahmgupta (578 AD) were the first Indian Mathematicianswho used variables for numbers and called them “Yavat-Tawat”. They illustrated the sum,difference, product and division of expressions using variables and even found their squares,cubes, square-roots, cube-roots. Aryabhatt and Brahmgupta worked on solving linear, quadraticand indeterminate equations also. They called the method of solving indeterminate equationsas “Chakrawal” and gave “Avyakat Ganit” to algebra. Bhaskaracharya and Mahaviracharya alsocontributed a lot to this, especially ratio and proportion and extended the works of previousMathematicians on equations and indices and surds. The name “Beejganit” was given toAlgebra by Bhaskaracharya. The credit of calling this as Algebra goes to Al-khowarizmi, thewise man of Al-Mansur.

In this module, we shall study about number system, polynomials, factorisation of algebraicexpressions, simplificaiton of rational expressions, solving linear and quadratic equations,indices and surds, Arithmetic and Geometric Progressions.

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1

Number Systems

1.1 INTRODUCTION

One of the greatest inventions in the history of civilization is the creation of NUMBERS. Youcan imagine the state of confusion, if you did not know about natural numbers or countingnumbers. The introduction of those numbers enabled us to answer the question ‘How many’?You may recall your familiarity with the concepts of natural numbers, whole numbers, integers,fractions and rational numbers. These helped us to count, to know about the number zerorepresenting nothingness, parts of a whole, describe opposites like profit and loss, rise and fall,going towards east and west etc. But one could have easily got along without the introductionof these numbers except the natural numbers. However, the working out of problems becameeasier with the introduction of numbers upto rational numbers. All these numbers mentionedabove are rational numbers. Recall that a natural number is a rational number, a whole numberis a rational number, a fraction is also a rational number and so also is an integer. At this stage,it is pertinent to ask ourselves a question. Are all numbers rational numbers ?

Is it possible to solve all mathematical and life problems with the help of rational numbers?The answer to this question is an emphatic ‘NO’. For example, given a unit of length, we arenot in a position to exactly measure all distances if we have with us rational numbers only.Likewise, it is not possible to find out the square root of an arbitrary natural number as a rationalnumber, as for example, you know that 2 is not a rational number. To overcome suchdifficulties, it is essential to extend to the system of real numbers.

In this lesson, you will be able to recall all that you may be knowing about rational numbersbut also extend the same to the system of real numbers. This is a big leap forward in thestudy of Mathematics.

1.2 OBJECTIVES

After studying this lesson, the learner will be able to :

illustrate the extension of system of numbers from natural numbers to real (rational andirrational) numbers.

identify different types of numbers.

express an integer as a rational number

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express a rational number as a terminating or non terminating but recurring decimal andvice-versa.

find a rational number between two given numbers.

represent a rational number on the number line

cite examples of irrational numbers

represent 2 3 5, , on the number line

find an irrational number between two given numbers.

round off a rational or irrational number to a given number of decimal places.

perform the four fundamental operations of Arithmetic i.e., addition, subtraction,multiplication and division on real numbers.

1.3 EXPECTED BACKGROUND KNOWLEDGE

Concept of Natural numbers, whole numbers, fractions, integers and rational numbers.

Prime and composite numbers and co-prime numbers.

HCF and LCM of two or more natural numbers.

1.4 RATIONAL NUMBERS

You may recall your familiarity with rational numbers. For example

1 2 3 12

45

47 0 9

4224, , , , , , , , , ...−

are all rational numbers.

1, 2, 3, 4, ... are counting numbers or Natural Numbers. With the help of theseyou are able to answer the question ‘How many’ ?

If we introduce another number ‘zero’ into the system of natural numbers, we get the systemof Whole Numbers as 0, 1, 2, 3, 4, ...

Notice that 0 is not a natural number but it is a whole number. All natural numbers are wholenumbers as well.

By introducing into the system of whole numbers the opposite (negatives) of natural numbers,we get the numbers

... –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, ...

called integers. Similarly, the numbers

12

23

64

57, , , , ...

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are called fractions. You may have a look at some numbers like

−− −

23

58

621, , , ...

These numbers do not fall into the categories mentioned above. If we include these numbersas well in our system of numbers, we get what is known as the system of ‘Rational Numbers’.Thus,

A number of the form pq where p and q are integers and q 0≠ is called a rational

numbers.

Here p is called the numerator and q the denominator of the rational number.

You may notice that all natural numbers, whole numbers, fractions, integers are rationalnumbers but a rational number may not be a natural number, a whole number, a fraction or

an integer. For example, the number 11

7− is not a natural number. It is not a whole number.

It is neither a fraction nor an integer. But certainly 11

7− is rational number. Can you justify

this statement ? Surely, you may see that 11

7− is a number of the form pq where p = 11,

q = –7 and q ≠ 0.

A rational number whose numerator and denominator are both positive integers or both negativeintegers is a positive rational number. Similarly, if in a rational number, one out of itsnumerator and the denominator is positive and the other is negative is a negative rationalnumber. Thus, for examples

23

711, −− are positive rational numbers

and −−

27

117, are negative rational numbers.

Notice that 0 01

= is the zero rational number.

It may be noted that

23 =

−−

= = −−

23

46

69 , and

117− =

− =−

= −117

2214

3321

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14.1 Rational Number in its Lowest Terms

Consider the rational number 2436 . Here its numerator and denominator are 24 and 36. But these

two natural numbers have a Highest Common Factor, namely 12 and so the number could aswell as be written as

2 123 12×× or simply

23

23 is said to be rational number in the lowest terms. Similarly, the rational number

−1248 , could

easily be rewritten as

− ××

= −1 124 12

14

The numerator and denominator of a rational number in lowest terms are co-prime numbers.

Note : The number −1248 is as well written as −

1248 .

CHECK YOUR PROGRESS 1.1

1. From the following numbers, pick out(a) natural numbers(b) integers which are not natural numbers(c) whole numbers which are not natural numbers(d) fractions which are not natural numbers(e) numbers which are neither whole numbers, nor integers not fractions

1 7 3 78

43 0 23 2

7153

116, , , , , , , , ,− − − − .

2. Express the following integers as rational numbers :(i) – 4 (ii) 23 (iii) 0 (iv) 1 (v) – 1

3. Express the following rational numbers in their lowest terms :

(i) − 1536 (ii)

4414 (iii)

−216 (iv)

812−

1.5 DECIMAL REPRESENTATION OF A RATIONAL NUMBER

You may be familiar with the process of representing a rational number in the form of a decimal.We illustrate this process with the help of a few examples.

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Example 1.1 : Represent the numbers 14

35

,− , in the decimal form.

Solution : By actual division, we know that

(i)14 = 0.25

(ii) − 35 = – 0.6

In the cases above, we see that in the process of repeated division by the denominator, we endup after a finite number of steps, when we get the remainder as zero. But this is not alwaysso. To illustrate this we consider another example.

Example 1.2 : Express the following rational numbers

(i) −23 (ii)

67

in the decimal form.

Solution : (i) −23 = – 0.6666 .....

Please note that in this case, the process ofrepeated division never comes to an end. In factat every step, we get the remainder 2.

(ii) 67 = 0.857142 857142 .....

In this example, we find that the remainder at eachstep keeps changing till we arrive at a stage thatthe remainder repeats and then on further divisionthe quotient will start repeating.

In examples, 1.1 and 1.2 above, we see that eitherthe remainder will be zero after a finite numberof steps or it will start repeating after a finitenumber of steps. In the first case we say that thedecimal is terminating and in the other we say thatthe decimal is non-terminating but repeating. Infact we have the following important statement.

4) 1.0 (0.25820200

5) 3.0 (0.6300

3) 2.0 (0.66...18

2018

20182

7) 6.0 (0.85714256

4035

5049

107302820146

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A rational number is either a terminating decimal or a non-terminating butrecurring (repeating) decimal.

In the following examples, we try to represent a decimal (terminating or non-terminating but

repeating decimal) in the form pq .

Examples 1.3 : Express (i) 0.72, (ii) 0.125 in the form pq .

Solution : (i) 0.72 = 72100

1825=

(ii) 0.125 = 1251000

18=

Example 1.4 : Express (i) 0.33....., (ii) 0.234 234 .... in the form pq .

Solution : (i) You may write

a = 0.33 .... ...(1)

Multiply both sides by 10 to get

10a = 3.33 .... ...(2)

Subtracting (1) from (2), we get

9a = 3.33.... – 0.33....= 3

∴ a = 13

Thus, 0.33 = 13

(ii) Let b = 0.234234.... ...(1)

Multiplying both sides by 1000 to get

1000 b = 234.234.... ...(2)

Subtracting (1) from (2), we get

999 b = 234

∴ b =234999

26111

= .

The examples help us to assign at a result which can be stated in the following form.

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A terminating decimal or a non-terminating but recurring decimal is a rationalnumber.

Note : Recurring decimals such as

0.66..., 0.234234..., 3.142857142857... are also written as

0 6. , 0 234. , 3 142857. .

In fact a digit or a group of digits which repeat are put below a bar to indicate that theserepeat again and again.


1. Represent the following rational numbers in the decimal form :

(i)2140 (ii)

1625 (iii)

138 (iv)

156 (v)

9135

2. Represent the following rational numbers in the decimal form :

(i)57 (ii) − 5

6 (iii)1511 (iv) − 27

13

3. Represent the following decimals in the form pq :

(i) 2.3 (ii) – 7.12 (iii) – 0.315 (iv) 8.146

4. Represent the following decimals in pq form :

(i) 0.333... (ii) 3 12. (iii) – 0.315315...

1.6 RATIONAL NUMBERS BETWEEN ANY TWO RATIONAL NUMBERS

Given two rational numbers, can you find a rational number between them ?

To answer this question, we consider the following examples :

Example 1.5 : Find a rational number between two rational numbers 12 and

34 .

Solution : Consider the number 12

34

2

+

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You may simplify this to get

2 342

+

i.e., 58

Surely, 58

12> and 3

458> . You may verify this by computing 5

812− and 3

458− which come

out to be both positive numbers, Thus, 58 is a rational number such that

12

58

34

< <

Think for a while, what we have done.

We added the two numbers and divided their sum by 2. The resulting number is between thetwo given numbers. It will be greater than the smaller of the two numbers and less than thelarger number.

Let us now consider the following question :

“How many rational numbers lie between two given rational numbers ?”

Can you guess the answer ? In fact you can find as many rational numbers between two givennumbers as you like. Can you think of as to how you may find many rational numbers lyingbetween two rational numbers ?

Example 1.6 : Find a rational number between 0.12 and 0.13.

Solution : Consider the number

012 0132

. .+

= 0.125

It is a number greater than 0.12 and less than 0.13.

Can you find some other rational numbers between 0.12 and 0.13 ? How many such likenumbers you can find ?

These are some of interesting aspects of the study of rational numbers.


1. Find a rational number between the following rational numbers

(i) 78 and 8

7 (ii) 2 and 3 (iii) − −34

13,

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2. Find two rational numbers between the following rational numbers :

(i) − 12 and 2

3 (ii) − 23 and − 1

4

3. Find two rational numbers between the following rational numbers

(i) 0.23 and 0.24 (ii) 7.31 and 7.32

1.7 THE NUMBER LINE

Rational numbers can be represented on a line which is known as the number line in thefollowing manner.

Consider a line as shown in Fig 1.1. We fix a point O on its line. We choose a convenientunit of length and mark points on the line on both sides of O at fixed distances equal to theunit of length chosen. These points represent the number 0, 1, 2, 3, 4, .... –1, –2, –3, –4, .....

as shown in the Fig 1.1. Suppose now we have to represent the number 74 on the line. We

divide the distance between 1 and 2 into four equal parts and mark the point after three equalparts as shown in the figure.

Fig. 1.1

In this way we can represent any positive rational numbers by a point on the number line tothe right of the point O.

Similarly if we were to represent the number −12 on the number line, we divide the line segment

between –1 and 0 into two equal parts and mark the point −12 as shown in the Fig 1.1. In

this manner you may see that any given rational number can be represented by a point on thenumber line. You may also note that the number 0 is represented by the point O on the numberline.

Example 1.7 : Represent the number 1.2 on the number line.

Solution :

Fig. 1.2

Consider the number line with points on it marked as –2, –1, 0, 1, 2 as shown in Fig. 1.2 above.Divide the line segment between 1 and 2 into 10 equal parts. Mark a point P as 1.2, as is shown

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in the Fig, 1.2, 2 steps ahead of the point representing the number 1. P is the point representingthe number 1.2 on the number line.

You may note that any rational number whether in the form pq or in the decimal form can

be represented by a point on the number line.


1. Represent the following numbers by points on the number line

(i)32 (ii) − 1

3 (iii) 1.5 (iv) – 1.3

2. Find the numbers corresponding to the points O, P, Q and R on the number line as shownin the Fig. 1.3.

Fig. 1.3

1.8 IRRATIONAL NUMBERS

In section 1.5, you have seen that when a rational number is represented as a decimal, theneither this decimal is terminating or it is a non-terminating but repeating decimal. The questionthat arises is this. Are there decimals which are neither terminating nor non-terminating butrepeating decimals ? The answer to this question is ‘Yes’. Consider for example the decimal

0.101001000100001 ...

This decimal has been written in such a manner that it has a definite pattern and so you cankeep on writing it indefinitely. But this pattern is such that no block of digits repeats againand again.

It is an example of a non-terminating and non-repeating decimal. Similarly, you may considerthe decimal

0.123456789 10 11 12 13 ...

Can you write the next nine digits in the decimal ?

Note that, all that has been done is that we have written successively all natural numbers inthe ascending order. The next six digits will be 14 15 16. Note once again you can continuethis process endlessly. Thus, there is a definite pattern to write the digits but no block of digitsis repeating again and again. This is another example of a non-terminating and non-repeatingdecimal.

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The two examples we have discussed above indicate that there are decimals which are notrational numbers. Why ? We conclude that if we have to possess in our fold all decimals, namely.

(i) terminating decimals

(ii) non-terminating but repeating decimals

(iii) non-terminating and non-repeating decimals

then the system of rational numbers is not adequate and so we must extend this system to includenumbers which are not rational i.e., irrational numbers.

Note that the name rational is derived from the word 'ratio', as a number pq , q ≠ 0 is the ratio

of two integers p and q. Also number which is not a ratio of integers p and q, q ≠ 0 is namedas an irrational number.

1.8.1 Inadequacy of rational numbers

You may recall that we would not have been able to answer the question 'how many' ?, if wedid not know counting numbers or natural numbers. Let us now try to examine what we shallnot be able to do if we have in our possession the rational numbers only. We put to ourselvesthe following question.

Can we measure the length of any given line segment in terms of a prescribed unit of length,with the help of rational numbers ?

The answer to this question is an emphatic ‘No’,as is clear from the following example.

Consider a square ABCD of each side of unitlength. The diagonal AC of this square has itslength 2 units. You also know that 2 is nota rational number as there is no rational numberwhose square is 2.

We conclude that we cannot exactly measure the length of the line segment AC in terms ofrational numbers, if the unit of length given to us is AB.

It is this inadequacy which necessitates the extension of the system of rational numbers to asystem which includes in it rational as well as irrational numbers.

There is yet another point which necessitates the extension of rational numbers. We discussthe same in the following paragraph.

Recall that you have seen in section 1.7, that to each rational number these corresponds a uniquepoint on the number line. Let us now consider the converse of it. Given a point on the numberline, will it always correspond to a rational number ? The answer to this question is also a'NO', as is clear from the following example.

Fig. 1.4

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On the number line mark the points corresponding to the number 0 and 1, as shown inFig 1.5.

Fig. 1.5

Construct a square on the line segment. The diagonal of the square has length 2 . With O

as centre and radius OA, draw an arc cutting the number line in P. Then OP = 2 . Thus, the

point P on the number line corresponds to the number 2 which is irrational. Thus, there isa point namely P, on the number line which does not correspond to a rational number.

How many such points are those on the line which do not correspond to rational numbers ?In fact there are many such points and their number is infinite i.e. there are infinitely manypoints on the number line which do not correspond to rational numbers.

In fact, if P corresponds to the number 2 as in Fig 1.6, we construct a rectangle with one

side OP and the other side 1. The diagonal of the rectangle will be OQ = 3 . With O as centreand OQ as radius draw an arc cutting the number line in the point R. Then the point Rcorresponds to the irrational number 3 .

Fig. 1.6

Similarly, by constructing the rectangle with sides 2 and 1 as shown in Fig 1.7 and with Oas centre and diagonal OA = 5 , draw an arc cutting the number line in P. This point P

corresponds to 5 , again an irrational number.

Fig. 1.7

Once you have the points on the number line corresponding to irrational number 2 3 5, , etc,

you can get points on the line corresponding to number like 12 2 2

3 3 34 5 7, , ,− p

q , etc.

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where pq is any rational numbers. All these points correspond to irrational numbers. This

discussion leads us to the following conclusion.

The number line consisting of points corresponding to rational numbers has gaps on it. In otherwords it is not complete. But the number line consisting of points corresponding to all rationaland irrational numbers is without any gaps and therefore is complete. This number line is calledthe real line and the one corresponding to rational numbers is known as the rational line. Wemay therefore say that the rational line is not complete whereas the real line is complete.

We have thus extended the system of rational numbers to include in it all irrational numbersas well. This system is called the Real Number System.

The system of numbers consisting of all rational and irrational numbers is calledthe system of real numbers or the Real Number System.


1. Write down the next three digits in the following numbers :

(i) 0.246 ...

(ii) 1.35 ...

(iii) 3.6912 ...

(iv) 0.11011110 ...

2. Represent the following numbers on the number line :

(i)13

2 (ii) 1 2+ (iii) − 3 (iv) − +2 3

1.9 IRRATIONAL NUMBER BETWEEN TWO GIVEN NUMBERS

Recall that you had found rational numbers between two given rational numbers. In thefollowing examples we illustrate how do we find irrational numbers between any two numbers.

Example 1.8 : Find an irrational number between the numbers 1 and 2.

Solution : Consider the number 1 2× . It is the square root of the product of the two numbers1 and 2.

You may refer to Fig 1.5 and see that the point P corresponding to 2 lies between the points

1 and 2 and so 2 is such that 2 is greater than 1 and less than 2. So 1 2× i.e., 2 isan irrational number lying between the number 1 and 2.

Number Systems 15

Example 1.9 : Find an irrational number lying between the number 2 and 3.

Solution : Consider the number 3 2

2+

We have 3 2

22+ − =

3 2 2 22

3 22

+ − = −

As 2 is less than 2, it is less than 3.

∴ 3 2

2−

is a positive real number

∴ 3 2

2+

is greater than 2 .

Also, you can verify than 3 is greater than the number 3 2

2+

.

∴ The required number is 3 2

2+

.


1. Find an irrational number between the following pairs of numbers :

(i) 2 and 3 (ii) 3 and 2

(iii) 2 and 3 (iv) 2 and 8

2. How many irrational numbers are there between 1 and 2 ? Give three examples of irrationalnumbers between these two numbers.

1.10 ROUNDING OFF NUMBERS TO A GIVEN NUMBER OF DECIMAL PLACES

Quite often it is convenient to write the approximate value of a real number to a specifiednumber of decimal places. We illustrate this process with the help of the following examples.

Example 1.10 : Express the number 2.178473

approximately by rounding it off to three places of decimal.

Solution : We look up the 4th place after the decimal point. In this case, it is 4 which is lessthan 5. So The approximate value of 2.178473, upto three places of decimal is 2.178.

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Example 1.11 : Find the approximate value of 2.178473... correct upto 4 places of decimal.

Solution : The fifth place of decimal (one after the fourth) is 7. So, we add to 4 the fourthplace 1, because 7 is greater than 5.

∴ The required approximate value of the number is 2.1785

Thus, we observe that, to round off a number to some decimal places, we observethe next digit in the decimal part of number.

(i) If the digit is less than 5, we ignore it and give the answer.

(ii) If the digit is 5 or more than 5, we add 1 to the preceding digit to get therequired number.


1. Write down the approximate value of the numbers, correct upto 4 places of decimals

(i) 0.7777 ... (ii) 7.325444 ...

(iii) 1.01101 ... (iv) 12.34567 ...

(iv) 3 142857. (iv) 3.14159 ...

LET US SUM UP

The Real Number System consists of rational and irrational numbers.

With the help of real numbers, we can measure exactly any length in terms of a unit oflength.

There are gaps on the rational line.

The real line is complete i.e., there are no gaps on it.

A rational number, when represented as a decimal is either terminating or non-terminatingbut repeating decimal.

An irrational number is a non-terminating and non-repeating decimal.

Between any two given numbers, there lie an infinite number of rational as well asirrational numbers.

TERMINAL EXERCISE

1. Express the decimal, in the form pq

(i) 0.3125 (ii) 0.1111 ... (iii) 0.9999 ...

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2. Express the following numbers as non-terminating but repeating decimals :

(i)27 (ii)

413 (iii)

611 (iv) 1

3. Find two rational and two irrational numbers between the following pairs of numbers :

(i) 2 and 3 (ii) 3 and 2

(iii) 1.2 and 1.3 (iv) 2 and 3

4. Represent the following numbers on the number line :

(i) − 32 (ii) 1.333...

(iii) 2 2 (iv) 2 5+

5. Find the value of the following numbers correct upto 3 places of decimal :

(i)23 (ii) 2 (iii) 1.7326

(iv) 3.142857 (iv) 0.99999...

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ANSWERS

Check Your Progress 1.1

1. (a) 1, 7 (b) –3, 0, –23, − 153 (c) 0 (d) 7

827

116, , (e) None

2. (i)−41 (ii) 23

1 (iii) 01 (iv) 1

1 (v) −11

3. (i)−512 (ii) 22

7 (iii) −72 (iv) −2

3


1. (i) 0.525 (ii) 0.64 (iii) 1.625 (iv) 2.5 (v) 2.6

2. (i) 0.714285 (ii) – 0.833 (iii) –1.363 (iv) – 2.076923

3. (i) 2310 (ii) − 168

25 (iii) −63200 (iv) 4073

500

4. (i) 13 (ii) 103

33 (iii) 3511


1. 0, 1.5, 2.5, –1.5


1. (i) 0.24681012...

(ii) 1.357911...

(iii) 3.6912151821...

(iv) 0.110 11110 1111110 111111110...


1. (i) 5 (ii) 3 22+ (iii) 2 3

2+ (iv) 3

2 2

2. Infinite number, 2 , 1 22+ , 2 2

2+

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1. (i) 0.7778 (ii) 7.3254 (iii) 1.0110 (iv) 12.3457

(v) 3.1429 (vi) 3.1416

Terminal Exercise

1. (i) 516 (ii) 1

9 (iii) 1

2. (i) 0.285714 (ii) 0.307692 (iii) 0.54 (iv) 1

5. (i) 0.667 (ii) 1.414 (iii) 1.733 (iv) 3.143 (v) 1.0

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2

Indices (Exponents)

2.1 INTRODUCTION

You have learnt, how to multiply two or more real numbers. You can easily write the product5 × 5 × 5 = 125 or 7 × 7 × 7 × 7 = 2401. In 5 × 5 × 5, five is being multiplied by itself threetimes. In 7 × 7 × 7 × 7, seven is being multiplied by itself 4 times. Think of the problem when11 is multiplied by itself 10 times or 15 is multiplied by itself 20 times. Isn’t it difficult towrite 15 × 15 × ... 20 times? In this lesson we shall try to overcome this difficulty by theintroduction of Exponential notation. We shall explain its meaning, state and derive laws ofexponents. We shall learn the application of laws of exponents. We shall express real numbersas product of powers of prime numbers. In the end of the lesson we shall also learn how toexpress very large and very small numbers using scientific notation.

2.2 OBJECTIVES


write a repeated multiplication in exponential notation and vice-versa.

identify the base and the exponent of a number in the exponential notation.

express a natural number as a product of powers of prime numbers uniquely.

state the laws of exponents.

explain the meaning of a0, a–m and apq .

express very large and very small numbers using scientific notation.

simplify expressions involving exponents using laws of exponents.


Multiplication of numbers

Division of numbers

Prime Numbers

Indices (Exponents) 21

Rational Numbers

Four fundamental operations on rational numbers.

2.4 EXPONENTIAL NOTATION

Let us consider the following products :

(i) 5 × 5

(ii) 3 × 3 × 3 × 3

(iii) 4 × 4 × 4 × 4 × 4 × 4

(iv) 6 × 6 × 6 × 6 × 6

In (i) we have to multiply 5 by itself 2 times,

in (ii) multiply 3 by itself 4 times,

in (iii) multiply 4 by itself 6 times,

and in (iv) multiply 6 by itself 5 times.

We write 5 × 5 as 52, 3 × 3 × 3 × 3 as 34, 4 × 4 × 4 × 4 × 4 × 4 as 46, and6 × 6 × 6 × 6 × 6 as 65.

52 is read as ‘five raised to the power 2’ or ‘second power of five’

In 52, the number 5 is called the base and 2 is called the exponent or index.

Similarly 34 is read as ‘three raised to the power 4, or ‘fourth power of three’. 3 is the baseand 4 is the exponent.

Again in 46, 4 is the base and 6 is the exponent and in 65, 6 is the base and 5 is the exponent.

Similarly (–2) × (–2) × (–2) × (–2) = (–2)4 where –2 is the base and 4 is the exponent.

The notation for writing the product of a number by itself several times is calledthe exponential notation.

You can now write 8 × 8 × 8 × ... 11 times as 811 and (–3) × (–3) × (–3) × ... 7 times as(–3)7.

Can you identify base and exponent in 811 ?

Of course, yes. In 811, the base is 8 and exponent is 11.

Write the base and exponent of (–3)7

Base : Exponent :

22 Mathematics

Exponential notation can also be used for writing the product of a rational number by itself

several times. For example, 34

34× is written as 3

42FH IK . It is read as ‘ 3

4 raised to the power

two’. Here 34 is base and 2 is the exponent.

Similarly 65

65

65× × = 6

53FH IK and −FH IK × −FH IK × −FH IK × −FH IK1

212

12

12 = −FH IK12

4

In general, if ‘a’ is any rational number, then

a × a × a × ..... m times = am, a is called the base and m is called the exponentor index.

a ×a = a2 is known as square of a and a × a × a = a3 is known as cube of a.

Note. The plural of the word index is indices.

Example 2.1 : Evaluate each of the following :

(a) 23

2FH IK (b) 35

4FH IK (c) −FH IK373.

Solution :(a) 23

2FH IK = 23

23

2 23 3

49× = ×

× =

(b) 35

4FH IK = 35

35

35

35× × × = 3 3 3 3

5 5 5 581625

× × ×× × × =

(c) −FH IK373

= −FH IK × −FH IK × −FH IK37

37

37 =

− × − × −× × = −

3 3 37 7 7

27343

b g b g b g

Example 2.2 : Express each of the following in the exponential form :

(a) 7 × 7 × 7 × 7 × 7 × 7 × 7

(b) 511

511

511× ×

(c) −FH IK × −FH IK × −FH IK × −FH IK × −FH IK49

49

49

49

49

Solution : (a) 7 × 7 × 7 × 7 × 7 × 7 × 7 = (7)7

(b) 511

511

511× × = 5

113FH IK

(c) −FH IK × −FH IK × −FH IK × −FH IK × −FH IK49

49

49

49

49 = −FH IK49

5


Example 2.3 : Express each of the following in exponential notation and write down the baseand exponent in each case :

(a) 2187 (b) 2581 (c) −8

27Solution : (a) 2187 = 3 × 3 × 3 × 3 × 3 × 3 × 3 = 37.

Base = 3, Exponent = 7.

(b) 2581 = 5 5

9 959

59

59

2×× = × = FH IK

Base = 59 and Exponent =2

(c) −827 =

− × − × −× × = −FH IK × −FH IK × −FH IK

2 2 23 3 3

23

23

23

b g b g b g = −FH IK23

3

Base = − 23 and Exponent = 3.

Example 2.4 : Express − 1243 in exponential form. Write down the base and the exponent.

Solution : − 1243 = −1

243

=− × − × − × − × −

× × × ×1 1 1 1 1

3 3 3 3 3b g b g b g b g b g

=−

= −FHGIKJ

13

13

5

5

5b g

Here Base = −13 and Exponent = 5.

Example 2.5 : Simplify 23

34

2 3FH IK × FH IK

Solution : 23

34

2 3FH IK × FH IK = 23

34

22

33×

= 49

2764× = 3

16Example 2.6 : Find the reciprocal of each of the following and express them in exponential

form :

(a) 52 (b) 37

4FH IK (c) −FH IK459

3 243

3 81

3 27

3 9

3

3 2187

3 729

3 243

3 81

3 27

3 9

3

24 Mathematics

Solution :(a) Reciprocal of 52 = 152 = 1

5 515

2

× = FH IK

(b) 37

4FH IK = 37

44

∴ Reciprocal of 37

4FH IK = 73

73

44

4= FH IK

(c) −FH IK459

= −45

9

9b g

∴ Reciprocal of −FH IK459

= 54

54

99

9

−= −FH IKb g .

From the above, we observe that

If pq is any non zero rational number and m is any positive integer, then the

reciprocal of pq

mFHIK is

qp

mFHIK .


1. Write the base and exponent in each of the following :

(a) 37 (b) 73 (c) 211

8FH IK (d) −FH IK5920

2. Evaluate each of the following :

(a) 813

3FH IK (b) −FH IK211

4(c) −FH IK52

5(d) −FH IK34

3.

3. Simplify each of the following :

(a) 25

53

2 3FH IK × FH IK

(b) −FH IK × −FH IK23

12

3 2

4. Find the reciprocal of each of the following :

(a) (3)4 (b) (–5)3 (c) −FH IK374

.

2.5 PRIME FACTORISATION

You may recall how to write the factors of a given natural number, in finding the HCF andLCM of numbers. Also, given a natural number, it is either the number 1 or it is a prime numberor it is a composite number. Any composite number can be expressed as a product of primenumbers. Let us consider the following examples.


(a) 2 242 122 6

3

(b) 2 9802 4905 2457 49

7

(c) 3 76233 25417 847

11 12111

From the examples given above, we see that any given natural number can be expressed asa product of powers of prime factors. This product is also unique apart from the order ofoccurrence of factors. This result is known as the Fundamental Theorem of Arithmetic whichis formally stated as follows.

Any natural number other than one, is expressible as a product of powers of primenumbers in a unique manner apart from the order of occurrence of factors.

Example 2.7 : Express each of the following in exponential form :

(a) 16200 (b) 49392

Solution : (a) 2 16200

2 8100

2 4050

3 2025

3 675

3 225

3 75

5 25

5

∴ 24 = 2 × 2 × 2 × 3= 23 × 31

∴ 980 = 2 × 2 × 5 × 7 × 7= 22 × 51 × 72.

∴ 7623 = 3 × 3 × 7 × 11 × 11= 32 × 7 × 112

∴ 16200 = 2 × 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

= 23 × 34 × 52

26 Mathematics

(b) 2 49392

2 24696

2 12348

2 6174

3 3087

3 1029

7 343

7 49

7


1. Express the following as products of powers of prime numbers :

(a) 39 (b) 36 (c) 91

(d) 320 (e) 216 (f) 3024

2. Express in the exponential form each of the following :

(a) 243 (b) 1024 (c) 1296

(d) − 72964 (e) 1331

4096

2.6 LAWS OF EXPONENTS

Let us study the following :

(a) 22 × 23 = (2 × 2) × (2 × 2 × 2) = 2 × 2 × 2 × 2 × 2 = 25 = 22+3

(b) (–3)3 × (–3)4 = [(–3) × (–3) × (–3)] × [(–3) × (–3) × (–3) × (–3)]

= (–3) × (–3) × (–3) × (–3) × (–3) × (–3) × (–3)

= (–3)7 = (–3)3+4

(c)25

25

2 4FHGIKJ × FHGIKJ =

25

25

25

25

25

25

×FHGIKJ × × × ×FHG

IKJ

= 25

25

25

25

25

25

× × × × ×

= 25

25

6 2 4FHGIKJ = FHGIKJ

+

∴ 49392 = 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 × 7

= 24 × 32 × 73.


(d) −FHGIKJ × −FHG

IKJ

25

25

3 4

= −FHGIKJ × −FHG

IKJ × −FHG

IKJ

LNM

OQP × −FHG

IKJ × −FHG

IKJ × −FHG

IKJ × −FHG

IKJ

LNM

OQP

25

25

25

25

25

25

25

= −FHGIKJ × −FHG

IKJ × −FHG

IKJ × −FHG

IKJ × −FHG

IKJ × −FHG

IKJ × −FHG

IKJ

25

25

25

25

25

25

25

= −FHGIKJ = −FHG

IKJ

+25

25

7 3 4

From the above, we observe that in each case the base is the same but the indices, get added.Thus, we have the following :

Law 1 : If a is any rational number and m and n are two positive integers, then

am × an = am+n

Example 2.8 : Find the value of −FH IK × −FH IK34

34

2 5

Solution : −FH IK342

and −FH IK345

have the same base − 34 . Therefore to find the product, we

add the exponents

∴ −FH IK × −FH IK34

34

2 5= −FH IK

+34

2 5 = −FH IK34

7 = − 2187

16384

Let us the study the following :

(a) 3 35 2÷ = 33

3 3 3 3 33 3

52 = × × × ×

× = 3 × 3 × 3 = 33 = 35–2

(b) (–2)7 ÷ (–2)3 = −

−

2

2

7

3b gb g =

− × − × − × − × − × − × −− × − × −

2 2 2 2 2 2 22 2 2

b g b g b g b g b g b g b gb g b g b g

= (–2) × (–2) × (–2) × (–2)

= (–2)4 = (–2)7–3

28 Mathematics

(c) 32

32

6 4FH IK ÷ FH IK =

3232

32

32

32

32

32

32

32

32

32

32

6

4

e je j

=× × × × ×

× × ×

= 32

32× = 3

232

2 6 4FH IK = FH IK−

(d) −FH IK ÷ −FH IK57

57

3 2 =

−

−=

− × − × −

− × −

5757

57

57

57

57

57

3

2

e je j

e j e j e je j e j

= −FH IK = −FH IK−5

757

1 3 2


Law 2 : If a is any non-zero rational number and m, n are positive integers such thatm > n, then

am ÷ an = am–n

Example 2.9 : Find the value of 75

75

11 8FH IK ÷ FH IK

Solution : 75

75

11 8FH IK ÷ FH IK = 75

75

11 8 3FH IK = FH IK−

= 343125


(a) 52 ÷ 53 = 5 5

5 5 5151

×× × = = 1

53 2−

(b) (–3)3 ÷ (–3)5 = − × − × −

− × − × − × − × −3 3 3

3 3 3 3 3b g b g b g

b g b g b g b g b g

=1

3 313 2− × −

=−b g b g b g =

13 5 3− −b g

(c) 32

32

2 4FH IK ÷ FH IK = 32

32

32

32

32

32

×

× × × = 1

32

32

132

2×=e j

= 132

4 2e j

−


(d) −FH IK ÷ −FH IK23

23

3 5 =

− × − × −

− × − × − × − × −

23

23

23

23

23

23

23

23

e j e j e je j e j e j e j e j

= 123

23

123

2− × −

=−e j e j e j

= 123

5 3−

−e j


Law 3 : If a is any non-zero rational number and m, n are positive integers such thatm < n, then

am ÷ an = 1an m−


511

7 9FH IK ÷ FH IK

Solution : 511

511

7 9FH IK ÷ FH IK = 1511

1511

9 7 2e j e j

− =

= 125

121

12125= .


(a) (22)3 = 22 × 22 × 22 = 22+2+2 = 26 = 22×3

(b) [(–5)3]4 = (–5)3 × (–5)3 × (–5)3 × (–5)3 = (–5)3+3+3+3 = (–5)12 = (–5)3×4.

(c) 29

3 5FH IKLNMOQP = 2

929

29

29

29

3 3 3 3 3FH IK × FH IK × FH IK × FH IK × FH IK

= 29

3 3 3 3 3FH IK+ + + +

= 29

29

15 3 5FH IK = FH IK×

(d) −FH IKLNM

OQP

43

2 4

= −FH IK × −FH IK × −FH IK × −FH IK43

43

43

43

2 2 2 2

= −FH IK = −FH IK+ + +4

343

2 2 2 2 8 = −FH IK

×43

2 4.

30 Mathematics

From the above we observe that

Law 4 : If a is any rational number and m and n are two positive integers, then

(am)n = am×n = amn


3 2FH IKLNMOQP .

Solution : 13

3 2FH IKLNMOQP = 1

313

3 2 6FH IK = FH IK×

= 13

17296 =

2.6.1 ZERO EXPONENT

You have seen that so far we have established four laws of exponents, which were for positiveintegers. Do you know, what will happen when the index is negative or zero? We shall takeup negative exponents at a later stage. We will now see what happens when exponent is zero.

You are aware that if a rational number is divided by itself, the result is 1.

i.e., 27

27÷ = 1

−FH IK ÷ −FH IK59

59 = 1

Again 23 ÷ 23 = 1 ...(i)

Now 23 ÷ 23 = 23–3 = 20 ...(ii)

From (i) and (ii), we get

20 = 1

Similarly,

(a) (–5)4 ÷ (–5)4 = − × − × − × −− × − × − × −

5 5 5 55 5 5 5b g b g b g b gb g b g b g b g = 1 = (–5)4–4 = (–5)0

(b) 313

313

3 3FH IK ÷ FH IK = 3

133

133

133

133

133

13

× ×

× × = 1 = 3

133

133 3 0FH IK = FH IK−

and (c) −FH IK ÷ −FH IK37

37

2 2 =

− × −

− × −

37

37

37

37

e j e je j e j = 1 = −FH IK = −FH IK

−37

37

2 2 0.


From the above, we get

Law 5 : If a is any rational number other than zero, then a0 = 1

Example 2.12 : Find the value of :

(a) 72


0

Solution : (a) 72

0FH IK = 1, (b) −FH IK519

0 = 1


1. Simplify and express the result in the exponential form :

(a) (3)3 × (3)4 (b) 23

23

4 5FH IK × FH IK

(c) −FH IK × −FH IK × −FH IK25

25

25

2 5


(a) (–11)8 ÷ (–11)6 (b) 57

57

8 5FH IK ÷ FH IK

(c) −FH IK ÷ −FH IK43

43

18 11


(a) (36)3 (b) 59

4 5FH IKLNMOQP (c) −FH IK

LNM

OQP

317

3 4


(a) 1113

1517

8 0FH IK × FH IK (b) −FH IK × −FH IK911

911

0 15

5. Which of the following are true statements ?

(a) 25

25

2 3FH IK × FH IK = 25

6FH IK (b) 311

311

3 3FH IK × FH IK = 311

6FH IK

(c) 49

5 4FH IKLNMOQP = 4

99FH IK (d) 9

196 5FH IK

LNMOQP = 9

1930FH IK

32 Mathematics

(e) 83

0FH IK = 0 (f) −FH IK452

= − 1625

(g) 825

719

4 0FH IK × FH IK = 825

4FH IK

2.7 NEGATIVE INTEGERS AS EXPONENTS

In the previous sections, we have been taking non-negative integers as exponents. Now weshall try to assign meaning to negative integers as exponents.

We know that :

The reciprocal of 3 is 13 . We write it as 3–1 and read it as 3 raised to the power (–1).

The reciprocal of –4 is 14− . We write it as (–4)–1 and read it as ‘–4 raised to the power

(–1)’.

Similarly, the reciprocals of 45 and − 7

9 are 14 5 and 1

7 9− respectively. We write them as

45

1FH IK−

and −FH IK−7

91 and read them as ‘ 4

5 raised to the power (–1)’ and ‘− 79 raised to the

power (–1)’ respectively.

Similarly reciprocal of 32 is 132 . We write it as 3–2 and read it as ‘3 raised to the power

(–2)’.

The reciprocal of (–5)4 is 15 4−b g

. We write it as (–5)–4 and read it as ‘–5 raised to the power

(–4)’.

The reciprocal of 718

5FH IK is 17 18 5b g . We write it as 7

185FH IK−

and read it as ‘ 718 raised to the

power (–5)’.

From the above we get,

If a is any non-zero rational number and m is any positive integer, then the reciprocal of am

i.e., 1am is written as a–m and is read as ‘a raised to the power (–m)’. Thus,

1am = a–m


Example 2.13 : Convert the negative exponents to positive exponents in the following :

(a) 47

3FH IK−

(b) −FH IK−5

67

.

Solution. (a) 47

3FH IK−

= 147

147

3 33e j

= = 74

74

33

3= FH IK

(b) −FH IK−5

67

= 156

15

6

7 7

7−

=−e j b g

= 65

65

77

7

−= −FH IKb g = −FH IK65

7

From the above example, we get the following result :

If pq is any non zero rational number and m is any positive integer, then

pq

mFHIK−

=qp

qp

m

m

m= FHIK

Example 2.14 : Simplify 23

53

3 2FH IK ÷ FH IK− −

Solution : 23

3FH IK−

= 32

3FH IK

and 53

2FH IK−

= 35

2FH IK

∴23

53

3 2FH IK ÷ FH IK− −

= 32

35

3 2FH IK ÷ FH IK

= 32

35

3 2FH IK ÷ FH IK = 3 3 32 2 2

5 53 3

× ×× × × ×

× = 758

2.8 LAWS OF EXPONENTS FOR INTEGRAL EXPONENTS

After attaching meaning to negative integers as exponents of rational numbers, let us seewhether the rules of exponents learnt earlier in this chapter hold for negative exponents alsoor there is any need to modify them.

Let us study the following examples :

34 Mathematics

Example 2.15 : Show that

(a) 35

35

5 4FH IK × FH IK−

= 35

5 4FH IK− +

(b) −FH IK × −FH IK− −2

323

2 3 = −FH IK

− −23

2 3

(c) −FH IK ÷ −FH IK− −3

434

3 7 = −FH IK

− +34

3 7(d) 3

112 3FH IK

LNM

OQP

− = 3

112 3FH IK− ×

(e) (5 × 3)–3 = 5–3 × 3–3.

Solution : (a) 35

35


= 135

355

4

e j× FH IK

= 135

355 4

5 4

e jb g

−

− −= FH IK = 3

55 4FH IK− +

(b) −FH IK × −FH IK− −2

323

2 3= 1

23

123

2 3−

×−e j e j

= 123

23

123

2 3 2 3− × −

=−

+e j e j e j

= −FH IK = −FH IK− + − −2

323

2 3 2 3b g

(c) −FH IK ÷ −FH IK− −3

434

3 7= 1

34

134

3 7−

÷−e j e j

= 134

343

7

−× −FH IK

e j

= −FH IK−3

47 3

= −FH IK− +3

43 7

(d) 311

2 3FH IKLNM

OQP

−= 11

3113

2 32

2

3FH IKLNMOQP = FH

IK

=113

2 3

2 3

×

× = 3

11

2 3

2 3

− ×

− × = 3

11

2 3FHGIKJ− ×

.


(e) (5 × 3)–3 =1

5 31

5 33 3 3×=

×b g = 5–3 × 3–3

From the above examples we see that Laws 1 to 5 hold good for negative exponents as well.In general, if

a, b be any non-zero rational numbers and m, n be any integers, then

1. am × an = am+n

2. am ÷ an = am–n

3. (am)n = amn

4. (a × b)m = am × bm


1. Express as a rational number : −FH IK−4

111.

2. Express as a power of a rational number with positive exponent :

(a) 29

3FH IK−

(b) 127 × 12–4 (c) 517

2 5FH IKLNM

OQP

−

3. Express as a power of a rational number with negative exponent

(a) 78

3FH IK (b) (53)4 (c) −FH IKLNM

OQP

74

2 5


(a) 32

32


(b) −FH IK × −FH IK−2

323

3 4(c) −FH IK ÷ −FH IK

− −35

35

5 3

5. Which of the following statements are true ?

(a) a–m × an = a–m–n

(b) (a–m)n = a–mn

(c) am × bm = (ab)m

(d) am ÷ an = am–n

(e) a–m × a0 = am

36 Mathematics

2.9 MEANING OF apq .

We know that

am × an = am+n for all integral values of m and n.

If a is a positive rational number, and q is a natural number, how should we define a1q .

Let us multiply a1q , q times.

i.e.,a a a a

q times

q q q q1 1 1 1× × × ×.....

= a aq q qqq

1 1 1+ + +=

..... q times = a

In other words, qth power of a1q = a. We state that a

1q is the qth root of a and is written as

aq . Let us consider an example to illustrate this point.

3 3 3 3 315

15

15

15

15× × × × = 3 3

15

15

15

15

15

55+ + + +

= = 31 = 3

or 315 is the fifth root of 3 which is written as 3

15 = 35

Now we define a rational power of a.

If a is a positive real number, p is an integer and q is a natural number, then

apq = apq

Now

a a a apq

pq

pq

pq

pq

pq× × × = = =

+ +...

...q factors a a

q times pqq p

∴ apq = apq .

Thus, apq means the qth root of ap.

We observe that if the exponent is a rational number, its numerator denotes the index and thedenominator denotes the root.

Thus, 423 means cube root of 42.


For rational exponents, the laws of exponents are as follows :

(i) am.an = am+n

(ii) a am n÷ = am–n

(iii) (am)n = amn

(iv) (ab)m = am b

m

(v)ab

mFHGIKJ =

ab

m

m

Let us take the following examples and verify the above laws :

Example 2.16 : Find the value of :

(a) 1614b g (b) 243

25b g (c) 16

81

34FH IK

−

Solution : (a) 1614b g = (2 × 2 × 2 × 2)

14

= (24)14 = 2

4 14× = 21 = 2.

(b) 24325b g = 3 3 3 3 3

25× × × ×b g = 3 35

25 5 2

5d i =×

= 32 = 9

(c) 1681

34FH IK

−= 2 2 2 2

3 3 3 3

34× × ×

× × ×FH IK

−

= 23

23

434 4 3

4FH IKLNMOQP = FH IK− × −e j

= 23

32

278

3 3FH IK = FH IK =−



(a) 823b g (b) 64

125

23F

HGIKJ−


(a) 125 2513

12b g− −

÷ (b) 1316

1316

34

74FH IK × FH IK

−

38 Mathematics

2.10 SCIENTIFIC NOTATION

Let us consider the following problem :

Can you calculate what powers of 10 should be multiplied to 1.7 to get

(i) 17 (ii) 170 (iii) 1700 (iv) 17000 ?

Clearly, in (i) 17 = 1.7 × 101

(ii) 170 = 1.7 × 102

(iii) 1700 = 1.7 × 103

(iv) 17000 = 1.7 × 104

Similarly, 170000 = 1.7 × 10 5

1700000 = 1.7 × 106 and so on.

Scientists and Engineers often find it convenient to write numbers in this form.

A number is said to be in scientific notation, when it is expressed as a numberbetween 1 and 10, multiplied by an integral power of 10.

Sometimes it is also referred to as writing a number in standard form. In symbols,A × 10n is in scientific notation, where A is terminating decimal lying between 1and 10 i.e. 1 < A < 10 and n is an integer.

Now we shall see how to write a number less than 1 in the scientific notation.

Let us write 0.00017 in the scientific notation.

17.0 = 1.7 × 10 = 1.7 × 101

1.7 = 1.7 × 1 = 1.7 × 100

0.17 = 1.7 × 1

10 = 1.7 × 10–1

0.017 = 1.7 × 1

100 = 1.7 × 10–2

0.0017 = 1.7 × 1

1000 = 1.7 × 10–3

0.00017 = 1.7 × 1

10000 = 1.7 × 10–4 and so on.

Similarly, 0.000031 = 3.1 × 1

105 = 3.1 × 10–5.


From the above, we observe that a number written in scientific notation has the followingcharacteristics :

(a) There is only one digit to the left of decimal point.

(b) The exponent of 10 is a positive integer equal to the number of places, thedecimal point has been moved when the given number is greater than or equalto 10

(c) The exponent of 10 is zero when the given number is greater than or equal to1 but less than 10

(d) The power of 10 is a negative integer equal to the number of places the decimalpoint has been moved when the given number is less than 1.

Example 2.17 : Write each of the following in scientific notation :

(a) Seven thousand (b) 143000

(b) 0.57 (d) 0.00031

Solution :(a) Seven thousand = 7000 = 7.0 × 103

(b) 143000 = 1.43000 × 105 = 1.43 × 105

(c) 0.57 = 5.7 × 10–1

(d) 0.00031 = 3.1 × 10–4

Example 2.18 : It is said that the distance of earth from the sun is 149000000 km. Expressit in the scientific notation.

Solution : The distance of earth from the sun = 149000000 km

= 1.49 × 108 km


1. Express each of the following numbers in scientific notation

(a) 15 lakh (b) 3730000

(b) 72 × 106 (d) 317 × 104

2. Express each of the following numbers in scientific notation :

(a) 0.00079 (b) 0.11

(c) 0.00000567 (d) 33 × 10–5

3. The diameter of the earth is said to be 13000 km. Write it in scientific notation.

4. It is said that distance of moon from the earth is 380000 km. Express it in scientificnotation.

40 Mathematics

5. An aeroplane flies at 7.0 × 105 metres per hour. How long will it take to cover a distanceof 3.5 × 104 km?

2.11 LET US SUM UP

a × a × a × ... m times = am is the exponential form, where a is the base and m is exponent.

Laws of exponents are

am × an = am+n

a am n÷ = am–n

(ab)m = am.bm

ab

mFH IK = ab

mm

(am)n = amn

a0 = 1

a–m = 1am

apq = apq

Very large and very small numbers in scientific notation are written as A × 10n, where1 < A < 10 and n is any integer

TERMINAL EXERCISE

1. Express each of the following in the exponential form :

(a) 3 × 3 × 3 × 5 × 5 × 7 × 7 × 7 × 7

(b) −FH IK × −FH IK × −FH IK × −FH IK511

511

511

511


(a) −FH IK × FH IK × −FH IK34

13

827

3 2(b) 3

73527

15

2 2FH IK × × −FH IK3. Simplify and express the result in the exponential form :

(a) (8)2 × (6)3 × (15)2 (b) −FH IK ÷ −FH IK3719

3719

26 20

(c) 943

5 7FH IKLNMOQP



(a) 50 + 30 + 170 – 31 (b) (70 – 30) (70 + 30).


(a) (35)12 ÷ (35)–3 (b) (101)6 × (101)–4

(c) −FH IK × −FH IK−2

929

3 4

6. Find x so that 78

78

78

3 11FH IK × FH IK = FH IK− x

7. Find x so that

313

313

313

2 9 2 1FH IK × FH IK = FH IK− − +x

8. Expressing as a product of primes, write the answers of each of the following inexponential form :

(a) 19440000 (b) 172872

(c) 605000 (d) 35591400

9. Write each of the following numbers in the scientific notation :

(a) 50 lakh (b) 3030000

(c) 4500000 (d) 720 × 106

10. Express each of the following numbers in the scientific notation :

(a) 0.00029 (b) 0.00000399

(c) 860 × 10–5 (d) 301 × 10–4.

11. The star sirus is about 8.1 × 1013 km from the earth. Assuming that light travels at3.0 × 105 km per second, find how long, light from sirus takes to reach the earth.

12. The number of haemoglobin molecules in a single red cell is 270000000. Express it inscientific notation.

42 Mathematics

ANSWERS


1. (a) Base = 3, Exponent = 7 (b) Base = 7, Exponent = 3

(c) Base = 211 , Exponent = 8 (d) Base =

−59 , Exponent = 20.

2. (a) 5122197 (b) 16

14641 (c) −312532 (d) − 27

64

3. (a) 2027 (b) − 2

27

4. (a) 13


(c) −FH IK734


1. (a) 3 × 13 (b) 22 × 32 (c) 7 × 13

(d) 26 × 5 (e) 23 × 33 (f) 24 × 33 × 7

2. (a) 35 (b) 210 (c) 24 × 34

(d) −FH IK326

(e)112

3

12


1. (a) 37 (b) 23


2. (a) (–11)2 (b) 57


3. (a) 318 (b) 59


12

4. (a) 1113


15

5.(b), (d) and (g)


1. − 114


2. (a) 92

3FH IK (b) 123 (c) 175

10FH IK

3. (a) 87

3FH IK−

(b) 15

12FH IK−

(c) −FH IK−4

710

4. (a) 94 (b) − 2

3 (c) 259

5. b, c and d


1. (a) 4 (b) 2516

2. (a) 1 (b) 1316


1. (a) 1.5 × 106 (b) 3.73 × 106 (c) 7.2 × 107 (d) 3.17 × 106

2. (a) 7.9 × 10–4 (d) 1.1 × 10–1 (c) 5.67 × 10–6 (d) 3.3 × 10–4

3. 1.3 × 104 km 4. 3.8 × 105 km 5. 5.0 × 101 hour.

Terminal Exercise

1. (a) 33 × 52 × 74 (b) −FH IK511

4

2. (a) 172 (b) 1

105

3. (a) 29 × 35 × 52 (b) −FH IK3719

6(c) 9

4335FH IK

4. (a) 0 (b) 0

5. (a) (35)15 (b) (101)2 (c) − 29

6. 8 7. – 6

8. (a) 54 × 35 × 27 (b) 23 × 32 × 74

(c) 23 × 54 × 112 (d) 23 × 34 × 52 × 133.

9. (a) 5.0 × 106 (b) 3.03 × 106 (c) 4.5 × 106 (d) 7.2 × 108

10. (a) 2.9 × 10–4 (b) 3.99 × 10–6 (c) 8.6 × 10–3 (d) 3.01 × 10–2

11. 7.5 × 104 hours 12. 2.7 × 108.

44 Mathematics

3

Radicals (Surds)

2.1 INTRODUCTION

In the last lesson, you have learnt the laws of exponents for positive integral indices and negative

integral indices. You have also learnt the meaning of numbers of the type apq .

In this lesson we shall study about a special type of numbers apq which are irrational numbers.

In the last lesson we gave meaning to the number a q1

as the qth root of a. In this lesson weshall call aq or xn a radical, q or n as index and a or x as the radicand. We shall also namethis as surd. We shall also discuss the laws of radicals. We shall find the simplest (lowest)form of a radical. We shall find the rationalising factor of a radical and rationalise thedenominator of a radical and also simplify expressions involving radicals.

3.2 OBJECTIVES


identify radicals from a given set of irrational numbers

identify index and radicand of a surd

state the laws of radicals (surds)

express a given surd in the simplest form

classify similar and non-similar surds

reduce surds of different orders to those of the same order

perform the four fundamental operations on surds

arrange the given surds in ascending/descending order of magnitude

find a rationalising factor of a given surd

rationalise the denominator of a given surd

simplify expressions involving surds.

Radicals (Surds) 45


Four fundamental operations on numbersPrime numbers, order relation in numbersLaws of exponents

Meaning of a0, a–m and apq .

3.4 SURD

You have already studied in lesson 1 that numbers of the type 2 3 5, , , ... are irrationalnumbers. Now we shall study irrational numbers of a particular type called radicals or surds.

Definition. A surd is defined as a positive irrational number of the type xn , where it is notpossible to find exactly the nth root of x, where x is a positive rational number.

Thus a number xn is a surd if and only if

(a) it is an irrational number

(b) it is a root of positive rational number

In the surd xn , the symbol is called the radical sign. The index n is called the orderof the surd and x the radicand. When the order is not mentioned it is taken as 2. For example,in the 53 , order of the surd is 3 and 5 is the radicand. In 7 , the order is 2 and the radicandis 7.

Similarly, 86 , 324 and 504 are all surds, but 83 is not a surd as its value 2 is rational.

Again 2 2+ , though it is an irrational number, is not a surd because it is the square notof an irrational number.

Similarly 5 3+ , π are not surds, as the radicands are not rational numbers. We mayrepeat, that surd is an irrational number in which the radicand is a positive rational number.

Note. If n is a positive integer and a be real number, then if a is irrational, an is not a surd.Again if an is rational, then also an is not a surd.

3.5 PURE AND MIXED SURDS

A surd which has unity as its rational factor, other factor being irrational is called a pure surd.

For example 274 , 1125 and 503 are pure surds.

A surd having rational factor other than unity along with irrational factor is called a mixedsurd.

For example 2 3 , 3 53 and 45 74 are mixed surds.

It may be noted that in the cases of conversion, type of the radicand should be kept unchanged.

46 Mathematics

3.6 ORDER OF SURDS

In the surd 7 23 , 7 is called coefficient of the surd, 3 is order of the surd and 2 is radicand.When there is no coefficient in a surd, it is assumed that coefficient is 1.

Example 3.1 : State which of the following are surds and which are not :

(a) 25 (b) 50

(c) 814 (d) 1283

Solution : 25 = 5 which is a rational number.

∴ 25 is not a surd.

(b) 50 = 2 5 5 5 2× × =

∴ 50 is an irrational number

Hence 50 is a surd.

(c) 814 = 3 3 3 34 × × × = 3, which is a rational number.

∴ 814 is not a surd

(d) 1283 = 4 4 4 23 × × ×

= 4 23 , which is an irrational number.

∴ 1283 is a surd.

Example 3.2 : Identify index and radicand of each of the following surds :

(a) 175 (b) 82

(c) 1236 (d) 51711

Solution : (a) Here index is 5 and radicand is 17

(b) Here index is 2 and radicand is 82

(c) Here index is 6 and radicand is 123

(d) Here index is 11 and radicand is 517.

Example 3.3 : Identify the pure and mixed surds from the following.

(i) 21 (ii) 183 (iii) 3 125

(iv) 2 1417 (v) 5 11255

Solution : Pure surds (i) and (ii); mixed surds (iii), (iv) and (v)

Radicals (Surds) 47

3.7 LAWS OF RADICALS

We state here some laws of radicals which are used to simplify surds :

(i) a an n=

(ii) a b abn n n=

(iii)ab

ab

n

nn=

where a and b are positive rational numbers and n is a positive integer.

Example 3.4 : Using laws of radicals, find which of the following are surds and which arenot :

(a) 10 40× (b) 15 104÷

(c) 2 43 3÷ (d) 48 27÷

Solution : (a) 10 40× = 400

= 20 20×

= 202d i = 20

which is a rational number.

∴ 10 40× is not a surd.

(b) 2 15 4 10÷

=2 154 10

=2 2 154 4 10× ×× ×

=60

160

=38

which is an irrational number.∴ 2 15 4 10÷ is a surd

48 Mathematics

(c) 2 43 3× = 2 43 ×

= 2 2 23 × ×

= 2, which is a rational number.

∴ 2 43 3× is not a surd

(d) 48 27÷ = 4827

= 4827

= 169

= 43 which is again a rational number

∴ 48 27÷ is not a surd.


1. Write the index (order) and the radicand in each of the following :

(a) 1255 (b) 3436 (c) 19

2. State which of the following are surds ?

(a) 493 (b) 6254 (c) 2166

(d) 500 (e) 5 45× (f) 3 2 5 6×

(g) 1 3+ (h)7

169

3. Identify pure and mixed surds, out of the following :

(i) 15 (ii) 2 12 (iii) 3 73 (iv) 35

3.8 LAWS OF SURDS

It has been seen that the surds can be expressed as numbers with fractional exponents. Lawsof indices studied in the last lesson are applicable to them also. We recall these laws here.

(i) x y xyn n n= or x y xyn n n1 1 1× = b g

Radicals (Surds) 49

(ii)xy

xy

n

nn= or x

yxy

nn

n11

1= FHIK

(iii) x x xnm mn mn= = or x x xn m mn m n1 1 1 1 1d i d i= =

(iv) x xmn m n= or x xm n m nd i1 =

(v) x xpm pnmn= or x x x xp m p m pn mn pn mnd i d i1 1= = =

In these results x and y are positive rational numbers and m, n and p are positive integers.

Let us illustrate these laws in the following examples :

(a) 2 73 3 = 2 7 2 713

13

13. ( )= × = =14 14

13 3

(b)5

9

1 7

1 7b gb g = 5

959

1 77F

HGIKJ =

(c) 335 = 3 3 3 3 3 31 35 1 3 1 5 1 15 15 5 3 53= = = = =×d i

(d) 435 = 4 4 4 4 4 43 1 5 3 5 9 15 9 1 15 915 3 33 5d i d i= = = = = ××

You must have observed from the above, that order of a surd can be changed by multiplyingthe index of the surd and index of the radicand by the same positive integer. For example

23 = 2 426 6= and 3 3 94 28 8= = .

3.9 SIMILAR (OR LIKE) SURDS

Two surds are similar (or like) if they can be reduced to the same irrational factors, whatevercoefficients they may have. For example 2 5 and 7 5 are similar surds. 75 and 12 can

be expressed as 5 3 and 2 3 OR as similar surds.

3.10 SIMPLEST (LOWEST FORM) OF A SURD

A surd is said to be in its simplest (lowest) form if it has

(a) the smallest possible index of the radical

(b) no fraction under the radical sign

50 Mathematics

(c) no factor of the form an where a is a positive integer is under the radical sign ofindex n.

Thus a pure surd of index n may be reduced to a rational multiple of nth root of a positiveinteger, no prime factor of which occurs to an exponent as high as n. Such a surd is said tobe in lowest term.

Thus for example

12518

3 = 125 1218 12

3 ××

= 56 123 is the simplest (lowest) form of the given surd.

Similarly, the simplest form of the surd 3 225 × is 125 .

Example 3.5 : Express as a pure surd in the simplified form.

(a) 2 7 (b) 3 54 (c) 34 32

Solution : (a) 2 7 = 2 7 4 7 282 × = × =

(b) 3 54 = 3 5 3 544 4 44. = ×

= 81 54 ×

= 4054

(c)34

32 = 34 32

2FH IK ×

=9

1632× = 18

Example 3.6 : Express as a mixed surd in the simplest form :

(a) 128 (b) 5674 (c) 3206 (d) 2503

Solution : (a) 128 = 64 2×

= 8 8 2 8 22× × = ×

= 8 2

(b) 5674 = 81 7 3 74 44× = ×

= 3 74

Radicals (Surds) 51

(c) 3206 = 64 5 2 56 66× = ×

= 2 56

(d) 2503 = 125 23 ×

= 5 233 ×

= 5 23

(e)2725

3 =27 525 5

35

53 3××

=


1. State which of the following are pairs of similar surds :

(a) 2 8, (b) 5 3 3 18,

(c) 75 48, (d) 20 125,

2. Express as a pure surd :

(a) 7 3 (b) 3 23 (c) 5 5 (d)52

12

3. Express as a mixed surd in the simplest form :

(a) 50 (b) 813 (c) 1285 (d) 2703 (e) 5124

3.11 ADDITION AND SUBTRACTION OF SURDS

We have studied addition and subtraction of rational numbers. Now we shall perform theseoperations on surds. We can add and subtract similar surds in the same way as we added andsubtracted like terms of an algebraic expression.

For example, 5 3 7 3+ = 5 7 3 12 3+ =b g and 8 5 3 5− = 8 3 5 5 5− =b gThus for adding or subtracting the surds we change them to similar surds before performingaddition and subtraction.

For example 50 72+

= 5 5 2 6 6 2× × + × ×

= 5 2 6 2 5 6 2+ = +b g= 11 2

52 Mathematics

And 48 1623 3+

= 2 2 2 6 3 3 3 63 3× × × + × × ×

= 2 6 3 6 2 3 63 3 3+ = +b g= 5 63

Similarly, 98 18−

= 7 7 2 3 3 2× × − × ×

= 7 2 3 2 7 3 2− = −b g= 4 2

Example 3.7 : Simplify each of the following :

(a) 4 3 6 27+

(b) 45 6 3 216−

Solution : (a) 4 3 6 27+

= 4 3 6 3 3 3+ × ×

= 4 3 6 3 3+ ×

= 4 3 18 3+

= 4 18 3+b g= 22 3

(b) 45 6 3 216−

= 45 6 3 6 6 6− × ×

= 45 6 3 6 6− ×

= 45 6 18 6−

= 45 18 6−b g= 27 6

Example 3.8 : Simplify : 2 250 8 16 3 54 323 3 3 4+ − +

Solution : 2 250 8 16 3 54 323 3 3 4+ − +

Radicals (Surds) 53

= 2 5 5 5 2 8 2 2 2 2 3 3 3 3 2 2 2 2 2 23 3 3 4× × × + × × × − × × × + × × × ×

= 2 5 2 8 2 2 3 3 2 2 23 3 3 4× + × − × +

= 10 2 16 2 9 2 2 23 3 3 4+ − +

= 10 16 9 2 2 23 4+ − +b g = 17 2 2 23 4+

Example 3.9 : Show that 8 45 8 20 245 3 125 0− + − =

Solution : 8 45 8 20 245 3 125− + −

= 8 3 3 5 8 2 2 5 5 7 7 3 5 5 5× × − × × + × × − × ×

= 8 3 5 8 2 5 7 5 3 5 5× − × + − ×

= 24 5 16 5 7 5 15 5− + −

= 24 7 5 16 15 5+ − +b g b g= 31 5 31 5−= 0


Simplify each of the following :

1. 175 28+

2. 32 50 128+ +

3. 3 50 4 18+

4. 108 75−

5. 500 80−

6. 24 81 2 33 3 3+ −

7. 3 50 4 8 7 18− +

8. 6 54 2 16 1283 3 3− +

9. 2 40 3 625 4 3203 3 3+ −

10. 12 18 6 20 6 147 3 50 8 45+ − + +

54 Mathematics

3.12 MULTIPLICATION AND DIVISION IN SURDS

In the last section, you have seen that operation of addition and subtraction can be performedwhen the surds are similar surds. Similarly two surds can be multiplied or divided if they areof the same order. You have also studied that order of a surd can be changed by multiplyingor dividing the index of the surd and index of the radicand by the same positive number. Thus,before multiplying or dividing, we change them to the surds of the same order. However ifthe given surds are of the same order, we can perform the operation directly.

For example 3 2 3 2 6× = × = Q 3 2and both are surds of order 2

and 12 2÷ =122

6= Q 12 2and are surds of order 2

But if we have to multiply 3 by 23 , we proceed as follows :

The order of the surds are 2 and 3.

LCM of 2 and 3 is 6

∴ We shall change both the surds to surds of order 6.

∴ 3 = 3 2736 6=

and 23 = 2 426 6=

∴ 3 23× = 27 46 6×

= 27 46 × = 1086

and 323 = 27

4274

6

66=

Example 3.10 : (a) Multiply 5 163 and 11 403

(b) Multiply 5 16 5 33 4and

Solution : (a) 5 16 11 403 3×

= 5 11 2 2 2 53 3× × ×

= 55 2 2 2 53 3× × × ×

= 220 2 53 ×

= 220 103

Radicals (Surds) 55

(b) 17 5 5 33 4×

The surds 17 53 and 5 34 are surds of order 3 and 4 respectively.

LCM of 3 and 4 = 12

∴ We change both the surds to the surds of order 12

Now 17 53 = 17 5 17 625412 12=

and 5 34 = 5 3 5 27312 12=

∴ 17 5 5 33 4× = 17 625 5 2712 12×

= 85 625 27 85 1687512 12× =

Example 3.11 : Divide 15 13 6 53 6by

Solution :15 13

6 5

3

6 =52

135

26

6.

=52

1695

52

1695

6

66=

Example 3.12 : Simplify and express in the simplest form 2 50 3 32 4 18× ×

Solution : 2 50 3 32 4 18× ×

= 2 5 5 2 3 4 4 2 4 3 3 2× × × × × × × ×

= 2 5 2 3 4 2 4 3 2× × × × ×

= 10 2 12 2 12 2× ×

= 1440 2 2 2× ×

= 2880 2

3.13 COMPARISON OF SURDS

It is difficult to calculate actual value of surds. That is why we cannot tell which of the twosurds is greater in value. To know this we change both the surds to surds of same order. Thenwe can compare them by the value of the radicands alongwith their coefficients.

56 Mathematics

Example 3.13 : Which is greater 12

13

3or ?

Solution : The two surds are of orders 2 and 3. LCM of 2 and 3 is 6

∴12

1 2FHGIKJ =

12

18

36 6FHGIKJ =

and13

1 3FHGIKJ =

13

19

26 6FHGIKJ =

Q18 >

19

∴18

6 >19

6

i.e.12

>13

3

Example 3.14 : Arrange in ascending order : 2 3 53 6, and

Solution : The surd 2 33 , and 56 are of orders 3, 2 and 6 respectively.

LCM of 3, 2 and 6 is 6.

∴ 23 = 226 = 46

3 = 336 = 276

and 56 = 56

Now 46 < 5 276 6<

∴ 23 < 5 36 <

∴ Ascending order of given surds is 2 5 33 6, ,


1. Multiply 32 5 43 3and

2. Multiply 3 53and

3. Divide 135 53 3by

4. Divide 24 3203by

5. Which is greater 5 44 3or ?

Radicals (Surds) 57

6. Which is smaller : 10 95 4or ?

7. Arrange in ascending order

2 3 43 6 3, ,

8. Arrange in descending order

2 3 43 4 3, ,

3.14 RATIONALISATION OF SURDS

Consider the following products

(a) 3 3 31 2 1 2. =

(b) 3 3 32 3 1 3. =

(c) 5 5 57 11 4 11. =

(d) 7 7 73 4 1 4. =

We observe that on multiplying the two surds we get the result as a rational number. In suchcases each surd is called a rationalising factor the other.

Thus (i) 3 is a rationalising factor of 3

(ii) 33 is a rationalising factor of 93 and vice versa

(iii) 5411 is a rationalising factor of 5711 and vice versa

and (iv) 74 is a rationalising factor of 734 and vice versa.

Thus we see here, how to multiply a surd by another surd in such a way that the product isa rational number. The process of converting the surds to rational numbers is calledrationalisation. Thus to rationalise, we search a factor which when multiplied by the givensurd gives us a rational number.

For example, the rationalising factor of x is x and rationalising factor of 3 2+ is

3 2−

Note : 1. The quantities x y− and x y+ are called conjugate surds. Their sum andproduct are always rational

2. Rationalisation is usually used to rationalise the denominator of a rationalexpression involving irrational surds.

58 Mathematics

Example 3.15 : Find the rationalising factor of 27

Solution : 27 = 3 3 3× ×

= 3 3

∴ Rationalising factor = 3

Example 3.16 : Find the rationalising factor of 8005

Solution : 8005 = 2 2 2 2 2 5 55 × × × × × ×

= 2 55 25 ×

= 2 525

∴ Rationalising factor is 535 or 1255

Example 3.17 : Rationalise the denominator of 3 53 5+−

Solution : 3 53 5+−

= 3 53 5

3 53 5

+−

× ++

[Multiplying and dividing by 3 5+ i.e.

by rationalising factor of 3 5− ]

=3 5

3 53 5 2 3 5

3 5

2

2 2

+

−= + + ×

−d id i d i

= 8 2 152 4 15+

− = − −

Example 3.18 : Rationalise 4 3 54 3 5+−

Solution : 4 3 54 3 5+−

= 4 3 54 3 5

4 3 54 3 5

+−

× +−

Radicals (Surds) 59

=4 3 5

4 3 5

2

2 2

+

−

d ib g d i

=16 45 24 5

16 45+ +

−

= 61 24 529

+−

= − −6129

2429 5

Example 3.19 : If a = 3 23 2−+ and b = 3 2

3 2+− , show that a + b = 10

Solution : a = 3 23 2−+

= 3 23 2

3 23 2

−+

× −−

= 3 2 2 63 2

+ −−

= 5 2 6− ...(i)

and b = 3 23 2+−

= 3 23 2

3 23 2

+−

× ++

=3 2 2 6

3 2+ +

−

= 5 2 6+ ...(ii)Adding (i) and (ii), we get

a + b = 10

Example 3.20 : Rationalise 1

3 2 1− +

Solution :1

3 2 1− + =1

3 2 1

3 2 1

3 2 1− +×

− −

− −d id id i

60 Mathematics

=3 2 1

3 2 12 2

− −

− −d i b g

=3 2 1

3 2 2 6 1− −

+ − −

=3 2 14 2 6− −−

=3 2 14 2 6

4 2 64 2 6

− −−

× ++

=4 3 4 2 4 2 18 2 12 2 6

4 2 62 2− − + − −

−b g d i

=4 3 4 2 4 6 2 4 3 2 6

16 24− − + − −

−

=2 2 4 2 6

8− −−

=2 2 6

4− −−

=2 6 2

4+ −

Example 3.21 : If 3 2 23 2

2+−

= +a b , find the values of a and b given, a and b are rational

numbers.

Solution : 3 2 23 2

3 2 23 2

3 23 2

+−

= +−

× ++

=9 6 2 3 2 4

3 22 2+ + +

−b g d i

=13 9 2

7+

Radicals (Surds) 61

=137

97

2+ = a b+ 2 ...(Given)

∴ a =137 and b =

97


1. Find the rationalising factor of each of the following :

(a) 493 (b) 2 1+

(c) 547 (d) x y xy23 23 3+ +

2. Simplify by rationalising the denominator in each of the following :

(a)25 (b)

317

(c)2 5

7

3

3

(d)11 511 5

−+

(e)3 13 1+−

3. Simplify :

2 32 3

2 32 3

+−

+ −+

4. Rationalise the denominator of

13 2 1− −

5. If a = 3 2 2+ , find the value of aa

+ 1

6. If 2 5 72 5 7

7+−

= +x y , find rational number x and y.

LET US SUM UP

An irrational number xn is called a surd if x is a rational number.

In xn , n is called index and x is called radicand

A surd having unity as its rational factor, other factor being irrational is called a pure surd

62 Mathematics

A surd having factor other than unity along with irrational factor is called a mixed surd

The order of the surd is the number that indicates the root.

The order of xn is n

Laws of radicals (a > 0, b > 0)

a an n=

a b abn n n=

ab

ab

n

nn=

Operations on surds

x y xyn n n1 1 1× = b g

xy

xy

n

n

n1

1

1

=FHGIKJ

x x xn m mn m n1 1 1 1 1d i d i= =

x xm n m nd i1 =

x x or x x x xam anmn a m a m an mn an mn= = = =d i d i1 1

Surds are similar if they have the same irrational factor.

In the simplest form of a surd, the radical has the smallest possible index with no fractionunder the radical sign and no factors of the form an under the radical sign

Similar surds can be added and subtracted

Orders of surds can be changed by multiplying index of the surd and index of the radicandbut the same number

Surds of the same order are multiplied and divided.

To compare surds, we change the surds to surds of the same order. Then they can becompared by their radicands

x y+ is called rationalising factor of x y− and vice versa.

If the product of two surds is rational, then each is called rationalising factor of the other.

Radicals (Surds) 63

TERMINAL EXERCISE

1. State which of the following are surds :

(a)25

289(b) 72912 (c) 5 13 + (d) 6255

2. Express as a pure surd :

(a) 2 53 (b) 3 44 (c) 5 25 (d) 7 73

3. Express as a mixed surd in the simplest form

(a) 4054 (b) 3205 (c) 1283 (d) 6863

4. Which of the following are pairs of similar surds :

(a) 112 343, (b) 125 500,

(c) 216 2506 , (d) 135 6253 3,


(a) 3 48 52

13 4 3− + (b) 63 28 175+ −

(c) 8 32 50+ − (d) 4 81 5 375 3 1923 3 3+ −

6. Which is greater ?

(a) 2 33or (b) 6 83 4or

7. Arrange in the descending order

(a) 3 4 53 4, , (b) 2 3 43, ,

8 Arrange in the ascending order :

16 12 3203 6, ,

9. Simplify by rationalising the denominator :

(a)5

6 5− (b)12

7 3+ (c) 3 13 1−+

(d) 3 2 23 2 2−+

10. Simplify each of the following by rationalising the denominator :

(a)1

1 2 3+ − (b)1

7 6 13+ −

11. If 5 2 37 4 3

3++

= +a b , find the values of a and b where a, b are rational numbers.

12. If x = 7 4 3+ , find the value of x x+ 1

64 Mathematics

ANSWERS


1. (a) Index 5, radicand 125

(b) Index 6, radicand 343

(c) Index 2, radicand 19

2. a, c, d, f and h

3. (i), (iv) are pure surds and (ii), (iii) are mixed surds.


1. a, c, d

2. (a) 147 (b) 543 (c) 125 (d) 75

3. (a) 5 2 (b) 3 33 (c) 2 45 (d) 3 103

(e) 4 24


1. 7 7 2. 17 2 3. 27 2

4. 3 5. 6 5 6. 3 33

7 28 2 8. 18 23 9. 3 53

10. 36 5 42 3 51 2− +


1. 10 43 2. 6756 3. 3

4. 27200

6 5. 43 6. 105

7. 3 2 46 3 3, , 8. 4 3 23 4 3, ,


1. (a) 73 (b) 2 1− (c) 53 7 (d) x y3 3−

Radicals (Surds) 65

2. (a) 2 55 (b) 51

17 (c) 2 2457

3(d) 8 55

3−

(e) 2 3+

3. 14 4. 2 2 64

+ +− 5. 6

6. x = −179171 , y = −20

171

Terminal Exercise

1. (b), (d)

2. (a) 403 (b) 3244 (c) 62505 (d) 24013

3. (a) 3 54. (b) 2 105 (c) 4 23 (d) 7 23

4. a, b, d

5. (a) 916 3 (b) 0 (c) 2 (d) 25 33

6. (a) 33 (b) 63

7. (a) 3 , 43 , 54

(b) 3 , 43 , 2

8. 163 , 3206 , 12

9. (a) 5 6 5+d i (b) 3 7 3−d i (c) 2 3− (d) 17 12 2−

10. (a) 2 2 64

+ +

(b) 7 6 6 7 54684

+ +

11. a = 11, b = –6 12. 14

66 Mathematics

4

Algebraic Expressions and Polynomials

4.1 INTRODUCTION

If you go to the market to purchase pens for your class and suppose each pen is available forRs 10, then

for one pen, you will have to pay Rs 10

two pens, you will have to pay Rs 10 × 2 or Rs 20

three pens, you will have to pay Rs 10 × 3 or Rs 30

In general for p pens, you will have to pay Rs 10 × p or Rs 10p.

Here p can take different values from one onwards. Therefore, p represents an unknownquantity, called a variable. Variables are generally denoted by alphabets x, y, z, a, b, c, p, qetc.

Thus, a variable p or x, takes different values which you assign to it and behaves like that (those)numbers (s).

The branch of mathematics which deals with variables and four fundamentaloperations on them is called Algebra.

In this lesson, you will learn about some basic concepts of algebra, algebraic expressions andpolynomials and four fundamental operations on them.

4.2 OBJECTIVES

After studying this lesson, the learner will be to able to :

identify a variablecite examples of algebraic expressionsclassify algebraic expressions into monomials, binomials and trinomialsperform four fundamental operations on algebraic expressionsevaluate an algebraic expression for given values of the variables

Algebraic Expressions and Polynomials 67

understand and identify a polynomial as a special case of an algebraic expression

write examples of polynomials in one and two variables

define a polynomial in one variable, say x

state the degree of a polynomial

perform four fundamental operations on polynomials


Different number systems and four fundamental operations on them

4.4 IDENTIFYING A VARIABLE

Consider numbers of the type

4 14 2 32

415 3 21

8, , , , , , ,− − x y z

You know that 4 14 2 32

415, , , ,− − are real numbers 3 21

8x y z, , contain unknown x, y and

z respectively and therefore do not have a fixed value like 4, –14 etc. Their values will dependon x, y and z respectively.

Thus, x, y and z are called variables.

Can you pick up variables from the following ?

13 14 154 4 3 4 7 17, , , , , , , ,− −p

x y z a b

You can see that p, x, y, z, a and b are variables and all others are constants, as their valuesdo not change.

A variable is a number which can have different values whereas a constant hasa fixed value.

4.5 ALGEBRAIC EXPRESSIONS

Consider expressions of the form

x + 4 , y – 3 , z4 , 3z + 6 , 2x + 3

y , etc.

The expressions of the above type, which involve variables and constants connected byoperations of addition, subtraction, multiplication and division are called Algebraic Expressions.

An algebraic expression is a number or a combination of numbers includingvariables joined by the four fundamental operations.

68 Mathematics

Term : When one or more of the symbols + or – occur in an algebraic expression, theyseparate the algebraic expression into parts, each of which is called a term.

Thus, x – 3y has two terms, namely, x and 3y;

x – 3y + 3 has three terms, namely, x, 3y and 3.

3x has only one term, namely 3x.

Similarly 4x2 + 3 has two terms 4x2 and 3 and x3 has only one term.

Thus a term contains either a variable or a constant or both variables(s) and constant connectedby the operation of multiplication.

Coefficient : The constant in a term is called the coefficient of the variable.

For example, in 4x + 3y, there are two terms 4x and 3y and the coefficient of x in 4x is 4and coefficient of y in 3y is 3.

Similarly, the coefficient of x2 in x2

3 is 13 because x x

2 23

13= × .


1. Identify variables and constants in the following :

2x, 41, 4y, 3t, 17

2. Write the terms and the coefficients of the variables in the terms, in the following algebraicexpressions :

(i) 3x2 – 4y (ii) 2

3 x y+

(iii) 3 4 52x y z+ − (iv) x2 + 2xy + y2

(v)x

y78−

4.6 TYPES OF ALGEBRAIC EXPRESSIONS

1. Monomial : An algebraic expression which has only one term is called a monomial.For example. 4x is a monomial.

2. Binomial : An algebraic expression which has two terms is called a binomial. Forexample 4x – 3y is a binomial :

3. Trinomial : An algebraic expression which has three terms is called a trinomial. Forexample, 2x – 3y + z is a trinomial.



Classify the following algebraic expressions into monomials, binomials and trinomials.

4y2 – 3y, – 3x, x + y + z, 32 x2 + 2x + 1, 5x – 2y2 , p

4.7 LIKE AND UNLIKE TERMS

The terms of the algebraic expression having the same variable (s) and the same exponent(s)of the variables are said to be like terms.

For example, in the algebraic expression 3xy – 9x + 8xy – 7x + 2x2, the terms 3xy and 8xy,–9x and –7x are like terms.

The terms of the algebraic expression 2x2 – 3xy + 9y2 – 7yz are all unlike i.e., there are notwo like terms in the given algebraic expression.


Write the like terms in the following algebraic expression :

(i) 3x2 – 4xy + 5x2 – 9

(ii) xyz – 5xy + 6zx + 15xyz – 1

(iii) x3 + y3 – 3xy + 6yx – 4y3

(iv) abc + bcd + cda

4.8 OPERATIONS ON ALGEBRAIC EXPRESSIONS

Let us now define the four fundamental operations of addition, subtraction, multiplication anddivision on algebraic expressions.

4.8.1 Addition

To add two or more algebraic expressions, the following steps are to be followed :

Step 1 : Group the like terms of the given algebraic expressions together.

Step 2 : Add the like terms grouped together, to get the sum.

Let us take some examples.

Example 4.1 : Add –3x + 4 and 2x2 – 7x – 2

Solution : (–3x + 4) + (2x2 – 7x – 2)

= 2x2 + (–3x – 7x) + (4 – 2)

70 Mathematics

= 2x2 – 10x + 2

∴ (–3x + 4) + (2x2 – 7x – 2) = 2x2 – 10x + 2

Algebraic expressions can be added more conveniently and accurately if :

(i) the given algebraic expressions are so arranged that the like terms are in one column; and

(ii) the coefficients of each column (i.e. of the group of like terms) are added.

Example 4.2 : Add 5x 3y 34+ − and − + +2x y 7

4

Solution : 5x 3y 34+ −

− + +2x y 74

+

(5 – 2)x + (3 + 1)y + − +FH IK34

74

= 3x + 4y + 1

∴ (5x + 3y – 34 ) + (–2x + y + 7

4 ) = 3x + 4y + 1

Example 4.3 : Add 17x 10y 9xy+ + and 3

7y73 x 18xy− −

Solution : 17x 10y 9xy+ +

+ − + −73 x 3

7y 18xy

17 73 x 10 3

71y 9 18 xy−FH IK + +FH IK + −b g

=443

737 9x y xy+ −

∴ 17 10 9x y xy+ +FH

IK + 3

773 18y x xy− −F

HIK

=443 x 73

7y 9xy+ −


Example 4.4 : Add 32 x x x 13 2+ + + and x x

2 3x 14 3− − +

Solution : 32 x x x 13 2+ + +

+ x x4 3

2− − +3 1x

x x x x4 3 232

12 1 3 2+ −FH IK + + − +b g

or x4 + x3 + x2 – 2x + 2


Add the following pairs of algebraic expressions :

(i) 23 12x x+ + and 3

714 52x x+ +

(ii) 75 13 2x x− + and 2x2 + x – 3

(iii) xy zy zx+ + and 3xy z

y−

(iv) 32

yxz

xy z+ + and − + −4 3

2y

xzxy z

4.8.2 Subtraction

In order to subtract one algebraic expression from another, we go through the following threesteps :

Step 1 : Arrange the given algebraic expressions so that the like terms are in one column.

Step 2 : Change the sign (from + to – and – to +) of each term of the algebraic expressionto be subtracted.

Step 3 : Add the coefficients of each column separately.

Let us understand the procedure by means of some examples.

Example 4.5 : Subtract :

(i) − + +4 3 23

2x x from 9 3 27

2x x− −

72 Mathematics

(ii) 5 2 12x xy+ − from 7 13 32x x

y− −

Solution : (i) 9 3 27

2x x− −

− + +4 3 23

2x x

+ – –

9 4 3 3 27

23

2+ + − − + − −FH IKb g b gx x

= 13 6 2021

2x x− −

∴ (9x2 – 3x – 27 ) – − + +FH IK4 3 2

32x x = 13x 6x 20

212 − −

(ii) 7 13 32x xy− −

5 1 22x xy− +

– + –

7 5 13 1 3 22− + − + − +b g b g b gx xy

= 2 12 52x xy− −


Subtract :

(i) 7x3 – 3x2 + 2 from x2 – 5x + 2

(ii) xy

yz+ − 3 from 3 2 7 2x

yyz x− + +

(iii) ax2 + 2hxy + by2 from cx2 + 2gxy + dy2

48.3 Multiplication

In order to multiply two algebraic expressions, each of the terms of one algebraic expressionis multiplied by each term of the other algebraic expression and the result simplified by addingthe like terms.


Let us take some examples :

Example 4.6 : Multiply 2n + 3 by n2 – 3n + 4

Solution : (2n + 3) (n2 – 3n + 4) = 2n(n2 – 3n + 4) + 3(n2 – 3n + 4)

= 2n × n2 + 2n (–3n) + 2n × 4 + 3 × n2 + 3(–3n) + 3 × 4

= 2n3 – 6n2 + 8n + 3n2 – 9n +12

= 2n3 – 3n2 – n + 12

∴ (2n + 3)(n2 – 3n + 4) = 2n3 – 3n2 – n + 12

Example 4.7 : Multiply 2x2 – 3x – 9x by − +x x

7 .

Solution : 2 3 9 72x x x x x− −FH IK − +FH IK = 2 7 3 7 9 72x x x x x x x x x− +FH IK − − +FH IK − − +FH IK

= 2 2 7 3 3 7 9 9 72 2x x x x x x x x x x x x× − + × − × − − × − × − − ×b g b g b g

= − + + − + −2 14 3 21 9 633 22x x x

x

= − + + − −2 3 14 12 633 22x x x

x

∴ 2 3 9 72x x x x x− −FH IK − +FH IK = − + + − −2 3 14 12 633 22x x x

x

4.8.4 Division

Try to recall the process when you divided 20 by 3

6 → QuotientDivisor ← 3)20 ( → Dividend

18

2 → Remainder

∴ 20 = 3 × 6 + 2

Algebraic expressions are also divided in the same way. The steps involved in the divisionprocess are explained with the help of the following example :

74 Mathematics

Example 4.8. Divide 2x2 + 5x + 3 by 2x + 3

Step 1 : Divide the first term of the dividend by the firstterm of the divisor and write the result as thefirst term of the quotient.

Step 2 : Multiply the divisor by the first term of thequotient and subtract the result from thedividend.

Step 3 : Divide the first term of the resultant dividendobtained in Step 2 by the first term of thedivisor and write the result as the second termof the quotient.

Step 4 : Multiply the divisor by the second term of thequotient and subtract the result from the resultantdividend of Step 3.

Step 5 : Repeat this process i.e. Step 3 and 4 till you getthe remainder zero or the highest exponent ofthe remainder is less than that of the highestexponent in the divisor in the above example,Remainder = zero, Quotient = x + 1

2x + 3) 2x2 + 5x + 3 (2x2 ÷ 2x = x

x2x + 3) 2x2 + 5x + 3 (

(2x + 3)x = 2x2 + 3x

2x + 3) 2x2 + 5x + 3(2x2 + 3x

2x + 3

x

– –

2x ÷ 2x = 1

∴ 2x + 3) 2x2 + 5x + 3(2x2 + 3x

– –

2x + 3

x + 1

–––––––––

2x + 3) 2x2 + 5x + 3(2x2 + 3x

– –2x + 3

x + 1

2x + 3

0– –


Let us take one more example.

Example 4.9 : Divide x4 + 2x3 + 23 x – 1

3 by x2 + 13

Solution : x2 + 2x – 13

x2 + 13

x4 + 2x3 + 2

3x – 1

3

x x4 213

+

– –

2 13

23

13

3 2x x x− + −

2x3 + 23 x

– –

− 13

2x – 13

− 13

2x – 19

+ +

– 29

Quotient = x2 + 2x – 13 and remainder = – 2

9

Thus you see that, on dividing an algebraic expression by another, the remainder may not alwaysbe zero.


1. Find the product of (x – 1) and (x2 + x + 1) :

2. Find the product of x x2 23

56+ +FH IK and x −FH IK74

3. Find the product of xy z2

34 7− + and y

x z2 2+

4. Find the remainder when x2 – x + 1 is divided by x + 1

5. Find the quotient and remainder when 6x2 – 5x + 1 is divided by 2x – 1.

IKJ

IKJ

76 Mathematics

4.9 EVALUATING AN ALGEBRAIC EXPRESSION

We can evaluate any algebraic expression for given values(s) of the variable(s) occurring init.

Let us understand the steps involved by evaluating 3x2 – x + 2 for x = 2

Step 1 : Substitute the given value(s) in place of the variable(s). i.e., 3 × 22 – 2 + 2

Step 2 : Simplify the numerical result obtained in Step 1.

3 × 22 – 2 + 2 = 3 × 4 – 2 + 2

= 12 – 2 + 2

= 12

or 3x2 – x + 2 = 12 when x = 2.

Let us take another example.

Example 4.10 : Evaluate :

(i) (3x2 – 3x + 1) (x – 1) for x = 3

(ii) (3x2 – 1) + (4x3 – 4x – 3) for x = –1

Solution : (i) Substituting 3 for x in (3x2 – 3x + 1)(x – 1) we get,

(3 × 32 – 3 × 3 + 1) (3 – 1)

= 2(3 × 9 – 9 + 1)

= 2[19] = 38

(ii) Substituting –1 for x in (3x2 – 1) + (4x3 – 4x – 3)

we get, 3 × (–1)2 – 1 + [4(–1)3 – 4(–1) – 3]

= 3 – 1 + [–4 + 4 – 3]

= 2 – 3 = –1.


Evaluate each of the following algebraic expression for the indicated values(s) of thevariables :

1. x2 + 3x – 5 at x = 4

2. 23

45

75

3 2x x+ − at x = –1

3. 23 4

722x

xx

+ − at x = 13


4. xy xy2

3 11+ − at x = 2 , y = 1

5. 3xyz – x3 – y3 + z3 at x = 2 , y = 1 , z = –3.

4.10 POLYNOMIALS

Polynomial : An algebraic expression in which the variable(s) does (do) not occurin the denominator and the exponents of the variable (or variables) are wholenumbers and the co-efficients of different terms are real numbers is called apolynomial.

For example 5 3 3, ,x y a b− − and x y3 2

3 2− are all polynomials whereas x x3 1− is not a

polynomial. Also x y+3 is not a polynomial.


1. Which of the following algebraic expressions are polynomials ?

23 13x + , 5 4 3 52 2

3x yx

y x y− − +, , , 3x3 – y3z, 3 4 13 22x y

z+ + .

4.11 GENERAL FORM OF A POLYNOMIAL IN ONE VARIABLE

An expression of the form

a0xn + a1xn–1 + a xn2

2− + .... + an–1x + an

where a0, a1, a2, ..., an are real constants, x is a variable and n is a non-negative integer is calleda polynomial in the variable x.

4.12 DEGREE OF A POLYNOMIAL :

The degree of a polynomial in one variable is the greatest exponent of the variableoccurring in the various terms of the polynomial.

For example, the degree of the polynomial 2x3 – 9x + 3 is 3. Similarly the degree of x4 9−

is 1 and the degree of 3 is zero.

The degree of a constant is taken as zero.

78 Mathematics

Note : A polynomial in one variable is written in decreasing powers of the variable and thisis called the standard form of the polynomial.

For example, 4 52

3 43 4x x x+ + + is written as

52

4 3 44 3x x x+ + + in standard form.


Write the following polynomials in standard form and specify their degree :

(i) 5x2 + 1 + 2x

(ii) 3x4 + x5 – 32 x2 + 1

(iii) 4x3 – 7x2 – x – 2

4.13 OPERATIONS ON POLYNOMIALS

We have already learnt four fundamental operations on algebraic expressions. As polynomialsare special types of algebraic expressions, the same rules and steps, stated in case of operationson algebraic expressions, hold good in case of polynomials also.

Let us illustrate these with the help of examples

Example 4.11 : Add 7x2 – 3x + 4 and 3x3 + 5x2 – 4x + 73

Solution : Writing the polynomials in standard form in columns, with like terms coming inthe same column, we have

+ 7x2 – 3x + 4

3x3 + 5x2 – 4x + 73

3x3 + (7 + 5)x2 – (3 + 4)x + 4 73

+FHGIKJ

= 3x3 + 12x2 – 7x + 193

Thus, 7 3 4 3 5 4 73 3 12 7 19

32 3 2 2 2x x x x x x x x− + + + − +FH IK = + − +d i


Example 4.12 : Subtract 3x – 5x2 + 7 + 3x3 from 2x2 – 5 + 11x – x3

Solution : Rewriting the polynomials in standard form in columns, we have

–x3 + 2x2 + 11x – 5

+ 3x3 – 5x2 + 3x + 7

– + – –

–(1+ 3)x3 + (2 + 5)x2 + (11 – 3)x – (5 + 7)

= – 4x3 + 7x2 + 8x – 12

Thus, (2x2 – 5 + 11x – x3) – (3x – 5x2 + 7 +3x3)

= –4x3 + 7x2 + 8x – 12

Example 4.13 : Find the product of

(2x2 + 7x + 2) and (3x – 1)

Solution : (2x2 + 7x + 2) (3x – 1)

= 3x (2x2 + 7x + 2) – 1(2x2 + 7x + 2)

= 6x3 + 21x2 + 6x – 2x2 – 7x – 2

= 6x3 + (21 – 2)x2 + (6 – 7)x – 2

= 6x3 + 19x2 – x – 2

Thus, (2x2 + 7x + 2) (3x – 1) = 6x3 + 19x2 – x – 2

Example 4.14 : Find the product of

(0.2x2 + 0.7x + 3) and (0.5x2 – 3x)

Solution : (0.2x2 + 0.7x + 3) (0.5x2 – 3x)

= 0.5x2 (0.2x2 + 0.7x + 3) – 3x(0.2 x2 + 0.7x + 3)

= 0.1x4 + 0.35 x3 + 1.5x2 – 0.6x3 – 2.1x2 – 9x

= 0.1x4 + (0.35 – 0.6)x3 + (1.5 – 2.1)x2 – 9x

= 0.1x4 – 0.25x3 – 0.6x2 – 9x

Thus, (0.2x2 + 0.7x + 3) (0.5x2 – 3x)

= 0.1x4 – 0.25 x3 – 0.6x2 – 9x

80 Mathematics

Example 4.15 : Divide x3 – 1 by x – 1

Solution : x2 + x + 1x – 1 ) x3 – 1

x3 – x2

– +x2 – 1x2 – x

– +x – 1x – 1

– +0

∴ Quotient is x2 + x + 1, Remainder = 0

Example 4.16 : Divide 2x3 – 3x2 + 5x – 11 by 2x – 5

Solution : x x2 72

454

+ +

2x – 5 ) 2x3 – 3x2 + 5x – 11 (2x3 – 10x2

– +

7x2 + 5x – 11

7x2 – 352 x

– +

452 x – 11

452 x – 225

4

– +

1814

∴ Quotient = x x2 72

454

+ +

Remainder =1814 .



1. Add 2x3 + 7x2 – 5x + 7 and –2x 2 + 7x3 – 3x – 7

2. Add 73

25

3 75

3 2x x x+ − + and 23

35

3 35

3 2x x x+ − +

3. Subtract 2x3 + 7x – 5x2 + 2 from 5x + 7 – 3x2 + 5x3

4. Subtract 12x3 – 3x2 – 11x +13 from 5x3 + 7x2 + 2x – 4

5. Find the product of (x2 + 1) and (x3 – 1)

6. Find the product of 23

54

32x x+ −FHG

IKJ and (3x2 + 4x + 1)

7. Find the remainder and quotient for each of the following :

(i) Divide x4 –1 by x – 1

(ii) Divide x3 – 3x2 + 5x – 8 by x – 2

LET US SUM UP

An unknown quantity, which can have various values, is called a variable.

An expression which has one or more terms involving variable joined by + or – sign iscalled an algebraic expression.

An algebraic expression with one term is called a monomial, with two terms, a binomialand with three terms is called a trinomial.

The terms of an expression having the same variable with same exponent are called liketerms. The terms which are not like are called unlike terms.

Four fundamental operations on algebraic expressions yield another algebraic expression.

The process of substituting a numerical value of the variable in the algebraic expressionis called evaluating the algebraic expression for that value of the variable.

An expression of the type a0xn + a1xn–1 + ... + an, where a0, a1, ..., an are real numbers,x is a variable and n is a non-negative integer is called a polynomial in x of degree nin standard form.

The degree of a polynomial in one variable is the greatest exponent of the variableoccurring in the various terms of the polynomial.

The degree of a constant is zero.

Four fundamental operations on polynomials yield another polynomial. In case of divisionof polynomial, the degree of remainder is less than that of divisor.

82 Mathematics

TERMINAL EXERCISE

1. Identify monomials, binomials and trinomials in the following :

2x + 3, 3z, 4x + 3z + 2, 3z2 – 9, 5x5, 3xyz

2. Identify like terms and unlike terms in the two algebraic expressions given below :

(i) 32 5 7 43 2x x x+ − + and 5 7 3

2 303 2x x x+ − −

(ii) 32

5 3 122 3

xxyz x xyz+ − − and 17 3

2 5 73

3xyz x xy x+ + −

3. Perform the indicated operations on the following algebraic expressions

(i) Add 3x2 + 7x + 4y and 6 32 72 3x x x y+ − −

(ii) Add 3 7 3

23 2x

yx x yz+ − and 2 2 53

22x x

yx yz− +

(iii) Subtract 3x3 – 5x4 + 6x2 – 2x + 1 form 2x + 3x4 – 2x3 + 5x2 –1

(iv) Multiply (x3 + 1) by x x

y

2 1+ +d i

4. Perform the indicated operation on the following polynomials and find the value of resultingpolynomial at x = 2

(i) Add 7x4 + 3x3 – 2x + 1 and 3x4 + 2x – 4x3 –1

(ii) Subtract 3x2 – 4x3 + 7x – 1 from 6x3 + 3x2 + 6x – 1

(iii) Multiply (3x2 + 2x + 1) by (2x + 3)

(iv) Multiply (0.5 x2 –3x – 0.7) by (0.7x – 0.3)

5. Find remainder and quotient for each of the following :

(i) Divide (x6 –1) by x – 1

(ii) Divide x3 + 1 by (x + 1)

(iii) Divide x6 + 5x4 + 3x2 – 1 by x2 – 2


ANSWERS


1. Variables : 2x, 4y, 3t ; constants : 41, 172. (i) Terms : 3x2, 4y ; coefficients are 3, 4 respectively

(ii) Terms : 23 x , y ; coefficients are 2

3 , 1 respectively

(iii) Terms : 3x, 4y, 52 z ; coefficients are 3, 4, 5

2 respectively

(iv) Terms : x2, 2xy, y2; coefficients are 1, 2, 1 respectively

(v) Terms : x

y78, ; coefficients are 1

7 , 8.


1. Monomials : –3x, pBinomials : 4y2 – 3y, 5x – 2y2

Trinomials : x + y + z, 32 x2 + 2x + 1


(i) 3x2, 5x2 (ii) xyz, 15xyz ; 6xz, 7zx(iii) y3, –4y3 ; –3xy, 6yx (iv) no like terms involved


(i) 2321

54 62x x+ + (ii) 7

5 23 2x x x+ + −

(iii) 4xy + zx (iv) − +yxz

xy

2


(i) –7x3 + 4x2 – 5x (ii) 2 3 10 2xy

yz x− + +

(iii) (c – a)x2 + 2(g – h)xy + (d – b)y2


1. x3 –1 2. x x x3 21312 3

3524− − −

3.14

38

72

32

142− + + − +yzx

yx

xzy

z z 4. 3

5. Quotient : 3x – 1, remainder ; 0 (zero)

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1. 23 2. − 1915 3. 2195

364. –1 5. –54


1. 23 13x + , 5x2 – y2, 3x3 – y3z


(i) 5x2 + 2x + 1; 2 (ii) x5 + 3x4 – 32 x2 + 1; 5

(iii) 4x3 – 7x2 – x – 2; 3


1. 9x3 + 5x2 – 8x 2. 3x3 + x2 –6x + 2

3. 3x3 + 2x2 – 2x + 5 4. –7x3 + 10x2 + 13x – 17

5. x5 + x3 – x2 – 1 6. 2 7712

103

434 34 3 2x x x x+ − − −

7. (i) Quotient : x3 + x2 + x + 1, remainder : 0 (zero)

(ii) Quotient : x2 – x + 3 ; remainder : –2.

Terminal Exercise1. Monomials : 3z, 5x5, 3xyz

Binomials : 2x + 3, 3z2 – 9Trinomial : 4x + 3z + 2

2. (i) Like terms : 32 73 3x x, ; 5x2, − 3

22x ; – 7x, 5x; 4, – 30.

(ii) Like terms : 32

73x x,− ; –3x3, 3

23x ; – 12xyz, 17 xyz

Unlike terms : 5 52x

yz xy,

3. (i) x x x y3 29 112 3+ + − (ii) 9 23 2 2x x

y x yz+ +

(iii) 8x4 – 5x3 – x2 + 4x – 2 (iv)x x x x x

y

5 4 3 2 1+ + + + +d i

4. (i) 10x4 – x3 (ii) 10x3 – x ; 78

(iii) 6x3 + 13x2 + 8x + 3; 119 (iv) 0.35 x3 – 0.36 x2 + 0.41 x + 0.21; 2.39

5. (i) Quotient : x5 + x4 + x3 + x2 + x + 1 ; remainder : 0(zero)

(ii) Quotient : x2 – x + 1 ; remainder : 0(zero)

(iii) Quotient : x4 + 7x2 + 17 ; remainder : 33

Special Products and Factorisation 85

5

Special Products and Factorisation

5.1 INTRODUCTION

You have already studied different number systems, algebraic expressions and polynomials ina previous lesson.

Try to recall how you would find the square of a number, say 1032. Simply you have to multiply103 with itself. Is this the only method ? No! Here is another method.

Method 1 : Direct Multiplication Method 2 : Breaking up 103

103 103 = 100 + 3

× 103 1032 = (100 + 3)2

309 = (100 + 3) (100 + 3)

000 = 100 × 100 + 100 × 3 + 3 × 100 + 3 × 3

103

10609 ∴ 1032 = 1002 + 2 × 100 × 3 + 32

or 1032 = 10609 or 1032 = 10000 + 600 + 9

∴ 1032 = 10609

The answer is the same in both cases.

Notice carefully the expression in the box in the second method. This is the most importantstep in this calculation.

i.e., 1032 = (100 + 3)2

= square of the first term + two times the product of the first and the second terms + squareof the second term.

This can easily be generalised to the square of any binomial (or an algebraic expression withtwo terms).

Now, how would you find the cube of a number, say, 993 ?

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Direct multiplication will surely make the calculations lengthy and difficult, but if you write,

99 = 100 – 1

∴ 993 = (100 – 1)3

Thus, you can easily calculate 993 if you know the formula for the cube of a binomial.

In this lesson, you will study these as well as many other useful results which will makecalculations simpler and also help you to understand polynomials better.

5.2 OBJECTIVES


write formulae for special products

calculate squares and cubes of numbers using special products

factorise a given algebraic expression and a polynomial in one variable

factorise a given quadratic polynomial by splitting the middle term

determine the HCF and LCM of two or more polynomials

cite examples of rational expressions in one or two variables

express a given rational expression in the simplest form

perform four fundamental operations on rational expressions.


Knowledge of four fundamental operations on numbers and algebraic expressions

GCD and LCM of numbers

Polynomials and operations on them

Area of squares and rectangles.

5.4 SQUARE OF A BINOMIAL

(a) Consider a binomial say, x + y

Let us find (x + y)2

(x + y)2 = (x + y) (x + y)

= x (x + y) + y(x + y)

= x.x + x.y + y.x + y.y

= x2 + 2xy + y2 ...(i)


The square of a binomial is equal to the sum of the square of the first term, twotimes the product of both the terms and the square of the second term.

or (x + y)2 = x2 + 2xy + y2 ...(i)

You can easily make a model and verify this formula geometrically.

Method

Step 1 : Take a square piece of cardboard ABCD and markpoints ‘P’ and ‘Q’ on sides AB and AD, respectively,in such a way that

AP = AQ = ‘x’ (say)

and PB = QD = ‘y’ (say) [See Fig. 5.1]

Step 2 : Through ‘P’ draw a line ‘l’ parallel to AD, intersecting‘DC’ at ‘R’.

Step 3 : Through ‘Q’ draw a line ‘m’ parallel to ‘AB’,intersecting ‘BC’ at S.

‘l’ and ‘m’ divide ABCD into two squares and tworectangles. (See Fig. 5.2)

∴ Area of ABCD = x2 + xy + y2 + xy

[Using the formulae for the areas of a square and a rectangle]

∴ (x + y)2 = x2 + 2xy + y2

(b) Consider the square of the binomial (x – y)

There are many ways of finding this product. Two methods are given below:

Method 1 : Direct Multiplication

(x – y)2 = (x – y) (x – y)

= x(x – y) – y(x – y)

= x.x + x (–y) – y.x –y. (–y)

= x2 – xy – xy + y2 ...(ii)

∴ (x– y)2 = x2 – 2xy + y2

Fig. 5.2

Fig. 5.1

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Method 2 : Using (x + y)2

x – y = x + (–y)

or (x – y)2 = [x + (–y)]2

= x2 + 2x(–y) + (–y)2

or (x – y)2 = x2 – 2xy + y2

∴ (x – y)2 = x2 – 2xy + y2

Deductions : What happens when you add the two sides of Formulae (i) and (ii)

(x + y)2 = x2 + 2xy + y2 ...(i)

+ (x – y)2 = x2 – 2xy + y2 ...(ii)

(x + y)2 + (x – y)2 = 2x2 + 2y2

or (x + y)2 + (x – y)2 = 2(x2 + y2)

On subtracting (ii) from (i), we get

(x + y)2 – (x – y)2 = 4xy

These formulae are very useful in finding the squares of binomials as well as of numbers.


Example 5.1 : Find each of the following products :

(i) (2a + 3b)2 (ii) (3x + 4y)2

(iii) (4x – 1)2 (iv) x x+FH IK12

Solution : (i) We know that (x + y)2 = x2 + 2xy + y2

Replacing ‘x’ by ‘2a’ and ‘y’ by ‘3b’, we get

(2a + 3b)2 = (2a)2 + 2(2a)(3b) + (3b)2

= 4a2 + 12ab + 9b2

(ii) Using Formula (i), we get

(3x + 4y)2 = (3x)2 + 2(3x)(4y) + (4y)2

= 9x2 + 24xy + 16y2

∴ (3x + 4y)2 = 9x2 + 24xy + 16y2.

(iii) Using Formula (ii) above, we get

(4x – 1)2 = (4x)2 – 2(4x).1 + 12


= 16x2 – 8x + 1∴ (4x – 1)2 = 16x2 – 8x + 1

(iv) Using Formula (i), we get

x x+FH IK12

= x x x x2

22 1 1+ + FH IK. .

= xx

222 1+ +

∴ x x+FH IK12

= xx

22

1 2+ +

Example 5.2 : Calculate each of the following without actual multiplication :

(a) 1012 (b) 982

Solution : (a) We can write 101 as

101 = 100 + 1∴ 1012 = (100 + 1)2

= 1002 + 2 × 100 × 1 + 12

= 10000 + 200 + 1∴ 1012 = 10201

(b) 982

Writing 98 = 90 + 8 and using Formula (i) will make the calculations a little lengthy, eventhough this method is correct.Let us write : 98 = 100 – 2

∴ 982 = (100 – 2)2

= 1002 – 2 × 100 × 2 + 22

= 10000 – 400 + 4∴ 982 = 9604

Example 5.3 : Evaluate xx

22

1+ , given that x x− 1 = 1

Solution : It is given that x x− 1 = 1

∴ x x−FH IK12

= 12

or xx

x x2

21 2 1+ − . . = 1

or xx

22

1 2+ − = 1

∴ xx

22

1+ = 3

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1. Find each of the following products :

(i) (5x + y)2 (ii) (x – 3)2 (iii) (ac + bd)2

(iv) (ab – c)2 (v) x x−FH IK12

(vi) (2x – 5y)2

(vii) x3 1

2+FH IK (viii) z

2 12

−FH IK (ix) (a2 + b2)2

(x) xy

yx+FHGIKJ

2

2. Without actual multiplication calculate each of the following :

(i) 972 (ii) 2012 (iii) 9982

(iv) 492 (v) 9992 (vi) 1022

3. Find each of the following :

(i) xx

22

1+ , given that xx

+ 1 = 2

(ii) aa

22

1+ , given that aa

− 1 = 2

(iii) xx

22

1+ , given that xx

− 1 = –3

(iv) x2 + y2, given that x + y = 5 and x – y = 3

(v) xy given that x + y = 3 and x – y = 1

(vi) x2 + y2 and xy, given that x + y = 10 and x – y = 6.

5.5 THE SPECIAL PRODUCT (x + y) (x – y)

Let us find the product (x + y) (x – y)

(x + y) (x – y) = x(x – y) + y(x – y)

= x.x + x(–y) + yx + y(–y)

= x2 – xy + xy – y2 = x2 – y2

∴ (x + y)(x – y) = x2 – y2

This is one of the most commonly used formula in algebra.


Let us consider some examples.

Examples 5.4 : Find each of the following products :

(i) (3x + y)(3x – y) (ii) (x2 – y2)(x2 + y2)

Solution : (i) (3x + y)(3x – y) = (3x)2 – y2

= 9x2 – y2

∴ (3x + y)(3x – y) = 9x2 – y2

(ii) (x2 – y2)(x2 + y2)

(x2 – y2)(x2 + y2) = (x2)2 – (y2)2

= x4 – y4

∴ (x2 – y2)(x2 + y2) = x4 – y4

Example 5.5 : Evaluate 47 × 53 without actual multiplication.

Solution : 47 × 53 = (50 – 3) (50 + 3)

= 502 – 32

= 2500 – 9

= 2491

∴ 47 × 53 = 2491

The answer has been found without acutal multiplication.


1. Calculate each of the following :

(i) (y + 2)(y – 2) (ii) (xy – 1)(xy + 1)

(iii) (a2 + 5)(a2 – 5) (iv) (x– y2) (x + y2)

(v) 98 × 102 (vi) 85 × 75

5.6 CUBES OF BINOMIALS

Consider a binomial ‘x + y’. Let us find its cube.

(x + y)3 = (x + y) (x + y)2

= (x + y) (x2 + 2xy + y2)

= x(x2 + 2xy + y2) + y(x2 + 2xy + y2)

= x.x2 + x.2xy + x.y2 + y.x2 + y.2xy + y.y2

= x3 + 2x2y + xy2 + x2y + 2xy2 + y3

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= x3 + 3x2y + 3xy2 + y3

∴ (x + y)3 = x3 + 3x2y + 3xy2 + y3

Observe that there is another way of writing (x + y)3 :

(x + y)3 = x3 + 3x2y + 3xy2 + y3

=x3 + 3xy(x + y) + y3

∴ (x + y)3 = x3 + 3xy(x + y) + y3

Consider (x – y)3 :

There are two commonly used ways of finding this product. Let us study them.

Method 1 : Direct Multiplication

(x – y)3 = (x – y) (x – y)2

= (x – y) (x2 – 2xy + y2)

= x(x2 – 2xy + y2) – y(x2 – 2xy + y2)

= x.x2 + x.(–2xy) + x.y2 – y.x2 – y.(–2xy) – y.y2

= x3 – 2x2y + xy2 – x2y + 2xy2 – y3

= x3 – 3x2y + 3xy2 – y3

∴ (x – y)3 = x3 – 3x2y + 3xy2 – y3

Method 2 : Using (x + y)3

x – y = x + (–y)

∴ (x – y)3 = [x + (–y)]3

or (x – y)3 = x3 – 3x2y + 3xy2 – y3

Writing this formula in another way, we get

(x – y)3 = x3 – 3xy(x – y) – y3



(i) (3x + 2y)3 (ii) (x – 4y)3 (iii) 1013

Solution. (i) (3x + 2y)3 = (3x)3 + 3(3x)2 2y + 3(3x)(2y)2 + (2y)3

= 27x3 + 3.9 x2.2y + 3.3x.4y2 + 8y3

= 27x3 + 54x2y + 36xy2 + 8y3

∴ (3x + 2y)3 = 27x3 + 54x2y + 36xy2 + 8y3


(ii) (x – 4y)3 = x3 – 3x2(4y) + 3x(4y)2 – (4y)3

= x3 – 12 x2y + 3x.16y2 – 64y3

∴ (x – 4y)3 = x3 – 12x2y + 48xy2 – 64 y3

(iii) 1013

We can write 101 = 100 + 1

∴ 1013 = (100 + 1)3

= 1003 + 3 × 1002 × 1 + 3 × 100 × 12 + 13

= 1000000 + 3 × 10000 + 3 × 100 + 1

= 1000000 + 30000 + 300 + 1

∴ 1013 = 1030301


Calculate each of the following :

(i) (2x + y)3 (ii) (5x – y)3 (iii) (xy – z)3

(iv) 993 (v) 973 (vi) 483

5.7 SOME OTHER SPECIAL PRODUCTS

Besides the squares and cubes of binomials, there are some other special products which haveuseful applications.

Let us now study some more special products.

(x + y)(x2 – xy + y2)

Now, (x + y)(x2 – xy + y2) = x(x2 – xy + y2) + y(x2 – xy + y2)= x.x2 + x(–xy) + x.y2 + y.x2 + y(–xy) + y.y2

= x3 – x2y + xy2 + x2y – xy2 + y3

= x3 + y3

∴ (x + y)(x2 – xy + y2) = x3 + y3

Similarly, on simplifying (x – y)(x2 + xy + y2), you will get x3 – y3. The reader is advisedto multiply the above and check the result

(x – y)(x2 + xy + y2) = x3 – y3

Consider the following product :

(x + y + z) (x2 + y2 + z2 – xy – yz – zx)

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Now, (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

= x(x2 + y2 + z2 – xy – yz – zx) + y(x2 + y2 + z2 – xy – yz – zx)

+ z(x2 + y2 + z2 – xy – yz – zx)

= x3 + xy2 + xz2 – x2y – xyz – zx2 + yx2 + y3 + yz2 – xy2 –yz2 – xyz

+ zx2 + zy2 + z3 – xyz – yz2 – xz2

= x3 + y3 + z3 – 3xyz

∴ (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = x3 + y3 + z3 – 3xyz



(i) (2x + y) (4x2 – 2xy + y2)

(ii) (x – 2) (x2 + 2x + 4)

(iii) (4x + y + z) (16x2 + y2 + z2 – 4xy – yz – 4zx)

Solution : (i) (2x + y) (4x2 – 2xy + y2)

= (2x + y) [(2x)2 – (2x)y + y2]

= (2x)3 + y3

= 8x3 + y3

∴ (2x + y) (4x2 – 2xy + y2) = 8x3 + y3

(ii) (x – 2) (x2 + 2x + 4)

= (x – 2) (x2 + x.2 + 22)

= x3 – 8

∴ (x – 2) (x2 + 2x + 4) = x3 – 8

(iii) (4x + y + z) (16x2 + y2 + z2 – 4xy – yz – 4zx)

= (4x + y + z) [(4x)2 + y2 + z2 – (4x)y – yz – (4x) z)]

= (4x)3 + y3 + z3 – 3(4x)yz

= 64x3 + y3 + z3 – 12xyz

∴ (4x + y + z) (16x2 + y2 + z2 – 4xy – yz – 4zx) = 64x3 + y3 + z3 – 12xyz



(i) (2x + 3y)3 (ii) (4x – 5)3 (iii) (xy – 1)3


(iv) (xy + z)3 (v) x x−FH IK13

(vi) (ax – by)3

(vii) (2x + 1) (4x2 – 2x + 1) (viii) (y – 3) (y2 + 3y + 9)

(ix) (1 + x) (1 – x + x2)

(x) (2x + y + 3z) (4x2 + y2 + 9z2 – 2xy – 6xz – 3yz)

2. Calculate each of the following without actual multiplication :

(i) 513 (ii) 1023 (iii) 953

5.8 FACTORISATION

You have studied some special products.

Let us take (x + y) (x – y) = x2 – y2

You can also write it as x2 – y2 = (x + y) (x – y)

In other words, the quadratic binomial x2 – y2 has been written as a product of two linearbinomials x + y and x – y. Thus each one of the two x + y and x – y is a factor of x2 – y2.

Note : If you divide x2 – y2 by x + y, you get x – y and vice versa.

Definition

The process of writing a polynomial as a product of two (or more) polynomialsis called factorisation.

Each polynomial in the product is called a factor of the given polynomial.

∴ (x + y) (x – y) is the factorisation of x2 – y2 and x + y and x – y are the factors ofx2 – y2.

Thus factorisation is the reverse of the process of finding products. The polynomial obtainedby multiplying two other polynomials can be factorised. Let us take the following :

1. Factorisation of difference of two squares

As (x + y) (x – y) = x2 – y2

∴ (x + y) and (x – y) are factors of x2 – y2

∴ x2 – y2 = (x + y) (x – y)

2. Factorisation of a perfect square trinomial

We know that

(x + a)2 = x2 + 2ax + a2

or x2 + 2ax + a2 = (x + a)2

Thus (x + a) (x + a) is the required factorisation of x2 + 2ax + a2

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(ii) Now, consider the expression (x – a)2

(x – a)2 = x2 – 2ax + a2

Thus, (x – a) (x – a) is the required factorization of x2 – 2ax + a2

Let us take some examples

Example 5.8 : Factorise each the following polynomials :

(i) x2 – 4 (ii) x2 – 2x + 1

(iii) 9x2 + 6x + 1 (iv) a2c2 – b2

Solution : (i) x2 – 4 = x2 – 22

This is of the form x2 – y2

∴ x2 – 4 = (x + 2) (x – 2)

(ii) x2 – 2x + 1 = x2 – 2.x.1 + 12

i.e. x2 – 2x + 1 = (x –1) (x – 1)

(iii) 9x2 + 6x + 1 = (3x)2 + 2.3x.1 + 12

= (3x + 1)2

i.e. 9x2 + 6x + 1 = (3x + 1) (3x + 1)

(iv) a2c2 – b2 = (ac)2 – b2

This is of the form x2 – y2

Hence a2c2 – b2 = (ac + b) (ac – b)


Factorise each of the following :

(i) 4a2 – 9 (ii) 4a2 – b2

(iii) a2c – b2c (Take ‘c’ common first) (iv) 9x2 – 25y2

(v) 25x2 + 40x + 16 (vi) x2 – 6x + 9

(vii) x2 + 12x + 36

We can also evaluate some numerical expressions by using factorisation.



Example 5.9 : Without actual multiplication, find the value of each of the following :

(i) 1022 – 982 (ii) 512 – 492 (iii) 272 – 732

Solution : (i) 1022 – 982 = (102 + 98) (102 – 98)

= 200 × 4

= 800

∴ 1022 – 982 = 800

(ii) 512 – 492 = (51 + 49) (51 – 49)

= 100 × 2

= 200

∴ 512 – 492 = 200

(iii) 272 – 732 = (27 + 73) (27 – 73)

= 100 × (–46) = –4600

∴ 272 – 732 = – 4600

3. FACTORISATION INVOLVING CUBES

Some of the cubic polynomials can be factorised using specials products. These are discussedbelow :

(a) Factorisation of perfect cubes

(i) Consider the special product (x + y)3

(x + y)3 = x3 + 3x2y + 3xy2 + y3

∴ x3 + 3x2y + 3xy2 + y3 = (x + y)(x + y) (x + y)

(ii) (x – y)3 = x3 – 3x2y + 3xy2 – y3

∴ x3 – 3x2y + 3xy2 – y3 = (x – y)(x – y) (x – y)

(b) Sum and difference of two cubes

(i) We know that

(x + y)(x2 + xy + y2) = x3 + y3

∴ x3 + y3 = (x + y)(x2 – xy + y2) is the factorisation of x3 + y3

(ii) We know that

(x – y)(x2 + xy + y2) = x3 – y3

∴ x3 – y3 = (x – y)(x2 + xy + y2) is the factorisation of x3 – y3

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Example 5.10 : Factorise each of the following polynomials :

(i) x3 + 64y3 + 12x2y + 48xy2

(ii) 8x3 – 36x2y + 54xy2 – 27y3

(iii) x3 + 27 (iv) 8x3 – 1 (v) 125x3 – 64y3

Solution : (i) x3 + 64y3 + 12x2y + 48xy2

= x3 + 3x2(4y) + 3x(4y)2 + (4y)3

= (x + 4y)3

∴ x3 + 64y3 + 12x2y + 48xy2 = (x + 4y)(x + 4y)(x + 4y)

(ii) 8x3 – 36x2y + 54xy2 – 27y3

= (2x)3 – 3(2x)2(3y) + 3 × 2x(3y)2 – (3y)3

= (2x – 3y)3

∴ 8x3 – 36x2y + 54xy2 – 27y3 = (2x – 3y)(2x – 3y)(2x – 3y)

(iii) x3 + 27 = x3 + 33

= (x + 3)(x2 – 3x + 32)

= (x + 3) (x2 – 3x + 9)

∴ x3 + 27 = (x + 3)(x2 – 3x + 9)

(iv) 125x3 – 64y3 = (5x)3 – (4y)3

= (5x – 4y) [(5x)2 + 5x.4y + (4y)2]

= (5x – 4y) (25x2 + 20xy + 16y2)

∴ 125x3 – 64y3 = (5x – 4y) (25x2 + 20xy + 16y2)

Example 5.11 : Evaluate a3 + b3 given that a + b = 7 and a.b = 12

Solution : a3 + b3 = (a + b) (a2 – ab + b2)

= (a + b)(a2 + b2 + 2ab – 2ab – ab)

= (a + b)[(a + b)2 – 3ab]

∴ a3 + b3 = 7(49 – 3 × 12)

= 7(49 – 36) = 7 × 13 = 91

∴ a3 + b3 = 91


1. Factorise each of the following polynomials :

(i) u2 – v2 (ii) 25a2 – 1 (iii) x4 – y4


(iv) x3 – 8y3 (v) a3 + 216 b3 (vi) a3 – 343

(vii) x8 – 1 (viii) a3 + 6a2 + 12a + 8

(ix) x3 – 33x2 + 363x – 1331 (x) x2 + 22x + 121

(xi) 9x2 – 30xy + 25y2 (xii) 81y2 – 36xy + 4x2

2. Evaluate each of the following without actual multiplication :

(i) 852 – 152 (ii) 1012 – 992 (iii) 832 – 172

(iv) a3 – b3 given that a – b = 1, a.b = 4

(v) aa

33

1+ given that aa

+ 1 = 2 (vi) 63 – 53

(vii) a3 – b3 given that a – b = –1, ab = 6

5.9 FACTORISING A QUADRATIC POLYNOMIAL BY SPLITTING THE MIDDLETERM

Let us consider the product (x + a) (x + b) = x(x + b) + a(x + b)

= x2 + (a + b)x + ab ...(i)We can write (i) as x2 + ax + bx + ab

and factorise by taking common factors from first two and last two terms separately

∴ x2 + ax + bx + ab = x(x + a) +b(x + a)= (x + a) (x + b)

This gives us a method to factorise a quadratic polynomial.

Let us take some examples and illustrate

x2 + 5x + 6

= x2 + 3x + 2x + 6

= x(x + 3) + 2(x + 3)

= (x + 3) (x + 2)

What have we done ? We have splitted the middle term 5x into two factor 3x and 2x suchthat their product is equal to 6x2, i.e. the product of the constant and the quadratic term.

We could have done in the following manner also :

Comparing x2 + 5x + 6 with (i), we get

a + b = 5 and ab = 6

Factors of 6 Sum of factors

1, 6 7

2, 3 5

∴ The required pair of values of a and b are 2 and 3 respectively.

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As x2 + (a + b)x + ab was factorised as (x + a)(x + b), therefore

x2 + (2 + 3)x + 2 × 3 can be written as

(x + 2) (x + 3)

∴ x2 + 5x + 6 = (x + 2)(x + 3)


(i) x2 + 3x + 2 (ii) x2 – 5x + 6

(iii) x2 + x – 6 (iv) x2 – x – 6

Solution : (i) x2 + 3x + 2

We need two factors of 2 whose sum is 3

Factors of 2 Sum of the factors

2 and 1 2 + 1 = 3

∴ x2 + 3x + 2 = x2 + (2 + 1) x + 2 × 1

= (x + 2)(x + 1)

∴ x2 + 3x + 2 = (x + 2)(x + 1)

(ii) x2 – 5x + 6

Here, we need two factors of 6 whose sum is negative and product is positive.

∴ Both the numbers should be negative

Factors of 6 Sum of the factors

–1, –6 (–1) + (–6) = –7

–2, –3 (–2) + (–3) = –5

∴ The required numbers are –2 and –3

Hence, x2 – 5x + 6 = x2 + (–2 – 3) x + (–2)(–3)

= (x – 2)(x – 3)

∴ x2 – 5x + 6 = (x – 2)(x – 3)

(iii) x2 + x – 6

Here the product of the numbers is negative and the sum is positive

∴ One of the two factors is negative and the other is positive

Factors of –6 Sum of the factors Right (√) or wrong (×)

–1, 6 (–1) + 6 = 5 ×

1, –6 1 + (–6) = –5 ×

–2, 3 (–2) + 3 = 1 √

2, –3 2 + (–3) = –1 ×

∴ The required pair of numbers is –2 and 3.


Hence, x2 + x – 6 = x2 + (–2 + 3) x + (–2) × 3

= (x – 2)(x + 3)

∴ x2 + x – 6 = (x – 2) (x + 3)

(iv) x2 – x – 6

Here the sum and product both are negative

∴ One number (smaller one) has to be positive and the other number has to be negative.

Factors of –6 Sum of the factors

1, –6 1 + (–6) = –5

2, –3 2 + (–3) = –1

∴ The required pair of numbers is 2 and –3.

∴ x2 – x – 6 = x2 + [2 + (–3)]x + 2 × (–3)= (x + 2)(x – 3)

∴ x2 – x – 6 = (x + 2)(x – 3)


Factorise each of the following polynomials by splitting the middle term :

(i) x2 + 7x + 6 (ii) x2 – 7x + 6 (iii) x2 – 10x + 21(iv) x2 – 4x – 5 (v) x2 – 2x – 8 (vi) z2 – 2z – 15

5.10 FACTORISATION OF QUADRATIC POLYNOMIALS OF THE TYPEax2 + bx + c, WHERE a ≠≠≠≠≠ 0 OR a ≠≠≠≠≠ 1, BY SPLITTING THE MIDDLE TERM

Consider the polynomial : 2x2 + 3x + 1. Here, the coefficient of x2 is 2 and not one. So, canwe follow the method of splitting the middle term ?

Yes, we can : but now we need to consider the factors of 2 × 1 (i.e. ‘a × c’) not one (i.e.,not just ‘c’).

The factors of 2 × 1 are 2 and 1 only.

∴ 2x2 + 3x + 1 = 2x2 + (2 + 1)x + 1 because 2 + 1 = 3

= 2x2 + 2x + x + 1

= 2x(x + 1) + 1(x + 1)

[By taking 2x common from the first two terms and 1 from the last two terms]

= (2x + 1) (x + 1)

∴ 2x2 + 3x + 1 = (2x + 1)(x + 1)

Let us take some more examples.

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(i) 3x2 + 5x + 2 (ii) 3x2 + 10x + 7 (iii) 2x2 – 11x + 14

Solution : (i) 3x2 + 5x + 2

The product of the coefficient of x2 and the constant term = 3 × 2 = 6

We need two factors of 6 whose sum is 5. Can you guess these ?

Yes, they are 2 and 3

∴ 3x2 + 5x + 2 = 3x2 + (2 + 3)x + 2

= 3x2 + 2x + 3x + 2

= x(3x + 2) + 1 (3x + 2)

= (3x + 2) (x + 1)

∴ 3x2 + 5x + 2 = (3x + 2) (x + 1)

Note that if you multiply 3x + 2 and x + 1, you get back 3x2 + 5x + 2.

(ii) 3x2 + 10x + 7

The product = 3 × 7 = 21

Factors of 21 Sum of the factors Right or wrong

1, 21 1 + 21 = 22 ×

3, 7 3 + 7 = 10 √

∴ 3x2 + 10x + 21 = 3x2 + (3 + 7)x + 7

= 3x2 + 3x + 7x + 7

= 3x(x + 1) + 7(x + 1)

= (3x + 7) (x + 1)

∴ 3x2 + 10x + 7 = (3x + 7) (x + 1)

(iii) 2x2 – 11x + 14

Product of 2 and 14 = 2 × 14 = 28. Also since the sum is negative and the product is positive,both the factors have to be negative.

Factors of 28 Sum of the factors Right or wrong

–1, –28 (–1) + (–28) = –29 ×

–2, –14 (–2) + (–14) = –16 ×

–4, –7 (–4) + (–7) = –11 √

2x2 – 11x + 14 = 2x2 – (4 + 7)x + 14


= 2x2 – 4x – 7x + 14

= 2x(x – 2) – 7(x – 2)

= (2x – 7) (x – 2)

∴ 2x2 – 11x + 14 = (2x –7)(x – 2)


Factorise each of the following polynomials by splitting the middle terms :

1. x2 + 9x + 20 2. x2 + x – 20

3. 6x2 + 7x – 10 4. x2 – 2x – 8

5. 4x2 + 12x + 5 6. x2 – x – 56

7. x2 + 4x – 60 8. 3x2 – 2x – 5

9. 9x2 – 12x + 4 10. 15x2 – 14x + 3

5.11 APPLICATIONS OF FACTORISATON

Factorisation helps us not only in expressing a given polynomial in terms of simplerpolynomials, but also enables us to generate new polynomials using the given polynomials.

5.11.1 H.C.F. of Polynomials

We already know the meaning of the term ‘H.C.F.’ of any given numbers. It is the largest numberwhich divides all the given numbers.

For instance, the H.C.F. of 12 and 18 is 6

The H.C.F. of two or more polynomials is the polynomial of the highest degreewhich divides all the given polynomials completely.

Let us recall how to find the H.C.F. of two numbers, say 24 and 36. We used the method ofprime factorisation.

24 = 23 × 3 and 36 = 22 × 32

The first common factor is 2 and it occurs at least twice in the factorisation of the two numbersand so it must occur twice in the H.C.F. Similarly, the next common factor is 3 and it occursat least once in both the numbers

∴ The H.C.F. of 24 and 36 = 22 × 3 = 12

Similarly, you can find the H.C.F. of two or more polynomials. Let us take some examplesand illustrate.

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Example 5.14 : Find the H.C.F. of 4x2y and x3y2

Solution : ‘x’ occurs as a factor at least twice in the two monomials and ‘y’ occurs at leastonce.

∴ The H.C.F. of 4x2y and x3y2 is x2y

Example 5.15 : Find the H.C.F. of (x – 2)3(2x – 3) and (x – 2)2(2x – 3)3

Solution : The first common factor is x – 2 and it occurs at least twice in the given polynomials.It must occur two times in the H.C.F.

Now 2x – 3 is the next common factor and it occurs only once in the first polynomial.

∴ H.C.F. of the given polynomials = (x – 2)2 (2x – 3)

Let us take some more examples.

Example 5.16 : Find the H.C.F. of each of the following pairs of polynomials :

(i) (3x – 1)4 (2x + 5)2 and (2x – 1) (2x + 5) (3x –1)3

(ii) x2 – 4 and (x + 2)2

Solution : (i) (3x – 1)4 (2x + 5)2 is the first polynomial.

The second polynomial can be written as (3x – 1)3 (2x + 5)(2x – 1)

The first common factor is 3x – 1 and occurs at least three times.

The next common factor is 2x + 5 and occurs at least once.

∴ H.C.F. of (3x –1)4 (2x + 5)2 and (3x –1)3 (2x + 5) (2x – 1)

= (3x – 1)3(2x + 5)

Note : (1) 2x – 1 does not, occur in the first polynomial and so does not occur in the H.C.F.

(2) Check that the H.C.F. divides both the polynomials completely.

(ii) x2 – 4 and (x + 2)2

Now, x2 – 4 = x2 – 22

= (x +2) (x – 2)

The only common factor is x + 2 and it occurs at least once

∴ H.C.F. of x2 – 4 and (x + 2)2 is x + 2.



Find the H.C.F. of each of the following pairs of polynomials :

(i) (x + 1)3 and (x + 1)2 (x – 1) (ii) x2 + 4x + 4 and x + 2

(iii) (x + 2)3 and x3 + 8 (iv) (x + 1)2 (x + 5)3 and x2 + 10x + 25

(v) (2x – 5)2 (x + 4)3 and (2x – 5)3(x – 4)

5.11.2 The L.C.M. of Polynomials

Recall how we found the L.C.M. of two numbers, say 36 and 40. We write the primefactorisation of the numbers and find the product of all the different prime factors of thenumbers, using each prime factor the maximum number of times it occurs in the factorisationof any of the numbers.

36 = 2 × 2 × 3 × 3

and 40 = 2 × 2 × 2 × 5

2 occurs three times, 3 occurs twice and 5 occurs once, at the most

∴ L.C.M. of 36 and 40 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.

Similarly, you can easily find the L.C.M. of polynomials.

The L.C.M. of two or more polynomials is defined as the polynomial of the leastdegree which is a multiple of all the given polynomials.

Let us take some examples to illustrate.

Example 5.17 : Find the L.C.M. of each of the following pairs of polynomials :

(i) (x – 2)3(2x – 3) and (x – 2)2(2x – 3)3

(ii) (3x – 1)4(2x + 5)2 and (2x – 1)(2x + 5)(3x – 1)3

(iii) x2 – 4 and (x + 2)2

Solution. (i) (x – 2)3 (2x – 3) and (x – 2)2 (2x – 3)3

x – 2 occurs 3 times at the most and 2x – 3 also occurs 3 times at the most.

∴ L.C.M. of (x – 2)3 (2x – 3) and (x – 2)2 (2x – 3)3 = (x – 2)3 (2x – 3)3

(ii) (3x – 1)4(2x + 5)2 and (2x – 1)(2x + 5)(3x – 1)3

The L.C.M. of the given polynomials = (3x – 1)4(2x + 5)2(2x – 1).

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Find the L.C.M. of each of the following pairs of polynomials :

(i) (x + 1)3 and (x + 1)2 (x – 1) (ii) x2 + 4x + 4 and x + 2

(iii) (x + 2)3 and x3 + 8

(iv) (x + 1)2( x + 5)3 and x2 + 10x + 25

(v) (2x – 5)2(x + 4)3 and (2x – 5)3 (x – 4)

5.12 THE RELATIONSHIP BETWEEN H.C.F. AND L.C.M.

What is the H.C.F. of 6 and 4 ? What is their L.C.M.? You know very well that the H.C.F.of 6 and 4 is 2, and their L.C.M. is 12. Now, what is 2 × 12 ?

Is it not the product 6 × 4 ?

Yes! The product of the H.C.F. and the L.C.M. is the same as the product of the numbers. Thesame result is true even for polynomials.

Let us verify this result with the help of some examples.

Example 5.18 : Is the result given above true for each of the following pairs ofpolynomials ?

(i) (x – 2)3 (2x – 3) and (x – 2)2(2x – 3)3

(ii) x2 – 1 and x3 – 1 (iii) (x – 1)3 and (x + 1)2

Solution : (i) (x – 2)3 (2x – 3) and (x – 2)2 (2x – 3)3

The product of (x – 2)3 (2x – 3) and (x – 2)2(2x – 3)3

= (x – 2)3 (2x – 3) (x – 2)2(2x – 3)3

= (x – 2)3 + 2 (3x – 3)1 + 3

= (x – 2)5(2x – 3)4

The H.C.F. of the given polynomials is (x – 2)2 (2x – 3) and the L.C.M. of the given polynomialsis (x – 2)3(2x – 3)3.

Hence H.C.F. × L.C.M. = (x – 2)2(2x – 3)(x – 2)3(2x – 3)3

= (x – 2)5(2x – 3)4

which is the same as the product of the given polynomials.

(ii) x2 – 1 and x3 – 1

x2 – 1 = (x + 1)(x – 1)


and x3 – 1 = (x – 1) (x2 + x + 1)

∴ (x2 – 1)(x3 – 1) = (x + 1)(x – 1) × (x – 1)(x2 + x + 1)= (x + 1)(x – 1)2 (x2 + x + 1)

The H.C.F. of the given polynomials is x – 1 and the L.C.M. of the given polynomials is(x + 1)(x – 1)(x2 + x + 1)

∴ H.C.F. × L.C.M. = (x – 1) (x + 1) (x – 1) (x2 + x + 1)= (x + 1)(x – 1)2 (x2 + x + 1)

which is the same as (x2 – 1)(x3 – 1).

(iii) There is no factor common to (x – 1)3 and (x + 1)2

∴ H.C.F of (x – 1)3 and (x + 1)2 is 1 and the L.C.M of (x – 1)3 and

(x + 1)2 is (x – 1)3(x + 1)2.

Hence H.C.F. × L.C.M. = 1 (x – 1)3 (x + 1)2

= (x – 1)3(x + 1)2

= Product of the given polynomials.In the other words.

Product of two given polynomials = Product of their H.C.F. and L.C.M.

i.e., First polynomial × Second polynomial = H.C.F. × L.C.M.

You can also use this method to find the L.C.M. of given polynomials.

Example 5.19 : Find The L.C.M. of x – 1 and x2 – 2x + 1.

Solution : x2 – 2x + 1 = (x – 1)2 and x –1 = x – 1

∴ H.C.F. is x – 1.

and L.C.M. =x 1 x 2x 1

x 1

2− − +

−b gd i

= x2 – 2x + 1.Example 5.20 : Find the L.C.M. of x2 – 1 and x2 – x – 2.Solution : x2 – 1 = (x – 1) (x + 1)

and x2 – x – 2 = x2 – 2x + x – 2= x (x – 2) + 1 (x – 2) = (x + 1) (x – 2)

∴ H.C.F. is x + 1.

∴ L.C.M. = x 1 x x 2

x 1

2 2− − −+

d id i=

x 1 x 1 x 2x 1

2− + −+

b gb g b g

= (x2 – 1)(x – 2) = x3 – 2x2 – x + 2.

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Thus finding the L.C.M. of two polynomials is not difficult and can be done in either of thetwo given ways besides many others.

Next given the H.C.F. and the L.C.M. of two polynomials and also given one of the polynomials,you can find the other.

Example 5.21 : The H.C.F. of an unknown polynomial and x3 + 8 is x + 2 and their L.C.M.is (x + 2)3 (x2 – 2x + 4). Find the unknown polynomial.

Solution : Let the polynomial be A.

The H.C.F. of A and x3 + 8 is x + 2 and their L.C.M is (x + 2)3 (x2 – 2x + 4).

∴ H.C.F. × L.C.M. = (x + 2)(x + 2)3(x2 – 2x + 4)= (x + 2)4 (x2 – 2x + 4)

Product of the polynomials = A(x3 + 8)= A(x + 2) (x2 – 2x + 4)

∴ A(x + 2)(x2 – 2x + 4) = (x + 2)4(x2 – 2x + 4)

∴ A =x 2 x 2x 4

x 2 x 2x 4

4 2

2

+ − +

+ − +

b g d ib gd i

Thus A = (x + 2)3.Example 5.22 : The H.C.F. of two polynomials is x – 2 and their L.C.M. isx4 + 2x3 – 8x – 16. If one of the polynomials is x3 – 8, find the other polynomial.

Solution : Let the other polynomial be A.

Then, product of the polynomials = A(x3 – 8)H.C.F. × L.C.M. = (x – 2) (x4 + 2x3 – 8x – 16)

= x5 – 4x3 – 8x2 + 32∴ A(x3 – 8) = x5 – 4x3 – 8x2 + 32∴ A = (x5 – 4x3 – 8x2 + 32) ÷ (x3 – 8)

= x2 – 4Thus A = x2 – 4.

Thus, the required polynomial is x2 – 4.


1. Find the H.C.F. and the L.C.M of each of the following pairs of polynomials :

(i) (x + 2)2 and (x + 2)3 (ii) (x + 3)2 and x3 + 27

(iii) (x + 3)3 (x + 1)3 and (x + 1)2 (x + 3)4

(iv) x2 + 4x – 60 and x2 + 5x – 50


(v) x2 – 2x – 35 and (x – 7)3 (vi) x2 – x + 1 and x3 + 1.

(vii) x3 + 1 and x3 – 1 (viii) x2 – x + 1 and (x + 1)2

(ix) 3x2 – 5x + 2 and x2 – 4 (x) x3 – y3 and x2 + xy + y2

2. The H.C.F. of two polynomials is 1 and their L.C.M. is (x – 1)2(x + 1). If one of thepolynomials is x2 – 2x + 1, find the other polynomial.

3. The H.C.F. of two polynomials is x + 1 and their L.C.M. is x6 – 1. If one of the polynomialsis x3 + 1, find the other polynomial.

5.13 RATIONAL EXPRESSIONS

An algebraic expression which is the quotient of two polynomials is called a rationalexpression.

For example, x 1x 1+− ;

x 3x 5x 5

2

2− +−

; x 3x 2x 72

−+ − ;

x y2x 3y

2 2−+ ;

2x y 3xy7x 3y

2 2

4 2− +− ;

x 1x 1

3

2−−

and p 2p

p 3

2 −− are all rational expressions in one or two variables. Are the polynomials,

x2 +1 and 3x2 + 4y rational expressions ?

Yes, a polynomial is a rational expression because x2 + 1 = x 1

1

2 + i.e., the denominator of

a polynomial can be written as 1, and so it is a rational expression.

But a rational expression need not be a polynomial.


1. Identify the rational expressions from the following :

(i) 2x 34x 1

−− (ii) 3x 4x 3

3x 112

3− ++

(iii) 2 3x x 5x 12

2− ++

(iv)3x y

x + y

2 2−

2. Cite three examples each of rational expressions in

(a) one variable, and

(b) two variables.

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5.14 REDUCTION OF RATIONAL EXPRESSIONS TO LOWEST FORM

Sometimes the polynomials in the numerator and denominator of a rational expression havecommon factors, which can be cancelled out and the resulting rational expression is said tobe in lowest terms.


Examples 5.23 : Reduce each of the following rational expressions to lowest form :

(i)x 2x 3x 4x 5

2

2+ −+ −

(ii)x 4 x 1

x 3x 2 x 2

2

2

− −

− + +d ib gd ib g .

Solution. (i) x2 + 2x – 3 = x2 + 3x – x – 3

= x (x + 3) – 1(x + 3)

= (x + 3)(x – 1)

and x2 + 4x – 5 = x2 + 5x – x – 5

= x(x + 5) – 1(x + 5)

= (x + 5) (x – 1)

∴x 2x 3x 4x 5

2

2+ −+ −

=x 3 x 1x 5 x 1+ −+ −b gb gb gb g =

x 3x 5++

(ii) (x2 – 4) = (x + 2) (x – 2)

and x2 – 3x + 2 = (x – 1) (x – 2)

∴x 4 x 1

x 3x 2 x 2

2

2

− −

− + +d ib gd ib g =

x 2 x 2 1x 1 x 2 x 2+ − −− − +b gb gb gb gb gb g

x

= 1.

Example 5.24 : Reduce each of the following rational expressions into lowest form :

(i) x 1x 3x 2

32

++ +

(ii)x 1

x 1 x 1

4

2−

− +b gd iSolution. (i) x3 + 1 = (x + 1) (x2 – x + 1)

and x2 + 3x + 2 = (x + 1) (x + 2)

∴ x 1x 3x 2

32

++ + =

x 1 x x 1

x 1 x 2

2+ − +

+ +b gd ib gb g

=x x 1

x 2

2 − ++


(ii) x4 – 1 = (x2 – 1) (x2 + 1)

= (x – 1) (x + 1) (x2 + 1)

∴x 1

x 1 x 1

4

2−

− +b gd i =x 1 x 1 x 1

x 1 x 1

2

2

− + +

− +

b gb gd ib gd i

= x + 1.


Reduce each of the following rational expressions into lowest form :

(i)3x 2x 56x x 15

2

2+ −+ −

(ii) 8 14 1

32

xx

+−

(iii)x 1

x x 1

3

4 2−

+ +(iv)

x yx y x y

8 8

4 4−

− +b gd i

(v)x 4 x 5x 6

x 7x 12 x 2

2

2

+ + +

+ + +

b gd id ib g

5.15 OPERATIONS ON RATIONAL EXPRESSIONS

The four fundamental operations on rational expressions are performed exactly in the sameway as in the case of rational numbers.


Example 5.25 : Add 2x 1x 1+− and

x 2x 1++ .

Solution : 2x 1x 1

x 2x 1

+−

+ ++ =

2x 1 x 1 x 1 x 2x 1 x 1

+ + + − +− +

b gb g b gb gb gb g

=2x 3x 1 x x 2

x 1

2 2

2

+ + + + −

−d i d i

=3x 4x 1

x 1

2

2+ −−

∴2x 1x 1

x 2x 1

+−

+ ++ =

3x 4x 1x 1

2

2+ −−

.

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Example 5.26 : Subtract x 1x 1−+ from

3x 23x 1

−+ .

Solution :3x 23x 1

x 1x 1

−+

− −+ =

3x 2 x 1 x 1 3x 13x 1 x 1

− + − − ++ +

b gb g b gb gb gb g

=3x x 2 3x 2x 1

3x 4x 1

2 2

2

+ − − − −

+ +d i d i

=3x x 2 3x 2x 1

3x 4x 1

2 2

2+ − − + +

+ +

=3 1

3 4 12x

x x−

+ +

∴3x 23x 1

x 1x 1

−+

− −+ =

3x 13x 4x 12

−+ + .

Example 5.27 : Multiply (i)5x 35x 1

+− with

2x 1x 1−+ .

(ii)2x 1x 1+− with

x 1x 3−+ .

Solution : (i)5x 35x 1

2x 1x 1

+−

× −+ =

5x 3 2x 15x 1 x 1+ −− +

b gb gb gb g

=10 5 6 3

5 5 1

2

2x x xx x x

− + −+ − −

=10 35 4 1

2

2x x

x x+ −+ −

∴5x 35x 1

2x 1x 1

+−

× −+ =

10x x 35x 4x 1

2

2+ −+ −

.

(ii)2x 1x 1

x 1x 3

+−

× −+ =

2x 1 x 1x 1 x 3+ −− +b gb gb gb g =

2x 1x 3++

∴2x 1x 1

x 1x 3

+−

× −+ =

2x 1x 3++ .


Example 5.28 : Divide (i)2x 1x 1+− by

3x 4x 1+−

(ii)x 1x 1

2 +−

by x 1x 2−+

Solution. (i)2x 1x 1

3x 4x 1

+−

÷ +− =

2x 1x 1

x 13x 4

+−

× −+

=2x 1 x 1x 1 3x 4

+ −− +b gb gb gb g

=2x 13x 4

++ .

∴2x 1x 1

3x 4x 1

+−

÷ +− =

2x 13x 4

++

(ii)x 1x 1

x 1x 2

2 +−

÷ −+

=x 1x 1

x 2x 1

2 +−

× +−

=x 1 x 2

x 1 x 1

2 + +

− −d ib gb gb g

=x 2x x 2

x 2x 1

3 2

2+ + +− +

∴x 1x 1

x 1x 2

2 +−

÷ −+

=x 2x x 2

x 2x 1

3 2

2+ + +− +

.


1. Add the following rational expressions :

(i) 2x 3x 2++ and x 2

2x 3+− (ii) 3x 1

x 12 ++ and 2x 3x 1

x 12 − +

−

2. Subtract the first rational expression from the second :

(i)2x 3x 2++ ,

2x 32x 1

−+ (ii)

2x 1x 1

2 +−

, 3x 2x 1

x 3

2 − ++

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3. Multiply the rational expressions in each case :

(i)7x 2

2x 3x 12+

+ + and

x 17x 5x 22

+− − (ii)

x 1x 1

3

4++

and x 1x 1

3

4−−

(iii)3x 15x 18

2x 4

2 − +−

and 17x 3

x 6x 92+

− +

4. Divide

(i)x 11x 26x 4x 45

2

2− −− −

by x 7x 10x 12x 13

2

2+ +− −

(ii)6x x 1

2x 7x 15

2

2+ −− −

by 4x 4x 1

4x 9

2

2+ +−

.

LET US SUM UP

(x + y)2 = x2 + 2xy + y2

(x – y)2 = x2 – 2xy + y2

x2 – y2 = (x + y) (x – y)

(x + y)3 = x3 + 3x2y + 3xy2 + y3

= x3 + 3xy(x + y) + y3

(x – y)3 = x3 – 3x2y + 3xy2 – y3

= x3 – 3xy (x – y) – y3

x3 + y3 = (x + y) (x2 – xy + y2)

x3 – y3 = (x – y) (x2 + xy + y2)

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 + z2 – xy – yz – zx)

x2 + (a + b)x + ab = (x + a) (x + b)

HCF × LCM = Product of Two Polynomials

HCF of two polynomials = Product of Two Polynomials

LCM of the Polynomials


LCM of two Polynomials = Product of Two Polynomials

HCF of the Polynomials

Unknown Polynomial = HCF LCM

Given Polynomial×

A rational expression is an expression obtained by dividing one polynomial by the other.

Reduction of rational expression into lowest forms means cancelling a common factor,if any, from the polynomials in the numerator and denominator.

TERMINAL EXERCISE


(i) (3x + 4)2 (ii) (4x – 3)2 (iii) (2ab + 3c)2

(iv) x 12

2−FH IK (v) 2x

3 52

+FH IK2. Calculate each of the following without actual multiplication :

(i) (105)2 (ii) 992

(iii) (10.2)2 (iv) 952

3. Find xx

22

1+ if

(a) x 1x+ = 5 (b) x 1

x− = 3


(i) (3x – 1) (3x + 1) (ii) (2x – 3y) (2x + 3y)

(iii) 97 × 103 (iv) 1001 × 999


(i) (x + 2)3 (ii) (2x – 1)3

(iii) (2ab – 3cd)3 (iv) 2x 12x

3+FHGIKJ

6. Evaluate each of the following, without actual multiplication :

(i) 493 (ii) 1053

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7. Factorise each of the following :

(i) 64x2 – 1 (ii) a3 + 343

(iii) x6 + 1 (iv) x6 – 1

(v) 25x2 – 90xy + 81y2 :

8. Factorise each of the following by splitting the middle term :

(i) 2x2 + 17x + 8 (ii) 3x2 + 2x – 8

(iii) 8x2 + 6x – 5 (iv) 8x2 – 6x – 5

9. Find the HCF and LCM of each of the following :

(i) x2 – 8x + 7 and x2 – 10x + 21

(ii) (2x + 1)2 (4x – 3)2 (x – 2)3 and (2x + 1) (4x – 3)4

(iii) 8(x – 2)2 (4x – 1)3 (3x + 1) and 12 (x – 2)3 (4x – 1) (3x + 1)4

10. The HCF of two polynomials x2 – 5x + 6 and x2 – 7x + 12 is x – 3. Find the LCM ofthe polynomials.

11. The LCM of two polynomials (2x + 1)3 (3x – 5)2 and (2x + 1)(3x – 5)3 is(2x + 1)3 (3x – 5)3. Find the HCF of the polynomials.

12. The HCF and LCM of two polynomials are (x2 – 1) and (x4 – 1) (x4 + x2 + 1) respectively.If one polynomial is x4 – 1, find the other polynomial.

13. Reduce each of the followng rational expressions to their lowest forms :

(i)6x 5x 6

12x 7x 10

2

2+ −+ −

(ii)4x 25

2x 11x 15

2

2−

+ +

(iii)x ax a

6 6

2 2++

(iv)x yx y

8 8

6 6−−

14. Perform the indicated operation :

(i) x 2x 3

x 3x 2

++ + +

+ (ii) 2x 1x 5

x 3x 2

−+ − −

+

(iii) x 13x 3

2x 1x 1

2 +− − +

−15. Perform the indicated operation :

(i)x 3 x 8x 15

x 92x 5x 5

2

2

+ − +

−× +

+b gd i

(ii)3x 6x 3

2 x 1

3x 3

x 5

2

2

2

2− ++

×+

−b gb gb g

(iii) 4x 12x 52x 3

6x x 354x 8x 3

2 22

+ ++ ÷ + −

+ + (iv)x yx y

x xy yx xy y

x yx y

3 3

3 3

2 2

2 2−+

÷ + +− +

FHG

IKJ ×

+−


ANSWERS


1. (i) 25x2 + 10xy + y2 (ii) x2 – 6x + 9

(ii) a2c2 + 2abcd + b2d2 (iv) a2b2 – 2abc + c2

(v) xx

22

1 2+ − (vi) 4x2 – 20xy + 25y2

(vii)x x2

923

1+ + (viii)z z

2

41− +

(ix) a4 + 2a2b2 + b4 (x)xy

yx

2

2

2

2 2+ +

2. (i) 9409 (ii) 40401 (iii) 996004

(iv) 2401 (v) 998001 (vi) 10404

3. (i) 2 (ii) 6 (iii) 11

(iv) 17 (v) 2 (vi) 68 and 16.


1. (i) y2 – 4 (ii) x2y2 – 1 (iii) a4 – 25

(iv) x2 – y4 (v) 9996 (iv) 6375


(i) 8x3 + 12x2y + 6xy2 + y3 (ii) 125x3 – 75x2y + 15xy2 – y3

(iii) x3y3 – 3x2y2z + 3xyz2 – z3 (iv) 970299

(v) 912673 (vi) 110592


(i) 8x3 + 36x2y + 54xy2 + 27y3 (ii) 64x3 – 240x2 + 300x – 125

(iii) x3y3 – 3x2y2 + 3xy – 1 (iv) x3y3 + 3x2y2z + 3xyz2 + z3

(v) x xx x

333 3 1− + − (vi) a3x3 – 3a2bx2y + 3ab2xy2 – b3y3

(vii) 8x3 + 1 (viii) y3 – 27

(ix) 1 + x3 (x) 8x3 + y3 + 9z3 – 18xyz

2. (i) 132651 (ii) 1061208 (iii) 857375

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(i) (2a + 3)(2a – 3) (ii) (2a + b) (2a – b)

(iii) c(a + b)(a – b) (iv) (3x + 5y)(3x –5y)

(v) (5x + 4)(5x + 4) (vi) (x – 3) (x – 3)

(vii) (x + 6)(x + 6)


1. (i) (u + v)(u –v) (ii) (5a + 1)(5a –1)

(iii) (x2 + y2) (x + y) (x – y) (iv) (x – 2y)(x2 + 2yx + 4y2)

(v) (a + 6b) (a2 – 6ab + 36b2) (vi) (a – 7) (a2 + 7a + 49)

(vii) (x4 + 1) (x2 + 1) (x + 1) (x – 1)

(viii)(a + 2)3 (ix) (x – 11)3

(x) (x + 11)2 (xi) (3x – 5y)2

(xii) (9y – 2x)2

2. (i) 7000 (ii) 400

(iii) 6600 (iv) 13

(v) 2 (vi) 91

(vii) –19


(i) (x + 6)(x + 1) (ii) (x – 6) (x – 1)

(iii) (x – 7)(x – 3) (iv) (x – 5)(x + 1)

(v) (x – 4)(x + 2) (vi) (z – 5) (z + 3)


1. (x + 5)(x + 4) 2. (x + 5) (x – 4)

3. (6x – 5)(x + 2) 4. (x – 4)(x + 2)

5. (2x + 5)(2x + 1) 6. (x – 8) (x + 7)

7. (x + 10)(x – 6) 8. (3x – 5) (x + 1)

9. (3x – 2)2 10. (5x – 3)(3x – 1)



(i) (x + 1)2 (ii) x + 2

(iii) x + 2 (iv) (x + 5)2

(v) (2x – 5)2


(i) (x + 1)3(x – 1) (ii) x2 + 4x + 4

(iii) (x + 2)3(x2 – 2x + 4) (iv) (x + 1)2(x + 5)3

(v) (2x – 5)3(x + 4)3(x – 4)


1. (i) (x + 2)2 ; (x + 2)3 (ii) x + 3 ; (x + 3)2(x2 – 3x + 9)

(iii) (x + 3)3(x + 1)2 ; (x + 3)4(x + 1)3

(iv) x + 10 ; (x2 + 4x – 60)(x – 5) (v) x –7 ; (x – 7)3(x + 5)

(vi) x2 – x + 1 ; x3 + 1 (vii) 1 ; x6 – 1

(viii)1 ; (x2 – x + 1)(x +1)2 (ix) 1 ; (3x2 – 5x + 2) (x2 – 4)

(x) x2 + xy + y2 ; x3 – y3

2. x + 1 3. x4 + x3 – x – 1


1. (i), (ii)

2. (a) For example xx2 1

1++ , x

xx

x2 1

12

1−+ −, etc.

(b)x y

x yx yx y

x yx y

2 2 3 323 5

++

++

+−, , etc


(i) xx−−1

2 3 (ii) 4 2 12 12x x

x− +−

(iii) xx x

−− +

112 (iv) x x y xy y3 2 2 3

1+ + +

(v) 1

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1. (i) 5 4 52 2 3

2x xx x

+ −+ −b gb g (ii) 5 4 2

13 2

2x x x

x− −

−

2. (i) − − −+ +

2 7 92 1 2

2x xx xb gb g (ii) x x x

x x3 211 2 4

3 1− + −+ −b gb g

3. (i) 12 1 1x x+ −b gb g (ii)

x x x x

x x

2 2

4 2

1 1

1 1

− + + +

+ +d id id id i

(iii) 51 92 6

xx+−

4. (i)x x

x x

+ −

− +

1 13

9 5

2

2b gb gb gb g (ii)

3 1 2 35 2 1

x xx x− −− +b gb gb gb g

Terminal Exercise

1. (i) 9x2 + 24x + 16 (ii) 16x2 – 24x + 9

(iii) 4a2b2 + 12abc + 9c2 (iv) x2 – x + 14

(v) 49

203 252x x+ +

2. (i) 11025 (ii) 9801

(iii) 104.04 (iv) 9025

3. (a) 23 (b) 11

4. (i) 9x2 – 1 (ii) 4x2 – 9y2

(iii) 9991 (iv) 999999

5. (i) x3 + 6x2 + 12x + 8

(ii) 8x3 – 12x2 + 6x – 1

(iii) 8a3b3 – 36a2b2cd + 54abc2d2 – 27c3d3

(iv) 8x3 + 6x + 32

18 3x x

+

6. (i) 117649 (ii) 1157625

7. (i) (8x + 1)(8x – 1) (ii) (a + 7) (a2 – 7a + 49)

(iii) (x2 + 1)(x4 – x2 + 1)


(iv) (x + 1)(x – 1)(x2 – x + 1)(x2 + x –1)

(v) (5x – 9y)2

8. (i) (2x + 1)(x + 8) (ii) (3x – 4) (x + 2)

(iii) (4x + 5)(2x – 1) (iv) (2x + 1)(4x – 5)

9. (i) x – 7 ; (x2 – 8x + 7)(x – 3)

(ii) (2x + 1)(4x – 3)2 ; (2x + 1)2 (4x – 3)4(x – 2)3

(iii) 4(x – 2)2(4x – 1)(3x + 1) ; 24(x –2)3(4x – 1)3 (3x + 1)4

10. (x – 2)(x2 –7x + 12)

11. (3x – 5)2(2x + 1)

12. (x2 – 1)(x4 – x2 + 1)

13. (i) 2 34 5

xx++ (ii) 2 5

3x

x−+

(iii) x4 – x2a2 + a4 (iv)x y x y

x xy y x xy y

2 2 4 4

2 2 2 2

+ +

+ + − +d id i

d id i

14. (i)2 10 13

3 22x x

x x+ ++ +b gb g (ii)

x xx x

2 135 2+ +

+ +b gb g

(iii)x x

x2 6 23 1− −−b g

15. (i)2 5 5

5x x

x+ −+

b gb g (ii)27 1

2 5

2

2x

x

−

−b gb g

(iii) 2 13 7

2xx+−

b g (iv) 1

122 Mathematics

6

Linear Equations

6.1 INTRODUCTION

Consider the following statements

(i) 6 + 5 = 11 (ii) 7 + 6 = 18

(iii) 8 × 7 = 56 (iv) 6 – 17 = –11

(v) 16 ÷ 4 = 4 (vi) 18 × 7 = 116

Of the above statements, some are true, whereas the others are false. Can you say which ofthem are true and which are false ?

Of course, your reply will be that (i), (iii), (iv) and (v) are true statements whereas (ii) and(vi) are false statements. We are able to call them true or false after finding their value. Allthe above statements (i) to (vi) are called numerical statements.

Now consider the statements :

(i) x + 10 = 22

(ii) x + y + 2 = 14

(iii) 2x – y + 4 = 20

Statements (i), (ii) and (iii) can be true for some specific value of the variables involved. In(i), if x equals 12, it is true. For all other values of x, (i) is false.

Similarly, (ii) is true for x = 10, y = 2. Can you see that (ii) is true for some other pairs ofvalues of x and y.

The same thing can be said for (iii) also. (i), (ii) and (iii) are all called equations, because theyinvolve two statements involving variables and constants connected by a sign of equality.

Further, (i) involves one variables x, whereas (ii) and (iii), involve two variables x and y.

Thus, (i) is an equation in one variable, whereas and (ii) and (iii) are equations in two variables.

Linear Equations 123

In this lesson, we shall study about linear equations in one and two variables and discuss aboutmethods of finding their solutions – algebraic as well as graphic. We shall also solve systemof two linear equations graphically and algebraically. We shall also solve word problemsinvolving one or two variables.

6.2 OBJECTIVES


identify a linear equation from a given collection of equations.

cite examples of linear equations.

write a linear equation in one variable and also give its solution.

write a linear equation in two variables

draw the graph of a linear equation in two variables.

find the solution of a system of two linear equations graphically.

check-the consistency or otherwise of systems of linear equations

solve system of equations algebraically.

translate a word problem into a linear equation in one or two variables.

solve real life problems involving linear equations in one or two variables.


Algebraic expressions and polynomials in one or two variables.

Four fundamental operations on algebraic expressions.

Linear polynomials in one or two variables.

Four fundamental operations on numbers.

Knowledge of exponents.

Plotting of points on a graph.

6.4 EQUATIONS IN ONE VARIABLE

If one equation has only one variable, it is said to be an equation in one variable or oneunknown.

There are equations having more than one unknown. You will learn about them later.

6.4.1 Degree of an Equation

Consider the following equations

(i) 3x – 4 = 8 (ii) (x – 4) – 8 = 2(x + 3)

(iii) x2 – 16 = 0 (iv) x3 – 3x = x.

Each of these is an equation in one unknown, namely x.

124 Mathematics

You will observe that the greatest exponent of the variable in (i) and (ii) is one. In equation(iii), the greatest exponent of x is 2 and in equation (iv) it is 3.

Equation (ii) can be simplified as follows :

x – 4 – 8 = 2x + 6

or x – 2x – 12 – 6 = 0

or –x –18 = 0

or x + 18 = 0

The greatest exponent of the variable in an equation in one unknown is called thedegree of the equation.

Thus, the degree of equations (i) and (ii) is one whereas, the degree of equation (iii) is 2 andof the equation (iv) is 3.

An equation in which the greatest exponent of the variable after simplification isone, is called a linear equation.

Thus, the degree of a linear equation is one.

Let us take some example to identify linear equations.

Example 6.1 : Which of the following equations are linear equations ?

(i) 5x – 3 = 5 (ii) 4x + 4 = 4

(ii) 8m – 3 = 12 (iv) 2(x + 3) – 1 = x2

Solution : (i) , (ii) and (iii) are equation of degree one and hence linear equations whereasequation (iv) is of degree 2 and hence is not a linear equation.

6.5 GENERAL FORM OF A LINEAR EQUATION

A linear equation in one unknown can be written as :

ax + b = 0 where ‘a’ and ‘b’ are real numbers and a ≠ 0.

6.6 SOLUTION OF A LINEAR EQUATION

The value (or values) of the variable (or variables) which make the equation a truestatement is (are) called solution(s) of the equation.

By solving an equation, we mean finding the solution (or solutions) of the equation.

For example, x = 2 is a solution of the equation x + 3 = 5 because on substituting 2 for xwe get,


2 + 3 = 5 which is a true statement.

Note. A solution is also called a root of the equation.

Let us now solve some linear equations in one unknown.

Example 6.2 : Solve each of the following equations :

(i) x – 12 = 7 (ii)y6

1+ = 7

(iii) 3(x + 5) = 24 (iv) 5(m – 3) = 19 + 3(m – 2).

Solution : (i) x – 12 = 7

Adding 12 on both sides, we get,

x – 12 + 12 = 7 + 12

or x = 19

Thus, x = 19 is a solution of the given equation.

Does it satisfy the given equation ?

This can easily be checked.

Check : Substituting 19 in place of x in the given equation, we get,

19 – 12 = 7

or 7 = 7 which is a true statement.

∴ x = 19 is a solution of the given equation.

(ii)y6

1+ = 7

ory6 = 7 – 1 (Subtracting 1 from both sides)

ory6 = 6

or y = 6 × 6 (Multiplying both sides by 6)

or y = 36

Check : Substituting 36 for y, we have

366

1+ = 7

or 6 + 1 = 7 which is true.

∴ y = 36 is the required solution.

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(iii) 3(x + 5) = 24

or 3x + 15 = 24

or 3x = 24 – 15 (Subtracting 15 from both sides)

or 3x = 9

or x =93 (Dividing both sides by 3)

or x = 3.

(iv) 5(m – 3) = 19 + 3(m – 2)

or 5m – 15 = 19 + 3m – 6

or 5m – 15 = 13 + 3m

or 5m – 15 – 3m = 13

or 2m – 15 = 13

or 2m = 13 + 15

or 2m = 28

or m = 28 ÷ 2

∴ m = 14

Thus, m = 14 is the required solution of the given equation.

From the above examples, we see that

A linear equation in one variable has only one solution.

Similarly, you can solve the general equation ax + b = 0, (a ≠ 0).

ax + b = 0

Subtracting b from both sides, we get

ax = 0 – bor ax = –b

Dividing both sides by a (a≠ 0). we get,

x = − ba

Check : Substituting − ba for x, we get

a × − ba + b = 0

or –b + b = 0 which is true for every value of b.

∴ x = − ba is the solution of the equation ax + b = 0.



1. Which one of the following statements is a numerical statement ?

(i) 65 + 1 = 70 (ii) 7 × 2 + x = 3

(iii) x + 2 = 3 (iv) x – 16 = 8

2. Which of the following is an open statement ?

(i) 7 + 7 = 14 (ii) 6 – x = 5

(iii) 3 × 4 + 1 = 13 (iv) 8 × 3 + 1 = 25

3. Which of the following open statements is true for x = –1 ?

(i) x – 2 = 4 (ii) x – 5 = – 6

(iii) 3x = 3 (iv) 2x – 7 = 0

4. Which of the following algebraic statements is false for p = 3 ?

(i) p – 7 = – 4 (ii) p + 3 > 0

(iii) 4p + 1 = 13 (iv) 5p – 2 = 5

5. The degree of a linear equation is :

(i) 1 (ii) 2 (iii) 3 (iv) 0.

6. Which of the following equations is of degree one ?

(i) x2 – 1 = 0 (ii) x3 – 1 = 0

(iii) 2x = 3 (iv) x0 = 1

7. Which of the following numbers is the root of the equation 2(x + 3) = 18 ?

(i) 21 (ii) 13 (iii) 12 (iv) 6

8. The root of 2m – 1 = 17 is :

(i) 18 (ii) 16 (iii) 9 (iv) 8

9. The degree of the equation 2z – 4 = 3(z – 4) is :

(i) 16 (ii) 8 (iii) 1 (iv) 0

10. The value of ‘p’ for which the equation 2p – (4 – p) = 5 – p becomes true is :

(i) 4.5 (ii) 3 (iii) 2.25 (iv) 0.5

11. Solve each of the following linear equations for x :

(i) 2x + 4 = 14 (ii) 2 63

x + = x (iii) 4 23

12

− + +x x = 1

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6.7 WORD PROBLEMS

In our day-to-day life, we come across many problems which can be solved using equations.To do so, follow the steps given below :

1. Represent the unknown by an alphabet say x,y,z,n,p,t etc.;

2. Translate the given statement into an equation

3. Solve the equation, and

4. Check the value found, by putting it in the original problem.

Let us now take some examples.

Example 6.3 : Which number when added to 6 gives 8 ?

Solution : Let x be the given number

6 added to x mean x + 6

According to the question

x + 6 = 8

This is a linear equation in x.

Solving it for x, we get x = 2

i.e., 2 added to 6 gives us 8.

Yes, this is true

∴ The required number is 2.

Example 6.4 : The perimeter of a square is 64 cm. Find its side.

Solution : Let the side of the square be x cm. Then,

Perimeter = 4x

∴ The equation becomes 4x = 64

or x = 64 ÷ 4

or x = 16

∴ The side of the square is 16 cm.

Example 6.5 : Renu is 20 years younger to her mother. After 10 years, her mother will betwice as old as Renu will be then. How old is Renu now ?

Solution : Let Renu’s present age be x years.

Her mother’s present age = (x + 20) years

After 10 years,


Renu’s age will be = (x + 10) years

and her mother’s age will be[(x + 20) + 10] years

According to the problem,

Renu’s mother will be 2 times as old as Renu after 10 years.

∴ (x + 20) + 10 = 2(x + 10)or x + 20 + 10 = 2x + 20or x + 30 = 2x + 20or x – 2x = 20 – 30or x = 10

i.e., Renu is presently 10 years old

Check : Renu’s present age = 10 years

Renu's mother's age = 30 years

After 10 years, Renu’s age = 10 + 10 = 20 years

After 10 years, Renu’s mother’s age = 30 + 10 = 40 years

and 40 is twice of 20

Thus, Renu’s present age is 10 years.

Example 6.6 : Twice a certain number increased by 10 equals 32. Find the number.

Solution : Let the number be x

Then, twice the number = 2x

∴ The equation is

2x + 10 = 32or 2x = 32 – 10or 2x = 22or x = 22 ÷ 2or x = 11

Thus, the required number is 11.


1. A man is 20 year older than his son. After 10 years his age becomes twice the age ofhis son. Find their present ages.

2. The denominator of a fraction is 2 more than its numerator. If 1 is added to each, the

fraction becomes 23 . Find the fraction.

130 Mathematics

3. The sum of three consecutive natural numbers is 42. Find the numbers.

4. 4 added to 5 times a number equals 2 subtracted from 7 times the number. Find the number.

5. The angles of a triangle are such that sum of two angles equals the third and the ratiobetween the acute angles is 2 : 3. Find the angles of the triangle.

6.8 A LINEAR EQUATION IN TWO VARIABLES

Recall that ax + b = 0 is the general form of a linear equation in one variable.

A linear equation in two unknowns is an equation which after simplificationcontains two unknowns, each one of them in a separate term and having theexponent one.

Given below are some linear equations in two unknowns :

1. 2x = 5y – 7

2.32

5 32

x y= −

3. 2 3 6x y+ =

4. x – y = π2

5. 2x – 1 = (y – 3) + (2y – x)A general linear equation in two unknowns is written as

ax + by + c = 0

where a, b and c are real numbers with a ≠ 0 and b≠ 0.

6.9 THE GRAPH OF A LINEAR EQUATION IN TWO UNKNOWNS

An equation in two unknowns represents a relationship between the unknowns, and its graphis the diagrammatic form of this representation.

(1) Representation of a Point in a plane

In order to draw the graph of any equation, your must first know how to plot points in a plane(i.e. on a sheet of paper). This plane is called the coordinate plane or cartesian plane.

You already know that a point on the number line represents a real number. Similarly, a pointon a plane represents an ordered pair of numbers i.e., a pair of numbers taken in a fixed order.

Let us now understand how to represent a point in a plane.

For that we go through the following steps :


Step 1 : Two perpendicular lines XOX′ and YOY′ are drawn intersecting each other at O. Theyare called the x-axis and y-axis respectively, and O is called the origin (See Fig. 6.1).

Fig. 6.1 Fig. 6.2

Step 2. These two coordinate axes divide the plane into four parts, each of which is calleda quadrant (See Fig. 6.2)

Step 3. The integers are marked along both the axes with O representing as zero on both theaxes. (See Fig. 6.3).

Fig. 6.3 Fig. 6.4

Along the x-axis, positive integers are marked to the right of O, and negative integersto the left of O.

Along the y-axis, positive integers are marked above O, and negative integersbelow O.

Thus, the two axes can be taken as two perpendicular number lines.

To represent on ordered pair say (2, 3), we proceed as follows :

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Step 4. Mark a point A on mark 2 on x-axis anddraw a line ‘l’ through A parallel to they-axis (See Fig. 6.4).

Step 5. Mark a point B on mark 3 on y-axis anddraw a line ‘m’ through B parallel to thex-axis (See Fig. 6.5).

‘l’ and ‘m’ intersect each other atpoint P.

P represents the point (2, 3) because Pis at a distance of 2 units measured alongthe x-axis and 3 units measured alongthe y-axis.

Let us take another example.

Example 6.7 : Represent the point (–3, 5) on the coordinate plane.

Solution. Let us first draw the x-axis and the y-axis and mark points on them representingintegers as in Fig. 6.6.

The first number in the ordered pair is –3, and so we will take 3 units along the x-axis to theleft of O. Similarly take 5 units along the y-axis above O and draw lines ‘l’ and ‘m’ parallelto the y-axis and the x-axis, respectively.

The point A represents (–3, 5) as in Fig. 6.7.

Fig. 6.6 Fig. 6.7Similarly, you can represent any ordered pair (x, y) by a point in the cartesian plane.Note that :

1. The first number in the ordered pair tells us the number of units to be taken along thex-axis and is thus called the x-coordinate or the abscissa of the point.

Fig. 6.5


2. The second number in the ordered pair tells us the number of units to be taken alongthe y-axis and is thus called the y-coordinate or the ordinate of the point.

3. If (x, y) is a point ‘P’ in the cartesian plane, then P lies in :

(i) the first quadrant if both ‘x’ and ‘y’ are positive real numbers;

(ii) the second quadrant if ‘x’ is negative and ‘y’ is positive;

(iii) the third quadrant if ‘x’ and ‘y’ both are negative;

(iv) the fourth quadrant if ‘x’ is positive and ‘y’ is negative.

“What happens to ‘P’ if y = 0 ?”

Obviously ‘P’ is then ‘x’ units along the x-axis and zero units along the y-axis and thus lieson the x-axis.

∴ Any point on the x-axis has the coordinates (a, 0).

Similarly, a point on the y-axis will have the x-coordinate equal to zero, and will thus havethe coordinates (0, b).

Thus, O has the coordinates (0, 0).

Therefore, given an ordered pair of numbers, you can easily find the point in the plane whichcorresponds to this ordered pair.


Represent the following ordered pairs as points in the cartesian plane :

(i) (5, 2), (3, –2), (3, –4), (2, 0) and (0, –5)

(ii) (–5, 2), (3, –2), (–3, –4), (–2, 0), (0, 5) and (–5, 4)

6.10 FINDING ORDERED PAIR CORRESPONDING TO A POINT IN THE PLANE

Given a point in a plane, can we associate an ordered pair of numbers with it ?

Yes, we can find its coordinates with respect to a given pair of axes.

Let us take an example to illustrate this.

Example 6.8 : ‘P’ is a point in a plane. Find its coordinates with respect to the coordinateaxes XOX′ and YOY′.

Solution :

Step 1 : Through ‘P’, draw a line ‘l’ parallel to the y-axis and mark A as the point of

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intersection of ‘l’ with the x-axis (See Fig. 6.8).

If OA = a units, the point A has the coordinates (a, 0).

Step 2 : Through ‘P’ draw a line ‘m’ parallel to the x-axis and mark B as the point ofintersection of ‘m’ with the y-axis (See Fig. 6.9).

Fig. 6.8 Fig. 6.9If B is ‘b’ units from O, along the y-axis,

then B has the coordinates (0, b).

∴ The y-coordinate of ‘P’ is ‘b’.

Hence the coordinates of ‘P’ are (a, b), i.e., the ordered pair associated with ‘P’ is (a, b).


Find the coordinates of the points P, Q, R and S in Fig. 6.10.

6.11. GRAPH OF A LINEAR EQUATION

Let us draw the graphs of some linear equations.

Remember that :

(i) If a point lies on the graph of an equation,its coordinates satisfy the equation.

(ii) If a point is not on the graph of an equation,its coordinates do not satisfy the equation.

Here are some examples of graphs of linearequations. Fig. 6.10


Example 6.9 : Draw the graphs of the following equations.

(i) x – 2y = 0 (ii) 2x – 5y – 10 = 0.

Solution : In order to draw the graph of a linear equation, we plot at least three ordered pairssatisfying the equation.

The steps followed are as given below.

Equation (i) x – 2y = 0

Step 1 : Transform the equation into thex-form or the y-form.

Step 2 : Substitute three values of x to get thecorresponding three values of y.

∴ Three ordered pairs satisfying theequation are (0, 0), (2, 1) and (6, 3)

Step 3 : Write the ordered pairs satisfying thegiven equation in the form of a table.

Step 4 : Plot the ordered pairs (0, 0) (2, 1) and (6, 3) in the same cartesian plane. (See Fig.6.11)

Step 5 : Join the point O, A and B and extend the line segment to get the line ‘l’ (asFig. 6.12). ‘l’ is the, required graph of the equation x – 2y = 0.

y-form of the equation is : y = x2

(i) x = 0 gives y = 02 = 0

(ii) x = 2 gives y = 22 = 1

(iii) x = 6 gives y = 62 = 3

(0, 0), (2, 1) and (6, 3) are threesolutions of x – 2y = 0

Tablex 0 2 6

9 0 1 3

Fig. 6.11 6.12

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Equation (ii) 2x – 5y – 10 = 0

or 2x = 5y + 10 or x = 52

5y +

This is the x-form of the equation.

Three of its ordered pairs are (10, 2), (5, 0) and (0, –2).

Plotting the ordered pairs (10, 2), (5, 0) and (0, –2), we get the three points P, Q and R.

Join the points P, Q and R to get the line ‘m’ (as shown in Fig. 6.13). ‘m’ is the required linewhich is the graph of the linear equation 2x – 5y – 10 = 0

Fig. 6.13

Notes 1. In fact, we an draw the graph of a linear equation by plotting two ordered pairs onlybecause two points determine a line uniquely. Plotting the third pair helps us to verifythe correctness of the graph.

2. The graph of a linear equation in two unknowns is a straight line which is neitherthe x-axis, nor the y-axis and is not parallel to either of the axis. This is becausefor a linear equation in two unknowns, ax + by + c = 0, neither ‘a’ nor ‘b’ is zero.

But what happens if out of ‘a’ and ‘b’ one of them is zero ?

Let us take an example.

Example 6.10 : Draw the graphs of each of the following equations :

(i) x = 2 (ii) 2y + 5 = 0

Solution : (i) The equation x = 2 can be written as x + 0.y = 2.

Hence, some of the ordered pairs satisfying it are

(2, 0), (2, 2), (2, –1), (2, 3), ...

We plot three ordered pairs, say (2, 0), (2, 2) and (2, –1)

Table

x 10 5 0

y 2 0 –2

Table

x 2 2 2

y 0 2 –1


We get three points A (2, 2), B (2, 0) andC (2, –1) on the cartesian plane. Join them to getthe line ‘l’ as in Fig. 6.14. Thus, ‘l’ is the requiredgraph of x = 2.

Note that ‘l’ is parallel to the y-axis.

(ii) 2y + 5 = 0

This equation can be written as

0.x + 2y + 5 = 0

and the y-form of this equation is

y =−52

or y = –2.5

∴ Any point whose y-coordinate is –2.5 is a solutionof this equation. Three of the ordered pairs satisfyingthe equation are (–1, –2.5), (0, –2.5) and (1, –2.5)

Plotting these ordered pairs, we get the three pointsM, N and R.

Join them to get the line ‘p’ (See Fig. 6.15)

p is the required graph of the equation 2y + 5 = 0

This is a line parallel to the x-axis.

Notes : 1.The graph of a linear equation in oneunknown is either the x-axis or the y-axis,or is a line parallel to either of the axes.

2. The y-axis is the graph of the equationx = 0.

3. The x-axis is the graph of the equationy = 0.


1. Draw the graphs of x = 0 and of y = 0.2. Draw the graphs of each of the following equations :

(i) 2x – 3y = –1 (ii) 3x + y = 4 (iii) y = 5(iv) 2x + 3 = 0 (v) x + y = 0 (vi) 3x + 3y = 6

(vii) x = –5 (viii) 2x – 3y = 8 (ix) x – y = 0

Table

x –1 0 1

y –2.5 –2.5 –2.5Fig. 6.14

Fig. 6.15

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3. Which of the following ordered pairs make the equation 2x + y = 10 a true statement?(i) (0, –4) (ii) (0, 10) (iii) (8, 8) (iv) (–3, 16)

4. State which of the following ordered pairs are not the solutions of the equation4x = 3y + 8.(i) (7, 2) (ii) (5, 4) (iii) (8, 8) (iv) (0, 0).

6.12 THE SYSTEM OF LINEAR EQUATIONS IN TWO UNKNOWNS

You know that ax + by + c = 0 (where a, b ≠ 0 and a, b, c are real numbers) is a linear equationin two unknowns (variables).

A system of linear equation in two unknowns is given as :

ax + by + c = 0

a'x + b'y + c' = 0

where a, b, c, a', b' and c' are all real numbers and a, b, a' and b' are non-zero.

Let us now learn to solve a system of linear equations in two unknowns.

6.13. GRAPHICAL SOLUTION OF A SYSTEM OF LINEAR EQUATIONS

In order to solve a system of linear equations in two unknowns graphically, we draw the graphsof the equations in the same cartesian plane and find the point(s) of intersection of their graphs.


Example 6.11 : Solve the following system of linear equations graphically :

x – 2y = 7

x + y = –2

Solution : You have already learnt how to draw the graph of a linear equation in two unknowns.

The graphs of these equations are straight lines.

x – 2y = 7 Table 1

x = 7 + 2y x – 2y = 7

Three of ordered pairs satisfying it are x 1 7 3

(1, –3), (7, 0) and (3, –2) y –3 0 –2

Also, x + y = –2 Table 2or y = –x – 2 x + y = –2

Three of ordered pairs satisfying it are x 0 –2 1

(0, –2), (–2, 0) and (1, –3) y – 2 0 –3

By plotting these ordered pairs we get a pair of straight lines as the graphs of the given equations(See Fig. 6.16).


Fig. 6.16.

Now, how many ordered pairs are the solutions of both the equations ?

In other words, how many points are common to both the lines ‘l’ and ‘m’ ?

From the graph, it is clear that there is only one such point, namely, ‘P’, whose coordinatesare (1, –3).

Thus, x = 1 and y = –3 is the solution of the given system of equations.

Example 6.12 : Solve graphically the following system of linear equations :

4x + 3y = 24

3y – 2x = 6

Solution : 4x + 3y = 24

or y = 24 43− x Table 1

4x + 3y = 24

Three of ordered pairs satisfying it are x 6 0 3

(6, 0), (0, 8) and (3, 4) y 0 8 4

Plotting these ordered pairs, we get three points A (6, 0), P (3, 4) and B (0, 8) and on joiningthem, we get line ‘l’ as the graph of 4x + 3y = 24.Also, 3y – 2x = 6

or y = 6 23+ x Table 2

3y – 2x = 6Three of ordered pairs satisfying it are x 0 –3 3(0, 2), (–3, 0) and (3, 4). y 2 0 4

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Plotting these ordered pairs in the same cartesian plane we get three points M(0, 2),N(–3, 0) and P (3, 4), and on joining them, we get the line ‘m’ as the graph of 3y – 2x = 6(See Fig. 6.17).

Fig. 6.17

To get the common solution of both the equations, we look for the points that are commonto both the lines ‘l’ and ‘m’.

The only point common to both the lines ‘l’ and ‘m’ is point ‘P’ with coordinates (3, 4).

In other words x = 3 and y = 4 is the only solution of the given system of linear equations

Example 6.13. Draw the graph of :

y – 2x = 32y – 4x = 10

and find their common solution, if any.Solution. (i) y – 2x = 3 Table

y – 2x = 3x 0 1 2

or y = 2x + 3 y 3 5 7∴ On plotting the ordered pairs (0, 3) (1, 5) and (2, 7), we get the straight line ‘p’ as the graphof y – 2x = 3(ii) 2y – 4x = 10 Table

or y = 2x + 5 2y – 4x = 10x 0 1 2y 5 7 9


∴ On plotting the ordered pairs (0, 5) (1, 7) and (2, 9), we get, the straight line ‘q’ as thegraph of 2y – 4x = 10 in the same plane, (See Fig. 6.18).

Fig. 6.18

It is very clear from the graph that the straight lines ‘p’ and ‘q’ are parallel, and hence do nothave any point common.

Thus, there is no solution common to both the equations.

Hence, the given system of equations has no solution.

Example 6.14 : Draw the graphs of :

y – 2x = 3

2y – 4x = 6

and obtain the common solutions, if any Table 1

y – 2x = 3

Solution : y – 2x = 3 x 0 1 2

or y = 2x + 3 y 3 5 7

Three of its solutions are the ordered pairs (0, 3), (1, 5) and (2, 7) and on plotting these weget three points P, Q and R. On joining them, we get the line ‘l’ as the graph of y – 2x = 3.

Similarly, 2y – 4x = 6

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or y = 2x + 3, and so the same ordered pairs satisfy the second equation also (See Fig. 6.19).

Fig. 6.19.

From Fig. 6.19, it is clear that the lines ‘l’ and ‘m’ coincide or are the same.This means that all the points of the line ‘l’ are also on the line ‘m’ and all the points of line‘m’ are also on the line ‘l’. Thus all points of the lines ‘l’ and ‘m’ are common.Thus, these two equations have an infinite number of common solutions.


1. Draw the graphs of each of the following systems of equations :

(i) 2x – y – 3 = 0 (ii) x + y = 7

4x – 2y – 10 = 0 2x + 6y = 14

2. Solve graphically each of the following system of equations :

(i) 2x – y = 5 (ii) x + y = 8

x + 3y = 6 2x – y = 1

3. Find graphically the vertices of a triangle whose sides have the equations 2y – x = 8,5y – x = 14 and y – 2x = 1 respectively.

(Hint : Draw the graphs of the three equations using the same cartesian plane and scale.1 cm = 1 unit on both axes and find the points of intersection of two lines at a time).


6.14 CONSISTENCY AND INCONSISTENCY OF A SYSTEM OF EQUATIONS

You now know, how to solve a system of equations graphically.

The system in Example 6.11 and 6.12 had a unique solution. In Example 6.13, the system ofequations had no solution, whereas the system in Example 6.14 had infinite solutions.

These three examples help you to conclude that :

If the graphs of the pair of equations ax + by + c = 0 and a'x + b'y + c' = 0:

(i) intersect each other, then the system has one and only one solution, i.e. the systemhas a unique solution and is said to be a consistent or compatible system.

(ii) are parallel straight lines, then the system has no solution, and hence is aninconsistent system.

(iii) are one and the same, i.e., are coincident straight lines, the system has infinitenumber of solutions and thus is a dependent system. This is also called aconsistent system of equations.

Notes : Any system of two linear equations in two unknowns has to be one of the typesof systems described above.

Solving the system graphically is not necessary if you want to determine whether the systemof equations is consistent or not. There are other algebraic methods to do so.

To understand some of these methods, let us study the examples solved in Section 6.14 of thislesson.

Method I. Comparison of y-forms

Let us compare the y-forms of the equation in Examples 6.12,6.13 and 6.14 and put the resultin the form of a table.

Equations y-forms Solutions Type of System

1. Example 6.12

4x + 3y = 24 y = 24 43− x Unique solution Consistent

3y – 2x = 6 y = 2 63

x +

2. Example 6.13y – 2x = 3 y = 2x + 3 No solution Inconsistent2y – 4x = 10 y = 2x + 5

3. Example 6.14y – 2x = 3 y = 2x + 3 Infinite number of solutions Dependent2y – 4x = 6 y = 2x + 3

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From the above table you can easily conclude that if in the y-forms of theequations :

1. The coefficients of x are different (even if the constant terms are the same),the system of equations has a unique solution.

2. The coefficients of x are the same but the constant terms are different, the systemof equations has no solution.

3. The coefficients of ‘x’ and the constant terms are the same, i.e., the two y-formsare identical, the system has infinite number solutions.

Method 2. The Ratio Method

This is another method to determine whether a system of equations is consistent or not.

A system of linear equations ax + by + c = 0 and a'x + b'y + c' = 0 is :

(i) consistent if aa

bb′ ≠ ′

(ii) inconsistent if aa

bb

cc′ = ′ ≠ ′

(iii) dependent if aa

bb

cc′ = ′ = ′

Let us take some examples and illustrate :

Example 6.15 : In each of the following, find whether the system is consistent, inconsistentor dependent :

(i) 5x + 2y = 16 (ii) 5x + 2y = 16 (iii) 5x + 2y = 15

7x – 4y = 2 3 65

2x y+ = 152

3 24x y+ =

Solution : (i) (a) Method 1 : Using y-formsThe y-forms of the two equations are :

y = − +52 8x and y = 7

412x −

Since the coefficient of ‘x’ are different in the two y-forms, the system of equations is consistent.(b) Method 2 : The Ratio Method

Here a = 5, b = 2 and c = –16 and a′ = 7, b′ = –4 and c’ = –2

∴ aa′ =

57 , b

b′ = −12 and c

c′ = 8

Since, aa

bb′ ≠ ′ , the system of equations is consistent.


(ii) Using y-forms of the two equations, we have

y = − +52 8x and y = − +5

253x

Since the coefficient of ‘x’ are the same and the constant terms are different, the system isinconsistent.(iii) The Ratio method

Here a = 5, b = 2 and c = –16, a′ = 152

b′ = 3 and c′ = –24

∴ aa′ =

515 2

23=

bb′ = 2

3

and −−

1624 = 2

3

∴ aa′ = b

bcc′ = ′ =

23

Hence, the system of equations is dependent.


1. In each case, draw the graphs of the system of equations and state whether the systemis consistent, inconsistent or dependent.

(i) x y2 6+ = (ii) x = 5 (iii) x + 3y = 32x + 4y = 24 2x – y = 9 2x – 5y = 0

2. Find without drawing the graphs, whether each of the following systems of equations has: (a) a unique solution, or (b) infinite number of solutions or (c) no solution.

(i) x + y = 1 (ii) y = 2x – 3 (iii) x y3 1 2− =

4 = – 4x – 4y x = y2

32+ x + y = 1

3. Using the Ratio Method find out whether each of the following system of equations isconsistent, inconsistent or dependent :

(i) 2x + 3y = 13 (ii) y x− = +27 5 (iii) 3y = –7x + 21

5x – 2y = 4 x y+ = −103 3 5y + 2x = –23

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4. Find the values of ‘k’ for which the system of equations given below has infinite numberof solutions :

3x + y = 0

x – ky = 2

6.15 ALGEBRAIC METHODS OF SOLVING A SYSTEM OF EQUATIONS

It is not always possible to find the exact solutions of a given system of equations graphically.Some algebraic methods are used to solve the system of equations and these are usually calledelimination methods. Some of these methods are discussed below.

6.15.1 Elimination by Comparison

Let us understand this method with the help of an example.

Example 6.16 : Solve : 2y – x = 5

4x – 3y = 5

Solution :

Method

1. Write the y-forms (or x-forms) of the givenequations.

2. Equate the right hand side (R.H.S.) of they-forms

3. Solve for equation for ‘x’

4. Substance the value of ‘x’ in one of theequations and get the value of ‘y’

Thus, x = 5 and y =5 is the solution of the system of equations : 2y – x =5 and 4x – 3y = 5

Solution of the Example

y-forms are :

2y – x = 5 or y = x2

52

+

4x – 3y = 5 or y = 43

53

x −

x2

52

+ = 43

53

x −

or 3x + 15 = 8x – 10

or 5x = 25

or x = 5

2y – x = 5

or 2y – 5 = 5

or 2y = 10

or y = 5


Check : Substitute the values ‘5’ for ‘x’ and ‘5’ for ‘y’ in both the equations and check whetherthe solution is correct or not.

2y – x = 5 gives 2 × 5 – 5 = 5 True

4x – 3y = 5 gives 4 × 5 – 3 × 5 = 5 True

Note : It is not necessary to write the equations in y-form only, you can write them in thex-form also and solve them.

Do it for Example 6.16 and check.


Solve each of the following pairs of equations by comparing the x-forms or y-forms.

1. x + 2y = 3 2. 2x + y = 1 3. 2x + 3y = 3

x – 2y = –1 3x – y = 4 3x + 2y = 2

6.15.2 Elimination by Addition or Subtraction

Let us take an example to understand this method.

Example 6.17. Solve : For x and y

3x – 7 = y

4x – 5y = 2

Solution :

Method

1. Arrange the equations so that the like termsare in the same column.

2. Multiply the equation by suitable constantsso that the coefficient of one of the unknownshas the same numerical co-efficient.

3. Eliminate the unknown whose coefficientshave the same numerical value by :

(a) adding if their signs are unlike

(b) subtracting if their signs are like.

4. Find the value of the unknown left.

Solution of the example

(i) 3x – 7 = y or 3x – y = 7

(ii) 4x – 5y = 2 or 4x – 5y = 2

(i) × 5 gives 15x – 5y = 35

(ii) gives 4x – 5y = 2

Eliminate y :

15x – 5y = 35

Subtract : 4x – 5y = 2– + –

11x = 33

11x = 33

∴ x = 3311

or x = 3

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5. Find the value of the other unknown bysubstituting the value of the unknown foundin either of the two given equations.

Thus, x = 3 and y = 2 is the solution of the given system of equations.Check : Substituting the values of ‘x’ and ‘y’ in the two given equations, we get

3 × 3 – 7 = 2 or 9 – 7 = 2 Trueand 4 × 3 – 5 × 2 = 2 or 12 –10 = 2 Truex = 3 and y = 2 i.e. the ordered pair (3, 2) is the required solution.

Here are some more solved examples.

Examples 6.18 : Solve by addition or subtraction each of the following system of equations:

(i) 10x + 4y = 20 (ii) 3r – 5s = 19

13x – 4y = 66 2r – 4s = 16

Solution : (i) 10x + 4y = 20 ...(1)

(ii) 13x – 4y = –66 ...(2)

Here, the coefficient of ‘y’ in (1) and (2) are numerically the same but opposite in sign.

Hence, ‘y’ can be eliminated by adding the two equations

10x + 4y = 20

+ 13x – 4y = – 66

23x = –46

or x = –2

Substituting the value of ‘x’ in (1), we get

10 × (–2) + 4y = 20

or 4y = 40

or y = 10

Thus, x = –2 and y = 10 is the solution required.

∴ x = –2 and y = 10 or the ordered pair (–2, 10) is the required solution.

(ii) 3r – 5s = 19 ...(1)

2r – 4s = 16 ...(2)

Here neither the coefficients of ‘r’ nor of ‘s’ are the same.

Multiplying (1) by 2 and (2) by 3, we get,

Find ‘y’ : (ii) is 4x – 5y = 2

or 4 × 3 – 5y = 2

or 12 – 5y = 2

or 5y = 10

or y = 2


6r – 10s = 38 ...(3)

6r – 12s = 48 ...(4)

Now, the coefficients of r are numerically the same and have the same sign.

Thus, subtracting (4) from (3), we get

6r – 10s = 38

6r – 12 s = 48

– + –

2s = –10

or s = –5

Substituting the value of ‘s’ in (2) we get

2r – 4(–5) = 16

or 2r + 20 = 16

or 2r = –4

or r = –2

Thus, r = –2 and s = –5 is the required solution.


Solve each of the following systems of equations and check your answers :

1. 3x + 2y = 11 2. x + y = 7 3. 7x – 2y = 1

2x + 3y = 4 3x – y = 11 3x + 4y = 15

6.15.3. Elimination by Substitution

Let us take an example to understand this method.

Example 6.19 : Solve for x and y :

x – 2y = 7

3x + y = 35 by elimination by substitution.

Solution : The equation are :

x – 2y = 7 ...(1)

3x + y = 35 ...(2)

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Method

1. Express one unknown in terms of the otherunknown form either of the equations.

2. Substitute the value of the first unknown inthe other equation.

3. Solve the resulting equation for the secondunknown.

4. Find the value of the first unknown bysubstituting the value of the second unknownfound in either of the two given equations.

Thus, x = 11 and y = 2.

∴ x = 11 and y = 2 is the solution of the given system of equations.

Example 6.20 : Solve by elimination by substitution the following system of equations :

3y = 2x – 14 and x – y = 10

Solution : The system of equations is

3y = 2x – 14 ...(1)

and x – y = 10 ...(2)

Expressing ‘y’ in terms of ‘x’ from equation (1) we get

y = 23

143x −

Substituting the value of ‘y’ in terms of ‘x’ in (2), we get

x x− −FH IK23

143 = 10

or x x− +23

143 = 10

Solution of the Example

Expressing ‘x’ in terms of ‘y’from (1), we get

x – 2y = 7

or x = 7 + 2y

Substituting for ‘x’ in (2), we get

3(7 + 2y) + y = 35

Solving for ‘y’, we get

3(7 + 2y) + y = 35

or 21 + 6y + y = 35

or 21 + 7y = 35

or 7y = 14

or y = 2

Solving for ‘x’ we get

x – 2 × 2 = 7 from (1)

or x – 4 = 7

or x = 11.


or 3x – 2x + 14 = 30

or x = 30 – 14

or x = 16.

Substituting the value of ‘x’ in (1), we get

3y = 2 × 16 – 14

or 3y = 18

or y = 6

Thus, x = 16 and y = 6 is the solution of the given system of equations.


Solve each of the following system of equations by substitution :

1. 5x + 3y = 17 2. y x3 2+ = 1

x + 3y =1 35 2 1x y+ =

6.16 EQUATIONS REDUCIBLE TOax + by + c = 0

a'x + b'y + c' = 0Consider the system of equations given below :

1 2u v− = 1

3 2u v+ = –5

This is not a system of linear equations, but it can be reduced to such a system and then solved.Example 6.21. Reduce

1 2u v− = 1

3 2u v+ = –5

to a system of linear equations and solve the same.

Solution. Let 1u = x and 1

v = y

∴ The system of equations becomes :

x – 2y = 1 ...(1)

3x + 2y = –5 ...(2)

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This is a system of linear equations which can easily be solved by the methods you have learntso far.

Adding (1) and (2) we get,

4x = – 4

or x = –1

Now –1 – 2y = 1 ...[Substituting for ‘x’ in (1)]

or –2y = 2 or y = –1

∴1u = –1 or u = –1

and y = –1

or1v = –1 or v = –1

∴ The solution is u = –1, v = –1.


1. Solve each of the following system of equations by comparison of x-forms or y-formsand check you answers.

(i) 4x – 3y = 8 (ii) 3y = 12 – 2x

x – 2y = –3 3x – 8y = –7

2. Solve each of the following system of equations by addition or subtraction and check youranswers :

(i) x + 2y = 24 (ii) 7x + 6y = 71

x – y = 4 5x – 8y = –23

(iii) 13x + 11y = 70 (iv) 13x – 4y = 57

11x + 13y = 74 5x + 2y = 29

3. Solve each of the following system of equations by substitution and check youranswers :

(i) y = 3x – 5 (ii) x + y = a

6x = 3y + 3 x – 2y = b, a and b are constants

(iii) y = 2x – 6 (iv) p = 2q –1

y = 0 q = 5 – 3p


4. If 2x + y = 23 and 4x – y = 19, find the value of A = 3x – 2y and B = 5y + 2x.

[Hint : Find ‘x’ and ‘y’ by solving the given equations and substitute their values in ‘A’and ‘B’]

5. Solve each of the following system of equations and check your answers.

(i) 4 6 15p q+ = (ii) x y3 4 4+ =

6 8 14p q− = 3x – y = 23

[Hint. substitute 1q = y in (i)]

(iii) x – y = 0.9

112 x y+b g = 1

[Hint. First simplify the second equation]

6. Find the solution for each of the following systems of equations and check youranswers :

(i) 3 3 1u v− = (ii) 8 9 1u v− = (iii) 2u + 3v = 6uv

1 1 1u v+ = 10 6 1u v+ = 5u – 2v = 3uv

6.17 WORD PROBLEMS

You already know that there are many problems in daily life which can easily be solved byusing the methods you have learnt in this lesson.

For solving these problems, you will have to (i) first express the statements given in the problemalgebraically, then (ii) find their common solution.

Let us take some examples to understand these two steps.

6.17.1 Expressing the Statements Algebraically

Example 6.22 : Express the following statements algebraically :

1. “The perimeter of a rectangle is 52 cm. The breadth of the rectangle is 2 cm more thanone third its length.

Solution : This is the problem discussed in the beginning of the lesson.

Let ‘x’ denote the length and ‘y’ denote the breadth of the rectangle

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Statement Equation

(i) The perimeter is 52 cm 2 (x + y) = 52

(ii) The breadth is 2 cm more y = 23

+ x

than one-third the length

Thus, the system of equations representing the statements in the problem is :

2(x + y) = 52 or x + y = 26

y = 23

+ x–x + 3y = 6

2. ‘The larger of two numbers is three times the smaller number. Their sum is 8 more thantwice the smaller number.

Solution. Let ‘x’ represent the smaller number and ‘y’ represent the larger number.

Then :

Statement Equation

(i) The larger number is ‘3’ times the smallernumber

(ii) The sum is 8 more than twice the smallernumber

Thus, the system of equations representing the statement given isy = 3x

x + y = 2x + 8

3. The sum of the digits of a two-digit number is 11. If the digits are reversed, the new numberis 20 less than twice the original number.

Solution; Let ‘t’ represent the digit in the ten’s place and ‘u’ represent the digit in the unit'splace of the number.

Then, the given number is 10 t + u.

When the digits are reversed, the new number is 10u + t

Statement Equation

(i) The sum of the digits is 11 u + t = 11

(ii) The new number is 20 less than twicethe given number

∴ The given statements can be expressed as :u + t = 11

and 10u + t = 2(10t + u) – 20.

10u + t = 2(10t + u) – 20

y = 3x

x + y = 8 + 2x



Express each of the following statements as a system of equations :

(i) The perimeter of a rectangle is 72 cm. Three times its breadth exceeds twice the lengthby 3 cm.

(ii) The sum of two numbers is 64 and the larger number is three times the smaller number.

6.17.2 Solving the Word Problems

Let us now find solutions to word-problems by taking some examples.

Examples 6.23 : Three years back (ago) Atul’s age was four times Parul’s age then. Five yearshence (from now) Atul’s age will be two times Parul’s age then. Find their present ages.

Solutions : Let the present age of Atul be ‘x’ years and Parul’s age be ‘y’ years

Then, the conditions in the problem can be expressed as follows :

x – 3 = 4(y – 3) ...[Ages 3 years ago]

and x + 5 = 2(y + 5) ...[Ages 5 years from now]

or x – 4y = –9 ...(1)

and x – 2y = 5 ...(2)

From (1), we have

x = 4y – 9

Substituting in (2), we get

4y – 9 – 2y = 5 or 2y = 14

or y = 7

Substituting y = 7 in (2), we get x – 2 × 7 = 5

or x = 19.

Hence, Atul’s present age is 19 years and Parul’s present age is 7 years.

Example 6.24 : The length of a rectangle is 5 cm less than twice its breadth. If the perimeteris 110 cm., find the area of the rectangle.

Solution : Let the length of the rectangle be ‘x cm’ and the breadth be ‘y’ cm.

Then, the statements in the problems can be expressed as :

x = 2y – 5 ...(1)

and 2(x + y) = 110 ...(2)

Substituting the value of ‘x’ in terms of ‘y’ from (1) in equation (2), we get

2(2y – 5 + y) = 110 or 2 (3y – 5) = 110

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or 6y – 10 = 110 or 6y = 120

∴ y = 20

Substituting the value of ‘y’ in (1), we get,

x = 2 × 20 – 5

or x = 35

Thus, the area of the rectangle = x.y cm2

= 35 × 20 cm2

= 700 cm2

Check : Here, length = 35 cm and breadth = 20 cm

2 × breadth – 5 = 2 × 20 –5

= 40 – 5 = 35 = length.

∴ The first condition is verified.

Also, perimeter = 2(35 + 20) = 2 × 55 = 110 cm.

Hence, the second condition is also verified.


1. Express each of the following statements as a system of equations

(a) The sum of two numbers is 12 and their difference is 2.

(b) Two numbers are in the ratio 4 : 7. If three times the larger number is added to twotimes the smaller, the sum is 59.

(c) 2 pens and 5 note-books cost Rs 16 and 3 pens and 4 note books cost Rs 17.

(d) The sum of the digits of a two-digit number is 10. The number obtained on reversingthe digits is one more than twice the given number.

(e) The perimeter of a rectangular plot of land is 32 m. If the length is increased by 2mand the breadth is reduced by 1m, the area of the plot remains unchanged (i.e., it isequal to the area of the plot given).

2. Solve each of the following problems by writing each of them as a system of equations:

(a) The sum of two numbers exceeds three times the smaller by 3. The difference of thenumbers is 5. Find the numbers.

(b) ‘A’ is five years older than ‘B’. Five years ago ‘A; was twice as old as ‘B’ was then.Find their present ages.


(c) The perimeter of a rectangle is 20 cm. If the difference between the length and thebreadth is 4 cm, find the length and breadth of the rectangle.

(d) The sum of the weights of a father and his son is 105 kg. The weight of the son isone-sixth that of the father. Find their respective weights.

(e) 400 people attended a school fete. Some people gave Rs 100 as a donation and somegave Rs 50. The total collection was Rs 34000. Find the number of people who gaveRs 100 each.

LET US SUM UP

An equation in one variable of degree one is called a linear equation in one variable.ax + b = 0, a≠ 0 and a, b are real numbers is the general form of a linear equation inone variable.Solution of a linear equation is also called its root.To solve a word problem, it is translated to algebraic statement(s) and then solved.The equation ax + by + c = 0, a≠ 0, b≠ 0 and a, b, c are real numbers is called thegeneral form of a linear equation in two variables.To draw the graph of a linear equation, we find three ordered pairs and plot them. Theline joining the points is the graph of the linear equation.The algebraic methods of solving a system of linear equations are(i) elimination by substitution

(ii) elimination by equating the co-efficients.To solve a word problem, we translate it into linear equations and solve them.

TERMINAL EXERCISE

1. Draw the graphs of each of the following system of equations and state whether they have

(i) a unique solution (ii) infinite number of solutions (iii) no solution

(a) x + y = 2 (b) 2x + 3y = 5 (c) x + 2y = 2

x – y = 4 3x – 2y = 1 2x + 4y = 6

(d) x + 2y = 2 (e) x + y = 0

2x + 4y = 4 2x – 3y = 0

2. Without solving the system of equations, find whether the system is consistent, inconsistentor dependent :

(a) 2x + 3y = 1 (b) 3x + y = 4y – x = 2 x + 3y = 4

(c) x – y = 1 (d)x y2

3+ = 2

3y – 3x = 6 x + 6y = 4

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3. Solve each of the following system of equations :

(a) 3x + y = 2 (b) x – 4y = –3x – 2y = 1 x + y = 2

(c) 52 x y− = 1 (d) x + y = 4

4x + y = 5 x – 4y = 44. Find the values of ‘k’ for which the system of equations

4x + y = 5

x + ky = 20

has (a) a unique solution, (b) infinite number of solutions.

5. Solve each of the following system of equations :

(i)ax

by+ = 2

ax

by− = 0 [a, b are non-zero constants, x, y≠ 0]

(ii) 3 1 4x y+ =

2 1 1x y− = , y≠ 0

(iii) 1 1 4u v+ =

2 1 2u v− = , u, v≠ 0

6. The sum of the digits of a two-digit number is 10. If 18 is added to the number, its digitsare reversed. Find the number.

7. The sum of the present ages of ‘A’ and ‘B’ is 85 years. Five years ago, ‘A’ was twiceas old as ‘B’ was then. Find their present ages.

8. Six chairs and two tables cost Rs 240. Also, the cost of a table is Rs 5 less than twicethe cost of a chair. Find the total cost of 10 chairs and 3 tables.

9. If the numerator of a fraction is decreased by one, it becomes 23 , but if the denominator

is increased by 5, the fraction becomes 12 . What is the fraction ?

10. The perimeter of a rectangular is 48 cm. If its length is 52 times the breadth, find out

the length of the rectangle. Also find out its area.


ANSWERS


1. (i) 2. (ii) 3. (iii) 4. (iv) 5. (i)

6. (iii) 7. (iv) 8. (iii) 9. (iii) 10. (iii)

11. (i) 5 (ii) 6 (iii) 5


1. Father : 30 years, Son : 10 year

2. 35 3. 13, 14, 15 4. 3 5. 36°, 54°, 90°


P(1, 1), Q(–2, 1), R(–2, –2) and S(2 –1)


3. (ii) and (iv) 4. (i) and (iv)


1. Graphs to be drawn.

2. (i) (3, 1) (ii) (3, 5) 3. (–4, 2), (1, 3) and (2, 5)


1. (i) Dependent (ii) Consistent (iii) Consistent

2. (i) No Solution (ii) Infinite number of solutions (iii) Unique solution.

3. (i) Consistent (ii) Consistent (iii) Consistent

(iv) Inconsistent.


1. x = 1, y = 1 2. x = 1, y = 1 3. x = 0, y = 1


1. x = 5, y = –2 2. x = 92 , y = 5 3. x = 1, y = 3

Check your progress 6.10

1. x = 4, y = –1 2. x = 103 , y = –2

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1. (i) x = 5, y = 4 (ii) x = 3, y = 2

2 (i) x = 323 , y = 20

3 (ii) x = 5, y = 6

(iii) x = 2, y = 4 (iv) x = 5, y = 2.

3. (i) x = 4, y = 7 (ii) x = 23

a b+ , y = a b−3

(iii) x = 3, y = 0 (iv) p = 97 , q = 8

7

4. A = 3, B = 59.

5. (i) p = 3, q = 2 (ii) x = 9, y = 4

(iii) x = 3.2, y = 2.3

6. (i) u = 32 , v = 3 (ii) u = 46

5 , v = – 69

(iii) u = 1924 , v = 19

21 .

Check your Progress 6.12

(i) 2(x + y) = 72, 3y = 2x + 3, where x is length and y is breadth

(ii) x + y = 64, x = 3y, where x is larger number.


1. (a) x + y = 12 and x – y = 2

(b) 7x – 4y = 0 and 3y + 2x = 59.

(c) 2x + 5y = 16 and 3x + 4y = 17.

(d) x + y = 10 and 8x – 19 y = 1, where ‘x’ is the unit’s digit and ‘y’ the ten’s digitof the number.

(e) x + y = 16 and 2y – x – 2 = 0, where ‘x’ is length and ‘y’ is breadth.

2. (a) Smaller number = 2, larger = 7.

(b) A’s age = 15 years, B’s age = 10 years.

(c) Length = 7 cm., Breadth = 3 cm.

(d) Father’s weight = 90 kg., son’s weight = 15 kg.

(e) 280 people.


Terminal Exercise

2. (a) Consistent (b) Consistent (c) Inconsistent (d) Dependent

3. (a) x = 57 , y = − 1

7 (b) x = 1, y = 1 (c) x = 1213 , y = 17

13

(d) x = 4, y = 0

4. (a) k ≠ 14 (b) k = 1

4

5. (i) x = a, y = b (ii) x = 1, y = 1

(iii) u = 12 , v = 1

2

6. 35 7. A : 55 years: B = 30 years 8. Rs 385

9. 79 10. Length : 120

7 cm, Area = 576049 cm2

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7

Quadratic Equations

7.1 INTRODUCTION

Many times in life we come across problems whose solution we find by hit and trial methodwith the help of available information. For example

(i) Suppose we are to find two consecutive natural numbers whose product is 12, we caneasily guess that the numbers are 3 and 4; similarly if the product is 56 again we canfind the number as 7 and 8, but if the product is 552, it becomes difficult to answer theproblem by a guess.

(ii) Similarly suppose we are to find the dimensions of a rectangle with area 168 square metresand the length exceeds breadth by 2 m. It is again difficult to answer the problem by makinga guess.

The above type of problems can be solved by a systematic method. For example in the firstcase if the numbers are x and x + 1, we have to solve the equation x (x + 1) = 552 orx2 + x – 552 = 0.

Again in 2nd case, when we take breadth as x, the length will be x + 2 and we will have tosolve the equation x (x + 2) = 168 or x2 + 2x – 168 = 0.

If we put different conditions on the length and breadth, we shall again get an equation of thetype ax2 + bx + c = 0 but with different values of a, b and c.

Such type of equation is called a quadratic equation, and in this lesson we shall deal withsolutions of such equations.

7.2 OBJECTIVES


identify a quadratic equation from a given collection of equations.

write quadratic equations in standard form.

solve a quadratic equation by (i) factorisation and (ii) using the quadratic formula.

form a quadratic equation with given roots.

Quadratic Equations 163

translate a word problem into a quadratic equation.

solve word problems, using quadratic equations.


identifying and solving Linear equations in one or two variables

finding square root of natural numbers

factorisation of polynomials.

7.4 QUADRATIC EQUATIONS

Recall that p(x) = ax2 + bx + c is the general form of a quadratic polynomial where a, b andc are real number and a ≠ 0.

In particular, consider a quadratic polynomial p(x) = 2x2 – 5x + 2.

You know that for every real value of x, p(x) has a real value. For example p(x) = –1 forx = 1 and p(x) = 5 for x = 3. Can you find a value of x for which p(x) = 0 ? By hit and trialmethod you can find that for x = 2, p(x) = 0. Can you find some more values of x for which

p(x) becomes zero ? After some more trials, you will find that for x = 12 , p(x) becomes zero.

Thus, x = 2 and x = 12 are the numbers for which p(x) = 0.

2 and 12 are called the zeroes of p(x).

The values of the variables for which a polynomial p(x) vanishes are known aszeroes of the polynomial p(x).

Quadratic Equation : Let p(x) = ax2 + bx + c, where a, b and c are real numbers anda ≠ 0 be a quadratic polynomial, then p(x) = 0 i.e. ax2 + bx + c = 0 is called a quadraticequation.

Thus,

An equation of the form ax2 + bx + c = 0, where a≠ 0 and a, b and c are realnumbers, is called a quadratic equation in x.

For example, 3x2 – 5x + 7 = 0 is a quadratic equation in x and 2y2 + 3y – 7 = 0 is a quadratic

equation in y. 3 5 02a + = , 23 x2 – 5x = 0, x2 = 0 are some other examples of quadratic

equations.Example 7.1 : Determine which of the following are quadratic equations in x.

(i) (x + 3) (x – 2) = 5 (ii) 2x2 + 3x = 2x(x – 7)

(iii) 3x2 – 5x = 5(x2 – x + 3) (iv) x x− 1 = 0, x≠ 0.

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Solution : (i) (x + 3)(x – 2) = 5

or x2 + 3x – 2x – 6 = 5or x2 + x – 11 = 0

which is of the form ax2 + bx + c = 0, a≠ 0

Hence (x + 3) (x –2) = 5 is a quadratic equation in x.

(ii) 2x2 + 3x = 2x (x – 7)or 2x2 + 3x = 2x2 – 14xor 17x = 0

This is not of the form ax2 + bx + c = 0 where a≠ 0.

Hence, 2x2 + 3x = 2x(x – 7) is not a quadratic equation.

(iii) 3x2 – 5x = 5(x2 – x + 3)or 3x2 – 5x = 5x2 – 5x + 15or 2x2 + 15 = 0

which is of the form ax2 + bx + c = 0, a≠ 0 as 2x2 + 15 = 0 can be rewritten as2x2 + 0.x + 15 = 0.

Hence 3x2 – 5x = 5(x2 – x + 3) is a quadratic equation in x.

(iv) x x− 1 = 0, x ≠ 0

or xx

2 1− = 0

or x2 – 1 = 0which is of the form ax2 + bx + c = 0, a≠ 0

Hence x x− 1 = 0, x≠ 0 is a quadratic equation.


Determine whether the following equations are quadratic equations :

1. x2 – 3x + 7 = 5 2. x x3 5 0− =d i3. 3 5 9 02x x− + = 4. 2 5 7 02x x+ + =

5. xx

2 1 0+ = 6. 3 13

0xx

− =

7. x2 + 3x + 1= x(x + 5) 8. x(2x + 5) = x2 + 5x + 7


7.5 SOLUTION OF A QUADRATIC EQUATION

If p(x) = ax2 + bx + c, a ≠ 0 and x = α is such that p(α ) = 0, then x = α is called the solutionof quadratic equation ax2 + bx + c = 0, a ≠ 0.

In other words, (x – α ) is a factor of the polynomial ax2 + bx + c if ‘α ’ is a solution of theequation ax2 + bx + c = 0, a≠ 0.

Example 7.2 : Determine whether x = 12 and x = 3

2 are solutions of the equation

2x2 – 5x + 3 = 0.

Solution : Putting x = 12 in 2x2 – 5x + 3 = 0, we get

L.H.S = 2 12 5 1

2 32FH IK − FH IK +

= 12

52

3− +

=1 5 6

2− +

= 1which is not true as L.H.S ≠ R.H.S

Hence x = 12 is not a solution of equation 2x2 – 5x + 3 = 0

Again, putting x = 32 in 2x2 – 5x +3 = 0, we get

L.H.S = 2 32 5 3

2 32FH IK − FH IK +

=92

152

3− +

=9 15 6

2− +

= 0 = R.H.S

Hence, x = 32 is the solution of the equation 2x2 – 5x + 3 = 0.

Note : L.H.S and R.H.S stands for Left Hand Side and Right Hand Side of the given equationrespectively.

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7.5.1 ROOTS OF THE QUADRATIC EQUATION

If α and β are two zeroes of the quadratic polynomial p(x), we say that α and β are theroots of the corresponding quadratic equation p(x) = 0.

For example 5 and –4 are the zeroes of the polynomial x2 – x – 20. Thus, 5 and –4 are theroots of the quadratic equation x2 – x – 20 = 0. x = 5 or x = –4 are also called the solutionsof x2 – x – 20 = 0.

The value (s) of the variable (x in this case) which satisfy an equation is (are) calledthe root(s) or solution(s) of the equation.

We can find the solutions of a quadratic equation by the following methods :

(A) FACTOR METHOD

Let us now study how to solve quadratic equations by the factor method (whenever possible).Here is an example :

Example 7.3 : Solve the following equations :

(i) (x – 5) (x + 3) = 0

(ii) x2 + 3x = 18.

Solution : (i) (x – 5) (x + 3) = 0

Either x – 5 = 0

or x + 3 = 0

∴ x = 5

or x = – 3

Thus the equation (x – 5) (x + 3) = 0 has two roots, x = 5 and x = –3.

(ii) x2 + 3x = 18

or x2 + 3x – 18 = 0

or (x + 6) (x – 3) = 0

Either x + 6 = 0 or x – 3= 0

∴ x = –6

or x = 3

Hence, x = –6 and x = 3 are the solutions of the given quadratic equation.

From the above example, we generalize the following steps to be followed in finding solutionof a quadratic equation by factorisation


(a) Write all the terms on one side by making R.H.S. zero.

(b) Resolve into linear factors of the type (ax + b)(cx +d) of the terms on the L.H.S.

(c) Equate each factor to zero.

(d) Get the required solution.

Remarks : If the product of two polynomials is zero, then either of them is zero or each ofthem is zero.

i.e. (ax + b) (cx + d) = 0

⇒ Either ax + b = 0

or cx + d = 0.

(B) THE QUADRATIC FORMULA

Sometimes, solving a quadratic equation by factor method is difficult or too lengthy. In suchcases, we solve the equation by completing the squares.

Let us solve some quadratic equations by completing the squares.

Example 7.4 : Solve the following equations by the method of completing the squares

(i) x2 – 4x – 5 = 0

(ii) 3x2 – 4x – 7 = 0

Solution. (i) x2 – 4x – 5 = 0

The first two terms on the L.H.S. are the first two terms in the expansion of (x – 2)2

∴ x2 – 4x – 5 = 0 can be rewritten as

x2 – 4x + 4 – 4 – 5 = 0 (Adding and subtracting 4)

or (x – 2)2 – 9 = 0

or (x – 2)2 – (3)2 = 0

(x – 2 – 3) (x – 2 + 3) = 0

(By factorising the polynomial on the L.H.S.)

or (x – 5) (x + 1) = 0

∴ Either x – 5 = 0

or x + 1 = 0

⇒ x = 5

or x = –1

Hence x = 5 and x = –1 are the two solutions of the given quadratic equation.

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(ii) 3x2 – 4x – 7 = 0

Multiplying the equation by 3 (The co-efficient of x2)

We get 9x2 – 12x – 21 = 0

On completing the square (by adding and subtracting 4 in the L.H.S.), we get

9x2 – 12x + 4 – 4 – 21 = 0

or 9x2 – 12x + 4 – 25 = 0

or (3x – 2)2 – (5)2 = 0

or (3x – 2 – 5) (3x – 2 + 5) = 0

or (3x – 7) (3x + 3) = 0

Either 3x – 7 = 0 or 3x + 3 = 0

or x = 73

or x = –1.

Hence the solutions of the given equation are x = 73 and x = –1.

Now let us solve the standard quadratic equation ax2 + bx + c = 0, a≠ 0 by the methodof completing square(s) and find its roots.

We have ax2 + bx + c = 0 , a ≠ 0

Multiplying both sides by 4a, we have

4a(ax2 + bx + c) = 4a × 0

or 4a2x2 + 4abx + 4ac = 0

or (4a2x2 + 4abx + 4ac) + b2 = 0 + b2 [Adding b2 to both sides]

or (4a2x2 + 4abx + b2) = b2 – 4ac

or (2ax + b)2 = b2 – 4ac

or (2ax + b)2 = ± −b ac22

4e j{ }

or 2ax + b = ± −b ac2 4

or 2ax = − ± −b b ac2 4

or x = − ± −b b aca2 4

2

This is known as quadratic formula and is the most convenient tool to solve a quadraticequation.


It is clear from above that a quadratic equation of the form ax2 + bx + c = 0, a ≠ 0 can havetwo roots which are

− + − − − −b b aca

b b aca

2 242

42,

Here b2 – 4ac is a factor on which the nature of roots depend. We denote (b2 – 4ac) by D andit is called DISCRIMINANT.

For a quadratic equation ax2 + bx + c = 0, a ≠ 0, if

(i) D > 0, the equation will have two real and unequal roots

(ii) If D = 0, the equation will have two equal roots both equal to −ba2 .

(iii) D < 0, the equation will not have any real roots.

Thus, a quadratic equation can have maximum two roots.

Example 7.5 : Find the roots of quadratic equation 6x2 + x – 15 = 0 using quadratic formula.

Solution : We have 6x2 + x – 15 = 0

Comparing with ax2 + bx + c = 0 we get

a = 6, b = 1 and c = –15.

Using the quadratic formula, the roots are

− ± −b b aca

2 42

=− ± − −1 1 4 6 15

2 6b gb gb g

= − ± = − ±1 36112

1 1912

= 32

53,−

Thus, the two roots of the equation are 32

53,− .

Example 7.6 : Solve the quadratic equation 4x2 + 4x + 1 = 0 using quadratic formula.

Solution : Here a = 4, b =4 and c = 1

Using x = − ± −b b aca

2 42

, we get

170 Mathematics

x =− ± −4 16 4 4 1

8b gb g

= − ± = −4 08

12

∴ x = – 12 is the solution of the quadratic equation.

7.6 TO FORM A QUADRATIC EQUATION WHEN ITS ROOTS ARE GIVEN

Let a and b be the roots of a quadratic equation in x, then (x – a) and (x – b) are the factorsof the quadratic equation

∴ We have(x – a) (x – b)= 0

or x2 – (a + b)x + ab = 0

or x2 – (sum of the roots)x + (product of roots) = 0

Example 7.7 : Form a quadratic equation whose roots are –3 and 5.

Solution : Here sum of roots = (–3 + 5) = 2

and product of roots = (–3).5 = –15

∴ The quadratic equation is

x2 – (2)x + (–15) = 0

i.e. x2 – 2x – 15 = 0

Example 7.8 : Form a quadratic equation with two roots as 2 3+d i and 2 3−d i .

Solution : Here, sum of roots is 2 3 2 3+ + −d i d i = 4

and product of roots = 2 3 2 3+ −d id i = 4 – 3 = 1.

∴ The equation is x2 – (4)x + 1 = 0

or x2 – 4x + 1 = 0.Example 7.9 : If 3 is one root of the equation x2 – ax + 6 = 0, find the value of ‘a’ and alsofind the other root of the equation.

Solution : Since 3 is a root of the equation x2 – ax + 6 = 0

∴ (3)2 – a(3) + 6 = 0⇒ a = 5

Thus, the equation becomesx2 – 5x + 6 = 0


Comparing with x2 – (sum of the roots)x + Product of roots = 0

We have sum of roots = 5

Since one root is 3, other root is 5 – 3 = 2.


1. Determine whether the value x = 3 and x = 12 are solutions of the quadratic equation

2x2 + 5x – 3 = 0.

2. Show that x = 2 is one of the roots of the equation 3x2 – 7x + 2 = 0.

3. Using factor method, solve the following equations :

(i) x2 – 5x + 6 = 0 (ii) 6x2 + 7x – 5 = 0

(iii) 4x2 + 8x – 5 = 0 (iv) 2x2 + 5x – 3 = 0

4. Solve the following equations by method of completing the squares :

(i) x2 – 4x – 5 = 0 (ii) 4x2 – 12x – 16 = 0

(iii) 4x2 – 8x + 3 = 0 (iv) 2x2 – x – 6 = 0.

5. Solve for x :

(i) 6x2 – 19x + 15 = 0 (ii) x2 + 3x – 5 = 0

(iii) 3x2 – 10x + 3 = 0 (iv) 3x2 + x – 2 = 0

(v) 21 + x = 2x2 (vi) 9x2 – 16 = 0.

6. Form the quadratic equation whose roots are given below :

(i) –3 and –5 (ii) –3 and 7

(iii) 3 2+ and 3 2− (iv) 3 and 13

(v) 6 and 5 (vi) 3 and –3

7. One root of the quadratic equation 3x2 – 10x + 3 = 0 is 13 . Find the other root.

8. If 13 is one root of the equation 3x2 + kx –3 = 0, find the value of k. Hence find the

other root.

7.7 WORD PROBLEMS

In the following discussion, you will learn to apply the methods you have studied in this lessonto solve some problems of daily life.

172 Mathematics

The steps in solving them are the same as the ones you have learnt earlier. They are :

(1) Translating the given problem into algebraic form.

(2) Solving the equation obtained.

(3) Interpreting (verifying) the solution obtained.

Example 7.10 : Find two successive natural numbers whose squares have the sum 221.

Solution : Let the two successive natural numbers be x and x + 1.

As per problem, x2 + (x + 1)2 = 221

or x2 + x2 + 2x + 1 – 221 = 0

or 2x2 + 2x – 220 = 0

or x2 + x – 110 = 0

or x2 + 11x – 10x – 110 = 0

or x(x + 11) – 10 (x + 11) = 0

or (x – 10) (x + 11) = 0

⇒ Either x + 11 = 0

or x – 10 = 0

∴ x = –11

or x = 10

Rejecting x = – 11, which is not a natural number.

we get x = 10

∴ Ist natural number = 10

2nd natural number = 11.

Example 7.11 : The sum of two numbers is 15. If the sum of their reciprocal is 3/10, Findthe two numbers.

Solution : Let one number = x

Other number = 15 – x

As per problem,

1 115x x

+− =

310

or15

15− +−x x

x xb g =3

10

or15

15x x−b g =3

10


or 150 = 3x (15 – x)

or 150 = 45x – 3x2

or 50 = 15x – x2

or x2 – 15x + 50 = 0

or x2 – 10x – 5x + 50 = 0

or x(x – 10) – 5(x – 10) = 0

or (x – 10) (x – 5) = 0

⇒ Either x – 10 = 0 or x – 5 = 0

∴ x = 10

or x = 5

When x = 5

Other number = 15 – 5 = 10

Or

When x = 10

Other number = 15 – 10 = 5

∴ The required numbers are 5, 10

Example 7.12 : The sum of a number and its reciprocal is 174 . Find the number.

Solution : Let the number = x

∴ Reciprocal of x = 1x

As per problem , x x+ 1 = 174

or xx

2 1+ = 174

or 4x2 + 4 = 17x

or 4x2 – 17x + 4 = 0

or 4x2 – 16x – x + 4 = 0

or 4x(x – 4) – 1(x – 4) = 0

or (4x – 1) (x – 4) = 0.

∴ Either 4x – 1 = 0

or x – 4 = 0

174 Mathematics

∴ x = 14

or x = 4

∴ The required number is 4 or 14

Example 7.13 : A two-digit number is such that the product of the digits is 12. When 36 isadded to this number the digits interchange their places. Determine the number.

Solution : Let digit at ten’s place be x

and digit at unit’s place be y

∴ The number = 10x + y

When digits are interchanged, the new number = 10y + x

As per problem,

product of digits = 12 i.e. xy = 12 ...(1)

Also, 10x + y + 36 = 10y + x

or 10x – x = 10y – y – 36

or 9x = 9y – 36

or x =9 4

9y −b g = y – 4 ...(2)

From (1) and (2)

xy = 12

or (y – 4)y = 12

or y2 – 4y – 12 = 0

or y2 – 6y + 2y – 12 = 0

or y(y – 6) + 2(y – 6) = 0

or (y + 2) (y – 6) = 0

Either y + 2 = 0 or y – 6 = 0

⇒ y = –2 or y = 6

Rejecting y = –2, we get y = 6

When y = 6, x = 2


∴ The required number = 10 × 2 + 6

= 26

Therefore, the required number is 26

Example 7.14 : Find two consecutive even positive integers whose squares have the sum 340.

Solution : Let the two consecutive even positive integers be x and x + 2.

As per problem,

x2 + (x + 2)2 = 340 [As sum of their squares is 340]

or 2x2 + 4x – 336 = 0

or x2 + 2x – 168 = 0

or x2 + 14x – 12x – 168 = 0

or x(x + 14) – 12(x + 14) = 0

or (x – 12)(x + 14) = 0.

or Either x + 14 = 0 or x – 12 = 0

⇒ x = –14 or x = 12

Since – 14 is not a positive integer, x = –14 is not possible.

∴ x = 12, x + 2 = 12 + 2 = 14

Hence, the two consecutive even positive integers are 12, 14.

Example 7.15 : The sides (in cm) of a right angled triangle are x – 1, x andx + 1. Find x and hence the sides of the triangle.

Solution : The sides (in cm) of a right angled triangle are x – 1, x and x + 1

Q x + 1 > x – 1 and x + 1 > x

∴ x + 1 is the hypotenuse (largest side)

By Pythagoras Theorem,

x2 + (x – 1)2 = (x + 1)2

or x2 + x2 – 2x + 1 = x2 + 2x + 1

or 2x2 – 2x + 1 – x2 – 2x – 1 = 0

or x2 – 4x = 0

or x(x – 4) = 0

∴ x – 4 = 0 Since x≠ 0

or x = 4

176 Mathematics

⇒ x – 1 = 4 – 1 = 3

⇒ x + 1 = 4 + 1 = 5

∴ The sides of the triangle are 3 cm, 4 cm and 5 cm.

Example 7.16 : The sum of ages of Ashu and her mother is 45. The product of their agesis 126. Find their ages.

Solution : Let Ashu’s age = x years

∴ Her mother’s age = (45 – x) years

The product of their ages = x(45 –x)

= 45x –x2

It is given that the product of their ages is 126.

∴ 45x – x2 = 126

or x2 – 45x + 126 = 0

or x2 – 42x – 3x + 126 = 0

or x(x – 42) – 3(x – 42) = 0

or (x – 42) (x – 3) = 0

∴ Either x – 42 = 0 or x – 3 = 0

⇒ x = 42

or x = 3

We reject x = 42 because Ashu’s age cannot be greater than her mother’s age.

∴ x = 3

∴ Ashu’s age is 3 years and her mother’s age = 45 – 3 = 42 years.

Example 7.17 : By increasing his average speed by 10 km/hr a motorist saves 36 minutesin travelling a distance of 120 km. Find out his actual speed.

Solution. Let the actual speed of the motorist be x km/hr.

He increases his vehicle's speed by 10 km/hr.

∴ His new speed = (x + 10) km/hr

speed = DistanceTime

∴ In the first case, time = 120xFH IK hours


In the second case, time = 120x +10FH IK hours

By the given condition we get,

120x

120x +10− =

35 (36 min. = 3

5 hr)

or120 1200 120

10x x

x x+ −

+b g = 35

or1200

10x x +b g = 35

or 6000 = 3x2 + 30xor x2 + 10x – 2000 = 0or x2 + 50x – 40x – 2000 = 0or x(x + 50) – 40(x + 50) = 0

or (x + 50) (x – 40) = 0

∴ Either x + 50 = 0 or x – 40 = 0

⇒ x = – 50 or x = 40Aliter

∴ x = − ±10 81002 = − ±10 90

2

∴ x = 802

or x = −1002

or x = 40 or x = –50Rejecting x = –50 as speed cannot be negative.

∴ x = 40

∴ The actual speed of the motorist is 40 km/hr.


1. Find two consecutive natural numbers whose product is 240.

2. Two successive natural numbers are such that the sum of their squares is 145. Find thenumbers.

178 Mathematics

3. The sum of a number and its reciprocal is 507 , Find the number .

4. A two digit number is such that the product of the digits is 12. When 9 is added to thenumber, the digits interchange their places. Determine the number.

5. Find two consecutive odd positive integers whose squares have the sum 290.

6. If the length of a rectangle exceeds the breadth by 4 cm and its area is 525 cm2, findits dimensions.

7. If one side of a right triangle exceeds the other by 7 cm and the hypotenuse is 13 cm.Find the sides.

8. The age of father is 10 times the age of his son. If the product of their ages is 160, findthe age of father.

LET US SUM UP

An equation of the form ax2 + bx + c = 0, a ≠ 0 and a, b and c are real numbers, is calleda quadratic equation in x.

Roots of the quadratic equation ax2 + bx + c = 0, a ≠ 0 are given by

− ± −b b aca2 4

2

b2 – 4ac is called discriminant of the quadratic equation. It is usually denoted by D.

(i) If D > 0 then the quadratic equation have two real and unequal roots.

(ii) If D = 0 then the quadratic equation have two equal roots.

(iii) If D < 0 then the quadratic equation have no real roots.

A quadratic equation whose roots are α, β is given by

x2 – (α + β)x + αβ = 0

The value (s) of the variable which satisfy an equation is (are) called a root(s) or solution(s)of the equation.

TERMINAL EXERCISE

Solve the following equations by factorisation method :

1. (x – 8) (x+ 4) = 13

2. 2x2 – 5x = 0

3. x2 – 8x + 15 = 0

4. 9x2 + 15 x – 14 = 0

5. 6x2 + x – 15 = 0


6. x x2 1 2 2 0− + + =d i7. ax2 + (a + b)x + b = 0

8. x2 – 11ax + 28a2 = 0

9. 3a2x2 + 2abx – b2 = 0

10. 2x2 – 3x + 1 = 0

11. Form a quadratic equation whose root are :

(a) –1 and 3 (b) –2 and 3 (c) –3 and 4.

12. Form a quadratic equation whose one root is 1 2+ and the sum of its roots is 2.

13. The sum of the ages of a father and his son is 60 years and the product of their ages is576. Find their ages.

14. The sum of a number and its reciprocal is 2512 . Find the numbers.

15. Solve for x : (a) (x + 3)2 + 5(x + 3) + 4 = 0.

(b) (x + 1)2 – 8 (x + 1) + 15 = 0.

180 Mathematics

ANSWERS


1. Yes 2. Yes 3. No 4. Yes5. No 6. Yes 7. No 8. Yes


1. x = 12

3. (i) x = 2, 3 (ii) x = 12

53,− , (iii) − 5

212, , (iv)

12

3,−

4. (i) x = 5, –1 (ii) x = 4, – 1 (iii) 32

12, (iv) x = 2 3

2,−

5. (i) x = 32

53, (ii) x = − ±3 29

2 (iii) a = 3 13, (iv) x = −1 2

3,

(v) x = − 3 72, (vi) 4

343,− .

6. (i) x2 + 8x + 15 (ii) x2 – 4x – 21 = 0 (iii) x2 – 6x + 7 = 0(iv) 3x2 – 10x + 3 = 0 (v) x2 – 11x + 30 = 0 (vi) x2 – 9 = 0.

7. 3 8. k = 8, the other root is –3.

Check Your progress 7.3

1. 15, 16 2. 8, 9 3. 7 or 17 4. 34

5. 11, 13 6. 21 × 25 cm 7. 5 cm, 12 cm 8. 40 years.

Terminal Exercise

1. x = 9, x = –5 2. x = 0, x = 52 3. x = 5; x = 3

4. x = 23 , x = − 7

3 5. x = 32 ; x = − 5

3 6. x = 1; x = 2

7. x = –1, x = − ba 8. x = 4a, x = 7a 9. x = − b

a3 , x = − ba

10. x = 12 , x = 1

11. (a) x2 – 2x – 3 = 0 (b) x2 – x – 6 = 0 (c) x2 – x – 12 = 012. x2 – 2x – 1 = 013. Age of Father = 48 years, Age of Son = 12 years

14.43 or

34

15. (a) x = –4 or x = –7 (b) x = 2 or x = 4.

Number Patterns 181

8

Number Patterns

8.1 INTRODUCTION

In our day-to-day life, we see patterns of geometric figures on clothes, pictures, posters etc.They make the learners motivated to form such new patterns. This becomes a topic of interestand knowledge to predict the next figure in a pattern.

Consider the following patterns :

(i)

(ii)

(iii)

(iv)

Fig. 8.1

Can you predict the next figures in (i), (ii), (iii) and (iv) ? A little careful study of the abovepatterns shows that the next figures in (i), (ii), (iii) and (iv) are

, , and respectively.

Think about these and try to find the reasons for those.

182 Mathematics

Likewise number patterns are also faced by learners in their study. Number patterns play animportant role in the field of Mathematics. Let us study the following number patterns :

(i) 2, 4, 6, 8, 10, ...

(ii) 1, 1 12 , 2, 2 1

2 , 3, ...

(iii) 10, 7, 4, 1, –2, ...

(iv) 2, 4, 8, 16, 32, ...

(v) 4, 12 , 1

16 , 1128 , ...

(vi) 1, 12 , 1

3 , 14 , ...

(vii) 1, 11, 111, 1111, 11111, ...

It is an interesting study to find whether some specific names have been given to some of theabove number patterns and the methods of finding some next terms of the given patterns.

In this lesson, you will study about number patterns called Arithmetic Progressions andGeometric Progressions. You will also study methods of finding general term and sum to ‘n’terms of an Arithmetic Progression.

8.2 OBJECTIVES


recognise number patterns and identify those which are Arithmetic or Geometricprogressions.

determine nth term of an Arithmetic progression (A.P)

find the sum to ‘n’ terms of an A.P.


Idea of numbers

Idea of the number line

Knowledge of various number systems

Four fundamental operations on numbers

8.4 RECOGNITION OF NUMBER PATTERNS

Suppose you want to purchase a handkerchief whose cost is Rs 5. If you want to purchasetwo handkerchiefs, then you have to pay Rs 10.

Number Patterns 183

Therefore, if the number of handkerchiefs is increased by one successively, the respective cost(in Rs) would increase by 5 every time, i.e., the respective costs of one, two, three, ...handkerchiefs would be 5, 10, 15, ...

Can you recognize the relationship between any consecutive numbers of the above pattern ?

If you observe the pattern carefully, you will find that each successive number, other than thefirst number, is obtained by adding a constant number to the preceding number.

Therefore, these numbers form a number pattern.

Each number of the number pattern is called a term.

Some examples of numbers patterns are given below :

(a) 1, 2, 3, 4, ...

(b) 4, 2, 0, –2, –4, ...

(c) 3, 6, 9, 12, ...

(d) 13, 9, 5, 1, –3, –7, ...

(e) 2 14 , 6 1

4 , 10 14 , 14 1

4 , ...

(f) 1, 2, 4, 8, 16, ...

(g) 16, 4, 1, 14 , ...

(h) 5, 25, 125, 625, ...

In number patterns (a), (c) and (e), each successive term, other than the first term, can beobtained by adding a constant number 1, 3 and 4 respectively to the preceding term.

In the number patterns (b) and (d), each successive term, other than the first term, can beobtained by subtracting a constant number 2 and 4 respectively from the preceding term.

But in the number patterns (f), (g) and (h), each term, other than the first term, can be obtained

by multiplying the preceding term by a constant number 2, 14 and 5 respectively.

8.5 ARITHMETIC PROGRESSION

You are all aware that salaries of employees are often calculated on the basis of their basicsalary, plus fixed increments (increases) for each year of service and all other usual allowances.

Suppose a person begins to work for a firm in the scale of Rs 4500-125-7000.

184 Mathematics

Then in successive years his basic salary (in Rs) will be 4500, 4625, 4750, 4875, 5000 andhe will get usual allowances on the basic salary.

The first term of this pattern is 4500 and each successive term, other than the first term, canbe obtained by adding a constant number (here increament in Rs) 125 to the preceding number.We can say that the terms of the pattern progress arithmetically. Such a number pattern is calledan Arithmetic Progression (abbreviated as A.P). The constant number is called commondifference.

A progression is said to be an Arithmetic Progression (abbreviated as A.P), if thedifference of each term, except the first, from its preceding term is always the same.

Probably the simplest arithmetic progression is of natural numbers :

1, 2, 3, 4, 5, 6, ...

The multiplication tables are all familiar A.P's,

2, 4, 6, 8, 10, ...

3, 6, 9, 12, 15, ...

4, 8, 12, 16, 20, ...

An arithmetic progression can start from any number, positive or negative for example :

5, 8, 11, 14, 17, ...

2 14 , 6 1

4 , 10 14 , 14 1

4 , ...

–7, –2, 3, 8, ...

Rational numbers can serve as the common difference, as in the A.P :

9, 11 12 , 14, 16 1

2 , 19, ...

and negative numbers, as in the A.P

10, 7, 4, 1, –2, –5, ...

An arithmetic progression can be represented on the number line by a series of points placedthe same distance apart. For example, the progression 13, 17, 21, 25, ... is shown in Fig. 8.2.

Fig. 8.2

Number Patterns 185

The successive terms of an A.P are usually denoted by t1, t2, t3, ... The commondifference is usually denoted by ‘d’.

Observe the following A.P :

3, 5, 7, 9, 11, ...

It is clear from the above A.P that its first term is 3 and the common difference is 2 (usuallydenoted by ‘d’)

You can rewrite the above A.P in the following manner :

In general, if the first term is denoted by ‘a’ and the common difference by ‘d’,the standard form of an Arithmetic Progression would be

a , a + d , a + 2d , a + 3d , ...

It should be noted that, the common difference may be negative in which case the terms ofthe progression would decrease.

For example : 15, 13 12 , 12, 10 1

2 , ...

The first term of this A.P is 15.

The common difference = 13 12 – 15 = 12 – 13 1

2 = 10 12 – 12 = –1 1

2

Let us take some examples to illustrate :

Example 8.1 : Which of the following progressions are A.P. ?

(a) 3, 6, 9, 12, ... (b) 9, 14, 19, 24, ...

(c) 13, 15, 18, 22, ... (d) 21, 17, 13, 9, ...

Solution : (a) The given progression is 3, 6, 9, 12, ...

186 Mathematics

Here the first term is 3 and the common difference

= 6 – 3 = 9 – 6 = 12 – 9 = 3

∴ The given progression is an A.P.

(b) The given progression is 9, 14, 19, 24, ...

Here the first term is 9 and the common difference is

= 14 – 9 = 19 – 14 = 24 – 19 = 5


(c) The given progression is 13, 15, 18, 22, ...

Here the first term is 13.

But its common difference is not the same as

15 – 13 ≠ 18 – 15 ≠ 22 – 18

∴ The given progression is not an A.P.

(d) The given progression is 21, 17, 13, 9, ...

Here the first term is 21 and the common difference is

17 – 21 = 13 – 17 = 9 – 13 = – 4


CHECK YOUR PROGRESSION 8.1

1. Which of the following progression are A.P’s ?

(a) 6, 12, 18, 24, ... (b) 7, 10, 12, 14, ...

(c) –1, –3, –5, –7, ... (d) 31, 29, 27, 25, ...

8.6 THE GENERAL TERM (OR nTH TERM) OF AN ARITHMETIC PROGRESSION

If we need to find a particular term of an A.P, for example 12th term whose first term is 3and common difference is 2, we can, of course, build up the A.P. as far as the required termby adding 2 successively to the first term, as

3, 5, 7, 9, ...

But this method would become rather laborious if you had to find, say 102nd term of theprogression. We can find an easier method if we look at the progression more closely. Thefirst term (a) for an A.P is 3 and the common difference (d) = 2. The successive terms canbe formed by adding 2 to the first term.

Number Patterns 187

↓ ↓1 st term = 3 = 3 + 0.2 = 3 + ( 1 – 1).2

↓ ↓2 nd term = 3 + 2 = 3 + 1.2 = 3 + ( 2 – 1).2

↓ ↓3 rd term = (3 + 2) + 2 = 3 + 2.2 = 3 + ( 3 – 1).2

↓ ↓4 th term = (3 + 2 + 2) + 2 = 3 + 3.2 = 3 + ( 4 – 1).2

... ... ... ...

Without completing all the step-by-step calculations, we can find straight away that the 12th

term by adding 2, eleven times to the first term.

↓ ↓i.e., 12 th term would be 3 + 11.2 = 3 + ( 12 – 1).2

This suggests us a formula for finding any term of the progression.

? th term = 3 + ( ? – 1).2

In particular writing number 12 in each box we can find the 12th term :i.e., 12th term = 3 + (12 – 1).2

12th term = 2 + 11 × 2We generally denote 12th term by t12 and write the above result as

t12 = 3 + (12 – 1) × 2Thus the general term i.e. nth term can be written as :

nth term = a + (n – 1)d or tn = a + (n – 1)d

where ‘a’ and ‘d’ denote the first term and the common difference respectively.We will get the same formula for the nth term if we use the standard form of A.P.We know that the standard form of an A.P. is

a, a + d, a + 2d, a + 3d, ...In the similar manner, we can write

1st term = t 1 = a = a + 0.d = a + ( 1 – 1)d

2nd term = t 2 = a + d = a + ( 2 – 1)d

3rd term = t 3 = a + 2d = a + ( 3 – 1)d

... ... ... ...

→ →

→→

→→

188 Mathematics

If you see the above pattern, the formula for the nth term or tn would be

a + (n – 1)d

Hence, we can rewrite the standard form of an A.P as

a, a + d, a + 2d, ....., tn – 1, a + (n – 1)d

where tn–1 denotes the (n – 1)th term of the A.P.

Example 8.2 : Identify the first term and the common difference of each of the followingA.P’s :

(i) 6, 10, 14, 18, ...

(ii) 12, 9, 6, ...

Solution : (i) The given A.P is 6, 10, 14, 18, ....

Here, the first term (a) is 6 and the common difference (d) = 10 – 6 = 4

(ii) The given A.P. is 12, 9, 6, ...

Here, the first term (a) is 12 and the common difference (d) = 9 – 12 = –3.

Example 8.3 : Write the expressions for nth terms of the following A.P's and find the 100th

term of each of the following :

(i) 5, 8, 11, 14, ...

(ii) –7, –11, –15, – 19, ...

Solution : (i) The given A.P is

5, 8, 11, 14, ...

We know that the nth term of an A.P, whose first term is ‘a’ and common difference is ‘d’,is given by

tn = a + (n – 1)dHere, a = 5 and d = 8 – 5 = 3

∴ tn = 5 + (n – 1).3= 5 + 3n – 3= 3n + 2

∴ 100th term = t100 = 3 × 100 + 2 = 302.(ii) The given A.P is –7, –11, –15, –19, ...

nth term of an A.P is given by

tn =a + (n – 1)dHere, a = –7 and d = –11 – (–7) = –11 + 7 = –4∴ tn = –7 + (n – 1).(–4)

Number Patterns 189

=–7 – 4n + 4=–3 – 4n

∴ 100th term = t100 = –3 – 4 × 100 = – 403.Example 8.4 : If the nth term of a progression is given by

tn = 2 + 3n,show that it is an A.P.Solution : Given that

nth term tn = 2 + 3nPutting n = 1, 2, 3, 4, ..., we get

∴ 1st term, t1 = 2 + 3 × 1 = 52nd term, t2 = 2 + 3 × 2 = 83rd term, t3 = 2 + 3 × 3 = 114th term, t4 = 2 + 3 × 4 = 14

... ... ... ...

∴ The progression is 5, 8, 11, 14, ...

Here, t1 = 5, t2 = 8, t3 = 11, t4 = 14, ...t2 – t1 = 8 – 5 = 3t3 – t2 = 11 – 8 = 3t4 – t3 = 14 – 11 = 3

Since the common difference is always the same, the above progression is an ArithmeticProgression.

Example 8.5 : Find the 21st term of an A.P whose first term is –3 and common differenceis 5.

Solution : Here, the first term, a = –3 and

common difference d = 5

∴ nth term = tn= a + (n – 1)d= –3 + (n – 1) × 5= –3 + 5n – 5= –8 + 5n

∴ 21st term = t21 = –8 + 5 × 21= –8 + 105= 97

∴ The 21st term of the given progression is 97.

190 Mathematics

Example 8.6 : The 6th term of an A.P is zero and the 2nd term is 4. Find the first term andthe common difference.

Solution : Let the first term and common difference of the A.P be ‘a’ and ‘d’ respectively.

It’s nth term is given by

tn = a + (n – 1)d

∴ 2nd term = t2 = a + (2 – 1)d = a + d

6th term = t6 = a + (6 – 1)d = a + 5d

By the given condition

a + 5d = 0 ...(i)

and a + d = 4 ...(ii)

Subtracting (ii) from (i), we have

4d = –4 or d = –1

Putting d = –1 in (i), we get

a + 5(–1) = 0

or a = 5

∴ The first term of the A.P. is 5 and the common difference is –1.

Example 8.7 : Which term of the A.P

14

12

34 1 5

4, , , , , ..... is 6 14 ?

Solution : The first term (a) = 14

Common difference (d) = 12

14

14− =

Let the nth term be 6 14

or 14 1 1

4+ −nb g = 254

or 1 + (n – 1) = 25

or n = 25

Hence, 25th term of the above A.P is 6 14 .

Number Patterns 191

Example 8.8 : The ages of three brothers are in A.P. The age of the youngest brother is one-third the age of the eldest. Find the ages of all, if the sum of their ages is 90 years.

Solution : Let the ages (in years) of three brothers be a – d , a , a + d ...(i)

By the given condition a – d = 13 (a + d)

or 3a – 3d = a + d

or 2a = 4d

or a = 2d ...(ii)

and (a – d) + a + (a + d) = 90

or 3a = 90

or a = 30

Putting a = 30 in (ii), we get

30 = 2d

or d = 15

Putting the values of a and d in (i), we get

a – d = 15

a = 30

and a + d = 45

∴ The required age of the youngest brother is 15 years, that of 2nd brother is 30 years andof the eldest brother is 45 years.

Example 8.9 : The fourth term of an A.P is equal to three times its first term and 7th termexceeds twice the third term by 1. Find the first term and common difference.

Solution : Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.

t4 = a + 3d

t7 = a + 6d

and t3 = a + 2d


a + 3d = 3a

or 2a – 3d = 0 ...(i)

and a + 6d = 2(a + 2d) + 1

or a – 2d + 1 = 0

or a = 2d – 1 ...(ii)

192 Mathematics

Putting the value of a in (i) we get,

2(2d – 1) – 3d = 0

or 4d – 2 – 3d = 0

or d = 2

Substituting d = 2 in (i) ,we get

2a = 6

or a = 3

∴ The required first term and common difference of the A.P. are 3 and 2 respectively.

Example 8.10 : Find three numbers in A.P whose sum is 21 and sum of their squares is 389.

Solution : Let the three terms of the A.P be a – d, a, a + d


(a – d) + a + (a + d) = 21

or 3a = 21

or a = 7 ...(i)

and (a – d)2 + a2 + (a + d)2 = 389

or a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 389

or 3a2 + 2d2 = 389

or 3 × 72 + 2d2 = 389 [using (i)]

or 147 + 2d2 = 389

or 2d2 = 242

or d2 = 121

or d = ± 11

When a = 7, d = +11, the three numbers are –4, 7, 18

When a = 7, d = –11, the three numbers are 18, 7, –4.

Example 8.11 : If m times the mth term of an A.P is equal to n times its nth term, prove thatits (m + n)th term is zero.

Solution : Let the first term and common difference of the A.P be ‘a’ and ‘d’ respectively

∴ tm = a + (m – 1)d

tn = a + (n – 1)d


m × tm = n × tn

Number Patterns 193

or m{a + (m – 1)d} = n{a + (n – 1)d}

or m{a + (m – 1)d} – n{a + (n – 1)d} = 0

or (m – n)a + {m (m – 1) – n (n – 1)}d = 0

or (m – n)a + {(m2 – n2) – (m – n)}d = 0

or (m – n)a + {(m + n) (m – n) – (m – n)}d = 0

Dividing throughout by m – n, we get

or a + (m + n – 1)d = 0

We know that tm+n = a + (m + n – 1)d

∴ tm+n = 0

Hence the result.


1. Find the next two terms of the following A.P.’s

(a) 11, 19, 27, ... (b) 54 , 1, 3

4 , ...

2. Write the formula for nth term, of the following A.P.’s :

(a) 6, 11, 16, 21, ... (b) 12

12

32

52, , ,− − − , ...

3. The nth term of a progression is given by the relation :

(a) Tn = 10 – n (b) Tn = 7n + 2

(c) Tn = 1 2

3− n

Show that each of them is an A.P.

4. The 5th term of an A.P is 23 and 12th term is 37. Find the first term and common differenceof the A.P.

5. Which term of the A.P :

(a) 100, 95, 90, 85, ... is –20 ?

(b) 4, 10, 16, 22, ... is 334 ?

6. The angles of a triangle are in A.P. If the least angle is one-third the largest angle, findthe angles.

7. If the mth, nth and rth terms of an A.P. are x, y and z respectively.

Prove that x(n – r) + y(r – m) + z (m – n) = 0.

194 Mathematics

8.7 THE SUM OF THE FIRST n TERMS OF AN A.P

We know that the standard form of an A.P is given by :

a , a + d , a + 2d , ... , a + (n – 2)d , a + (n – 1)d

tn–1 tnwhere ‘a’ and ‘d’ are the first term and the common difference respectively, and tn–1 and tndenote the (n – 1)th and nth terms of the A.P.

Let Sn denote the sum of first n terms of the A.P.

∴ We can write

Sn = a + ( a + d) + (a + 2d) + ... + {a + (n – 2)d}+ {a + (n – 1)d} ...(1)

Writing the terms in reverse order, we get

Sn = {a+ (n – 1)d} + {a+ (n – 2)d} + ... + (a + 2d) + (a + d) + a ...(2)

Adding corresponding terms of (1) and (2), we get

2Sn = {2a + (n – 1)d} + {2a + (n – 1)d} + {2a + (n – 1)d} + ... + {2a + (n – 1)d}

= {2a + (n – 1)d} {1 + 1 + 1 + ... + 1}

= n {2a + (n – 1)d}

2Sn = n{2a + (n – 1)d}

Sn = n2 {2a + (n – 1)d} ...(3)

which denotes the general formula for finding the sum of the first n terms of the A.P.

(3) can also be rewritten as Sn = n a a n d2

1+ + −b gm r

= n a tn2 +b g [Q nth term = tn = a + (n – 1)d]

Sometimes nth term is named as the last term and is denoted by the letter ‘l’.

∴ Sn = n a2 + lb g ...(4)

Example 8.12 : Find the sum of the first 10 terms of the following A.P’s :

(a) 1, 6, 11, 16, ...

(b) –193, –189, –185, ...

Number Patterns 195

Solution : (a) The given A.P. is :

1, 6, 11, 16, ...

Here a = 1, d = 6 – 1 = 5

Number of terms = n = 10

We know that the sum of the first n terms of an A.P whose first term is ‘a’ and commondifference is d , is given by

Sn = n2 {2a + (n – 1)d}

∴ S10 = 102 2 1 10 1 5× + −b gm r

= 5(2 + 45)

= 235

∴ The required sum is 235

(b) The given A.P. is :

–193, –189, –185, ...

Here a = –193, d = –189 – (–193) = 4

and n = 10.

We know that,

Sn = n2 {2a + (n – 1)d}

= 102 2 193 10 1 4− + −b g b gm r= 5(–386 + 36)

= 5 × (–350)

= – 1750

The required sum is –1750.

Example 8.13 : Find the sum of each of the following A.P.’s :

(i) 2 + 4 + 6 + 8 + ... + 146

(ii) 5 14 4 1

2 3 34+ + + ... + (– 3)

Solution : (i) The given A.P. is :

2 + 4 + 6 + 8 + ... + 146

Here a = 2, d = (4 – 2) = 2 and the nth term = tn = 146

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We know that

tn = a + (n – 1)d = 146

or 2 + (n – 1).2 = 146

or 2n = 146

or n = 73

∴ Number of terms in the above A.P. is 73

The required sum of the first 73 terms is given by

S73 = 732 2 73 1 2+ −b gm r.

= 732 2 72 2+ ×l q

= 732 146×

= 5329(ii) The given A.P. is

5 14 4 1

2 3 34+ + + ... + (–3)

Here a = 5 14 = 21

4 , d = 4 12 5 1

4− = − 34

Let us find out which term of this A.P. is –3Here nth term is –3.We know that

tn = a + (n –1)d = –3

or –3 = 214 1 3

4+ − −FH IKnb g

or21 3 1

4− −nb g = –3

or 21 – 3n + 3 = –12or 24 + 12 = 3nor n = 12

The required sum of the first 12 terms is given by

S12 = 122

214 3−FH IK u g S n a tn nsin , = +LNM OQP2 l q

= 6 94

272× = .

Number Patterns 197

Example 8.14 : Find the sum of all multiples of 5 upto 1017.

Solution : The multiples of 5 upto 1017 are :

5, 10, 15, 20, ..., 1015

We have to find the above sum

i.e. 5 + 10 + 15 + 20 + ... + 1015

Here a = 5, d = 10 – 5 = 5 and tn = 1015

We can write

a + (n – 1)d = 1015or 5 + (n – 1).5 = 1015

or 5n = 1015or n = 203

∴ The required sum

S203 = 2032 5 1015+l q U g S n a tn nsin , = +LNM OQP2 l q

= 2032 1020×

= 103530


1. Find the following sums :

(a) 7 + 12 + 17 + 22 + ... + 1002

(b) 25 + 28 + 31 + ... + 328

2. How many terms of the A.P. :

(a) 1, 4, 7, 10, ... are needed to get the sum 715 ?

(b) –10. –7, –4, –1, ... are needs to get the sum 104 ?

3. Find the sum of the first 10 odd natural numbers.

4. Sum of first 6 terms of an AP is –45. If the Ist term is –5, find its last term and commondifference.

8.8 IDEA OF A GEOMETRIC PROGRESSION

So far we have learnt that an Arithmetic Progression can be formed by repeatedly adding/subtracting a constant number to/from a given first number (term). A progression can also

198 Mathematics

be formed by repeatedly multiplying or dividing its preceding term by a non-zero constant.

If we consider that the starting prize in a game of ‘Triple Your Money’ is 3, then repeatedtripling (multiplication by 3) would form the progression of possible prizes as

3, 9, 27, 81, 243, ... ...(i)

If you observe each term of the above pattern/progression, you will find that there is a commonratio between two consecutive terms. In this case it is 3 : 1 as

93

279

8127

24381

31= = = =

It is usually denoted by ‘r’

Such a progression is called a Geometric progression (abbreviated as G.P.)

Let us take another example to consolidate :

Consider that a person reads from the newspaper that the Government is stressing on moresecurity for the railway passengers.

Consider that he/she conveys this message to two persons and both of them pass on this messageto two other persons each and so on.

It is shown by following figure :

Step 1 : ..................................

Step 0 : ...........................................................

Step 2 : ..................

Step 3 : .........

Number Patterns 199

Generally rumours spreads in this manner. If you calculate the number of persons informedat each step, you will get the following progression.

1, 2, 4, 8, 16, ... ...(ii)

This type of progression are called a Geometric Progression.

A progression is said to be a Geometric Progression (abbreviated as G.P.), if theratio of any two successive terms is always the same.

You can verify that common ratio in (ii) is always the same as 21

42

84

168 2= = = =

The ratio of any two successive terms of a Geometric Progression is called itscommon ratio.

Now observe the G.P’s in (i) and (ii)

3, 9, 27, 81, ...

and 1, 2, 4, 8, 16, ...

Both can be rewritten as as follows :

In general if ‘a’ be the first term and ‘r’ be the common ratio, the standard formor general form of a Geometric progressions would be

a , ar , ar2 , ar3 , ...

The following progressions are all examples of G.P’s :

(i) 3, 12, 48, 192, ... (ii) 5, 10, 20, 40, ...

200 Mathematics

(iii) 12

18

132

1128, , , , ... (iv) 1 1

41

161

64, , ,− − , ...

(v) 4, 16, 64, 256, ...

Example 8.15 : Identify G.P’s from the following progressions :

(a) 1, 5, 25, 125, ...

(b) 7, 14, 21, 42, ...

(c) 1 12

14

18, , ,− − , ...

Solution : (a) The given progression is 1, 5, 25, 125, ...

Here the first term is 1 and the common ratio = 51

255

12525 5= = =

∴ The progression is a geometric progression.

(b) The given progression is 7, 14, 21, 42, ...

Here the first term is 7 but the common ratio is not the same

as 147

2114≠

Since, the common ratio is not the same, this is not a geometric progression.

(c) The given progression is

1 12

14

18, , ,− − , ...

Here the first term is 1 and the Common ratio is = − 12 .

∴ The progression is a geometric progression.

Example 8.16 : Find the common ratio of the following G.P.’s :

(a) 11, 121, 1331, ...

(b) –3, 15, –75, ...

Solution : (a) The given progression is 11, 121, 1331, ...

Here the first term = 11

Common ratio = 12111

1331121 11= =

Number Patterns 201

(b) The given progression is –3, 15, –75, ...

Here the first term = –3

Common ratio = +−

= −+

= −153

7515

5 .


1. Identify Geometric Progressions from the following progressions :

(a) 4, 20, 100, 500, ... (b) 24, 12, 6, ...

(c) 11, 14, 17, 20, ... (d) 45, 15, 5, ...

2. Show that the following progressions are in G.P’s :

(a) 16, 8, 4, 2, 1, ...

(b) 1 13

19

127, , , , ...

(c) 125, 25, 5, 1, ...

3. Find the common ratio of each of the progressions given in Question Number 2.

LET US SUM UP

A progression is said to be an Arithmetic Progression (abbreviated as A.P.), if thedifference of each term, except the first, from its preceding term is always the same.

The successive terms of an A.P. are usually denoted by t1, t2, t3, ...

The common difference in usually denoted by ‘d’.

The standard form of an A.P is given by

a , a + d , a + 2d , ..., tn–1 , tn

where ‘a’ and ‘d’ are the first term and common difference respectively.

The common difference may be negative also. In that case, the terms of an A.P willdecrease.

The general term (nth term) of an A.P is given by tn = a + (n – 1)d.

202 Mathematics

The sum of the first n terms of an A.P whose first term is ‘a’ and common difference

is ‘d’ is given by Sn = n a n d2 2 1+ −b gm r

= n2 a + lb g , where l is the last term.

A progression is said to be a Geometric Progression (abbreviation as G.P.)

if the ratio of any two successive terms is always the same.

The ratio of any two successive terms of a G.P is called its common ratio and is usuallydenoted by ‘r’.

The standard form of a Geometric Progression is given by a , ar , ar2 , ...

where ‘a’ and ‘r’ are the first term and common ratio respectively.

TERMINAL EXERCISE

1. Which of the following progressions are A.P.'s ?

(a) 132 , 8 , 19

2 , ... (b) –1, 1, –1, 1, ...

(c) 227 , 3 , 20

7 , ... (d) 1, 12 , 1

3 , 14 , ...

(e) 22, 52, 82, ...

2. Find the general term (or nth term) of the following A.P’s :

(a) –1, –3, –5, ... (b) 157 , 2 , 13

7 , ...

(c) 32 , 2, 5

2 , ... (d) 53 , 4

3 , 1 , ...

3. Show that the following progressions are in A.P whose nth term is given by

(a) n (b) 3 – 5n

(c) 2n +13 (d) – n

2 +1

4. Which term of the following A.P. :

(a) 5, 8, 11, ... is 56 ?

Number Patterns 203

(b) 1, 6, 11, ... is 506 ?

(c) –190, –186, –182, ... is –14 ?

5. How many terms are needed to make the sum of the A.P. 25, 28, 31, ... as 1625 ?

6. Find the sum of the following A.P’s for given number of terms :

(a) 7 + 3 – 1 – 5 ...; 20 terms

(b) 15 + 13 + 11 + ...; 16 terms

(c) 1 + 9 + 17 + ...; 20 terms

7. The sum of three consecutive terms of an A.P is 36 and their product is 1428. Find theterms.

8. If 5th and 9th terms of an A.P are 8 and 14 respectively, find the first term and the commondifference.

9. The 14th term of an A.P is twice its 8th term. If its 6th term is –8, find the first termand common difference.

10. Find the sum of all even numbers upto 1001.

11. A man saved Rs 20,000 in 10 years. If the saving of each year is Rs 100 more than thesaving in the preceding year, find his saving in the first year.

12. Show that the following progressions are in G.P. :

(a) 1, 7, 49, ...

(b) 1 14

116, ,− , ...

(c) 16, 12, 9, ...

(d) 1125

125

15, , , ...

204 Mathematics

ANSWERS


1. (a), (c) and (d) are in A.P


1. (a) 35, 43 (b) 12

14,

2. (a) tn = 5n + 1 (b) tn = 3 22− n 4. 15, 2

5. (a) 25th term (b) 56th term 6. 30°, 60°, and 90°


1. (a) 100900 (b) 18003

2. (a) 22 terms (b) 13 terms 3. 100

4. –10,–1


1. (a), (b) and (d) 3. (a) 12 (b)

13 (c)

15

Terminal Exercise

1. (a) and (c)

2. (a) 1 – 2n (b) 167− n (c) 2

2+ n (d) 6

3− n

4. (a) 18th term (b) 102th term (c) 45th term 5. 26

6. (a) –620 (b) 0 (c) 1540

7. 7, 12 and 17 or 17, 12 and 7

8. 2, 32 9. 2, –2 10. 250500 11. Rs 1550

208 Mathematics

Module 2Commercial Mathematics

It is a common saying by elders “spend within your limits” i.e., keep your expenditure lessthan your income. The latent meaning of this is to save something for difficult times. You musthave seen birds and animals saving eatables for rainy season, in their nest or caves. Takingthe lead from this, the students have been told about the importance and need of savings inthis module.

Many Indian mathematicians have worked on the topic of commercial Mathematics. Yodoksu(370 B.C.) worked on fractions and ratio and proportion. In the reigns of Ashoka andChandragupta, there is a description of levying taxes. There is a description of manymathematicians working on practice and proportion (like Aryabhatt, Mahavira, Brahmgupta,Sridharacharya). In 900 A.D., Bakshali Manuscript was discovered which had a number ofproblems on Arithmetic.

To keep your savings safe is another tough task. Banks and other financial institutions keepthe money of their customers and on the expiry of the period pay extra money, called interest,in addition to the money deposited. This encourages citizens to save and keep the money safe.This is why calculation of interest on deposits in banks is included for teaching.

The Government provides a number of facilities to the citizens. For that they levy certain taxeson citizens. Two of these taxes are sales tax and income tax to which the learners are introducedin this module. Financial transactions about buying and selling are generally done for profit.Due to greater supply of goods or sub-standard goods they are to be sold on loss. The learnersare, therefore, introduced to percentage and profit and loss. Sometimes we have to buy articleson instalments because of non-availability of adequate funds. Due to this the students are taughtto calculate interest when they buy articles on instalment plan. Sometimes when we are notable to return loaned money on time, the financer starts charging interest on interest also, whichis called compound interest. Due to this the study of compound interest has been included inthis module. The formulae of compound interest is also used in finding increase or decreasein prices of things. This is also taught under “Appreciation and Depreciation” of value.

Ratio and Proportion 209

9

Ratio and Proportion

9.1 INTRODUCTION

You are aware of the concept of comparison of numbers in two different ways. One way isto see which number is larger and which is smaller. Another way is to see one number is howmany times the other number. In the first case, we compare the numbers by difference and inthe second we compare the numbers by division. In the second case, we talk of the ratio oftwo numbers. In this lesson we shall study to write a ratio in the simplest form, introduce theconcept of proportion, direct and inverse proportion (variation) and use these concepts to solvereal life problems, pertaining to time and distance, time and work, work and wages andpartnership.

9.2 OBJECTIVES


write a ratio in the simplest form

determine whether given four numbers are in proportion

illustrate the concept of direct and inverse proportion (variation)

solve real life problems pertaining to time and distance, partnership, time and work, workand wages etc.


Four Fundamental operations on numbers

Comparison by division

Knowledge of different units of measures and their conversions.

9.4 CONCEPT OF RATIO

When we compare the two quantities by division, we say that we have formed a ratio of thetwo quantities.

210 Mathematics

Consider an example of a bag of wheat with weight 90 kg and a bag of rice with weight40 kg. Here, we can say that the ratio of the weight of a bag of wheat to the weight of a bagof rice is 90 : 40 (read as 90 is to 40)

The symbol ‘:’ is used to denote a ratio.

Thus, the ratio of any number ‘a’ and ‘b’ is written as a : b. This is also written as ab .

In the ratio 90 : 40, 90 and 40 are called terms of the ratio. The first term (90) is called theantecedent and the second term (40) is called the consequent.

Remark : We cannot compare two quantities if they are not of the same kind i.e., we do notcompare 5 boys and 6 cows or 10 kg and 500 m or 5 toys and 3 fruits.

To compare two quantities as a ratio, it is necessary to express the two quantities in the sameunit. For example, to compare 3 months and 37 days, we compare (3 × 30) days and 37 days.The ratio, thus, is 90 : 37. It may be noted that in the above examples ratio is not 3 : 37.

Example 9.1 : In a class, there are 28 girls and 16 boys. Find

(i) the ratio of the number of boys to that of girls

(ii) the ratio of the number of girls to that of boys.

Solution : (i) The required ratio is 16 : 28.

(ii) The required ratio is 28 : 16.

Remarks : From Example 1 given above, the ratio a : b is different from the ratio b : a. Hence,the order of two terms in a ratio is very important.

Example 9.2 : A labourer earns Rs 3200 a month and spends Rs 2500. Find the ratio of his

(i) expenditure to income

(ii) savings to income

(iii) savings to expenditure

Solution : Here

Income = Rs 3200; Expenditure = Rs 2500

∴ Savings = Income – Expenditure

= Rs 3200 – Rs 2500

= Rs 700

(i) Ratio of expenditure to income = 2500 : 3200

(ii) Ratio of savings to income = 700 : 3200

(iii) Ratio of savings to expenditure = 700 : 2500.


Example 9.3 : Find the ratio of

(i) 5 days to 1 week

(ii) 7 km to 750 m

(iii) 250 ml to 1 l.

Solution : (i) Since the given quantities are in different units, we convert 1 week into days,getting 7 days. Hence, the required ratio is 5 : 7.

(ii) The two quantities are not in the same units. We, therefore first convert 7 km into m,getting 7000 m.

Hence, the required ratio is 7000 : 750.

(iii) The two quantities are not in same units, we therefore convert 1 litre into ml, getting1000 ml.

Hence, the required ratio is 250 : 1000.

9.4.1 Equal Ratios

A ratio remains unchanged when both of its terms are multiplied or divided by the some non-zero number. Hence,

20 : 50 is equal to 10 : 25

or 4 : 10

or 2 : 5

Note that in the ratio 2 : 5, the two terms 2 and 5 do not have a common factor otherthan 1.

9.4.2 Ratio in the Simplest Form

A ratio a : b is said to be in the simplest form if its two terms do not have a common factor,other than 1, i.e. HCF of the two terms is 1.

For example, the ratio 20 : 50 is not in the simplest form, because 10 is a common factor ofits two terms, The simplest form of the ratio 20 : 50 is 2 : 5.

Note : A ratio can be expressed in several ways, as seen above. The ratio 20 : 50 can be writtenas 10 : 25, or 4 : 10 or 2 : 5, etc.

How to obtain the simplest form of a given ratio ?

For a given ratio, say a : b, find the H.C.F. of its two terms a and b. Divide each of the termsby the H.C.F. obtained. The ratio formed of the two new numbers obtained is the simplest formof the given ratio.

212 Mathematics

Example 9.4 : Express each of the following ratios in the simplest form :

(i) 24 : 30 (ii) 150 : 400

(ii) 85 : 225 (iv) 480 : 576

Solution : (i) The H.C.F. of 24 and 30 is 6.

∴ 24 : 30 =246

306

: = 4 : 5

(ii) The H.C.F. of 150 and 400 is 50.

∴ 150 : 400 =15050

40050

: = 3 : 8

Thus, 3 : 8 is the simplest form of the ratio 150 : 400.

(iii) The H.C.F.of 85 and 225 is 5

∴ 85 : 225 =855

2255

: = 17 : 45


(iv) The H.C.F.of 480 and 576 is 96

∴ 480 : 576 =48096

57696

: = 5 : 6


Example 9.5 : Find the simplest form of the ratio of each of the following two quantities:

(i) 65 km to 91 km

(ii) 45 seconds to 2 minutes

(iii) 6 hours to 1 day

Solution : (i) The ratio of 65 km to 91 km is 65 : 91

The H.C.F. of 65 and 91 is 13.

∴ 65 : 91 = 6513

9113: = 5 : 7

Thus, the simplest form of the ratio is 5 :7.

(ii) The ratio of 45 seconds to 2 minutes is 45 : 120, because 2 minutes = 120 seconds.



∴ 45 : 120 = 4515

12015: = 3 : 8


(iii) The ratio of 6 hours to 1 day is the same as the ratio for 6 hours to 24 hours(1 day = 24 hours),

So, the ratio is 6 : 24


∴ 6 : 24 = 66

246: = 1 : 4



1. Fill in the blanks to make each of the following a true statement :

(i) In the ratio 3 : 7, the antecedent is and consequent is

(ii) The ratio 3 : 5 is from the ratio 5 : 3.

(iii) The simplest form of the ratio 15 : 20 is .

(iv) The ratio of 2 months to 2 days, in the simplest form, is .

(v) The ratio 5 : 4 is of the ratio 100 : 80.

2. Express each of the following ratios in the simplest form :

(i) 39 : 65 (ii) 200 : 350 (iii) 172 : 528

(iv) 120 : 144 (v) 0.32 : 1.2 (vi) 4860 : 8370

3. Find the ratio (in the simplest form) of :

(i) 45 cm to 5 m (ii) 200 g to 5 kg

(iii) 90 paise to Rs 3 (iv) 35 minutes to 45 seconds

(v) 1 hour to 15 seconds (vi) Rs 38 to Rs 9.50.

4. In a year, Narmita earned Rs 84000 and paid Rs 1200 as income tax. Find the ratio (inthe simplest form) of her

(i) income to income tax (ii) income tax to income.

5. Total number of workers in a factory is 196. 49 of them are men and the rest are women.Find the ratio (in the simplest form) of :

(i) the number of men to that of women

(ii) the number of women to the total number of workers

214 Mathematics

6. A rectangular sheet of paper is 30 cm long and 21 cm wide. Find the ratio (in the simplestform) of its

(i) width to length

(ii) length to perimeter.

9.5 DIVISION OF A NUMBER IN THE GIVEN RATIO

To divide a given number x in the given ratio l : m, we follow the procedure given below :

Step 1 : Find the sum of the two terms of the ratio, i.e. l + m.

Step 2 : Use the following formula to obtain the two parts of the given number x.

First part = ll m x+ ×

Second part = ml m x+ ×

Remark : The same procedure is followed even if the given number is to be divided in morethan two parts in the given ratio.

Example 9.6 : Divide 144 in two parts in the ratio 7 : 9.

Solution : We have

Sum of the two terms of the ratio = 7 + 9 = 16

∴ First part = 716 144× = 63

and second part = 916 144× = 81

Thus, the two parts of 144 with given ratio are 63 and 81.

Example 9.7 : Divide Rs 1050 among A, B, C in the ratio 3 : 5 : 7.

Solution : We have

Sum of the terms of the ratio = 3 + 5 + 7 = 15

∴ A’s share = Rs 315 1050×FH IK = Rs 210

B’s share = Rs 515 1050×FH IK = Rs 350

and C’s share = Rs 715 1050×FH IK = Rs 490.



1. (i) Divide 15 in the ratio 2 : 3.

(ii) Divide 184 in the ratio 3 : 5

(iii) Divide 7780 in the ratio 7 : 8 : 5.

2. Divide :

(i) Rs 140 in the ratio 2 : 5

(ii) Rs 154 in the ratio 3 : 4

(iii) 9 cm 8 mm in the ratio 2 : 5

(iv) 10 cm 5 mm in the ratio 1 : 4.

3. The ratio between two quantities is 2 : 7. If the second quantity is 9.8 kg, how much isthe first ?

4. The angles of a triangle are in the ratio 3 : 5 : 7. Find the angles.

[Hint : Sum of the three angles of a triangle is 180°]

5. In a camp, the ratio of the number of teachers to the number of students is 3 : 50. If thetotal number of persons in the camp is 159, then find the number of teachers.

6. In a bank, the ratio of the number of clerks to the number of officers is 17 : 2. If thetotal number of persons working in the bank is 57, find the number of clerks in the bank.

7. The sides of a triangle are in the ratio 2 : 3 : 4. If the perimeter is 36 cm, then find itssides.

9.6. PROPORTION

An equality of two ratios constitutes a proportion. Consider two ratios 8 : 14 and 20 : 35.On expressing these ratios in the simplest form, we find that

8 : 14 = 4 : 7

and 20 : 35 = 4 : 7

Therefore, 8 : 14 = 20 : 35

Thus, 8 : 14 = 20 : 35 is a proportion.

Similarly,

20 : 70 = 2 : 7 is a proportion.

Thus, we define proportion as follows :

“Four numbers a, b, c, d (in order) are said to be in proportion, if the ratio of the first twois equal to ratio of the last two i.e. a : b = c : d”.

216 Mathematics

When four numbers a, b, c, d (in order) are in proportion, then we write

a : b : : c : d

which is read as “a is to be as c is to d” or “a to b as c to d”.

In a proportion a : b : : c : d, a, b, c and d are the first, second, third and fourth terms of theproportion. The first and fourth terms are called extremes, the second and third terms are calledmeans.

a : b = c : d ⇒ ab

cd= or ad = bc.

Note : In a proportion, the product of extremes is equal to the product of means.

In other words

a : b : : c : d if and only if ad = bc.

Example 9.8 : Which of the following four numbers constitute a proportion ?

(i) 3, 5, 15, 25

(ii) 3, 15, 25, 5

Solution : (i) Here,

The product of extremes = 3 × 25 = 75

The product of means = 5 × 15 = 75

Since the two products are equal, the four numbers are in proportion.

(ii) Here,

The product of extremes = 3 × 5 = 15

The product of means = 15 × 25 = 375

Since the two products are not equal, the four numbers are not in proportion.

Remark : From Example 9.8 above, we find that the order of four numbers for a proportionis important.

Example 9.9 : Are, 5, 10, 30, 60 in proportion ?

Solution : Ratio of first two terms is 5 : 10, which is equal to 1 : 2

Ratio of last two terms is 30 : 60, which is equal to 1 : 2

∴ 5 : 10 = 30 : 60Hence, 5, 10, 30, 60 are in proportion.

Aliter

Here, Product of extremes = 5 × 60 = 300

Product of means = 10 × 30 = 300

Since the two products are equal, then given numbers are in proportion.


Example 9.10 : The first, second and fourth terms of a proportion are 5, 10 and 30 respectively.Find the third therm.

Solution : Let the third term be x so that 5, 10, x, 30 are in proportion.

⇒ 5 × 30 = 10 × x

⇒ x =5 30

10×

= 15

Thus, the third term is 15.

Example 9.11 : The two extremes of a proportion are 12 and 28. If one of means is 24, thenfind the other mean.

Solution : Let the other mean be x.

We know that for a proportion,

product of means = product of extremes

⇒ 24 × x = 12 × 28

⇒ x =12 28

24×

⇒ x = 14

Thus, the other mean is 14.

Example 9.12 : Are 4, 8, 8, 16 in proportion ?

Solution : Here,

Product of means = 8 × 8 = 64

Product of extremes = 4 × 16 = 64

Since the two products are equal, the four numbers 4, 8, 8, 16 are in proportion.

Remark : In Example 5 above, the means are the same. In this case, we say that 4, 8, 16 arein continued proportion.

Thus, if a, b, b, c are in proportion, we say that a, b, c are in continued proportion. In sucha case, b2 = ac and b is called the mean proportional of a and c.


1. Which of the following statements are true ?

(i) 4 : 6 = 8 : 12 (ii) 4 : 28 = 7 : 1

(iii) 16 : 24 : : 30 : 20 (iv) 6.0 : 4.5 : : 4.8 : 3.6

218 Mathematics

2. Determine which of the following numbers are in proportion :

(i) 18, 27, 12, 18 (ii) 12, 10, 16, 10

(iii) 2, 3, 14, 15 (iv) 40, 30, 60, 45

3. Find the value of x, if x : 6 : : 5 : 3.

4. If 18, x, x, 50 are in proportion, find the value of x.

5. If 3, x, 12 are in continued proportion, find x.

6. Set up all possible proportions from the numbers 15, 18, 35 and 42.

7. Find the mean proportional between 5 and 125.

9.7 DIRECT PROPORTION (VARIATION)

If you buy 6 pens for Rs 12, you will have to pay Rs 16 for 8 pens and Rs 24 for 12 pens.Thus, you see that as the number of pens purchased increases, the price you have to pay alsoincreases. Again, if you buy 4 pens, you will have to pay only Rs 8. Here you see that

more the number of pens purchased, more money you have to pay; less the number of penspurchased, less money you have pay.

To indicate the above situation, we say that the number of pens purchased and the price paidare in direct proportion.

In general, two quantities a and b are said to be in direct proportion if increase/decrease inone quantity results in the corresponding increase/decrease in the other.

We also say that a varies directly as b and write a = kb, where k is a constant, called the constantof variation or proportionality.

9.7.1 Inverse Proportion (Variation)

If two persons, employed to construct a wall, take 24 days to complete it, then 4 personsemployed to do the same job will take 12 days to complete it. Likewise, 6 persons will takeonly 8 days to complete the job. We may note that as the number of persons employed increases,the number of days required to complete the job decreases.

Thus, we see that

lesser the number of persons employed, more the number of days

Or more the number of persons employed, lesser the number of days.

We say that the number of persons employed to complete a job and the number of days takenby them are in inverse proportion (variation).

In general, two quantities a and b are said to be in inverse proportion if increase/decrease inone quantity results in corresponding decrease/increase in the other quantity.

We say that a varies inversely (indirectly) to b, if a = kb or ab = k, where k is a constant of

proportionality.


9.8 APPLICATIONS OF DIRECT PROPORTION

We will take some examples to illustrate the application of direct proportion (variation) in dailylife situations.

Example 9.13 : If 5 books cost Rs 70, how much will 8 books cost ?

Solution : Let Rs x denote the cost of 8 books, we get the following :

Quantity Cost (in rupees)5 708 x

Since the cost varies directly as quantity. Equating the two ratios we get5 : 8 = 70 : x

⇒58 =

70x

⇒ x =8 70

5×

= 112

Thus, the cost of 8 books is Rs 112

Example 9.14 : A typist types 70 pages in 5 days. How many pages can he type in 7 days?

Solution : Let y denote the number of pages that the typist would type in 7 days

Since the number of pages vary directly as the number of days, we get thefollowing :

Number of pages Number of days70 5y 7

Equating the two ratios, we get5 : 7 = 70 : y

⇒57 =

70y

⇒ y =70 7

5×

= 98

Thus, the typist will type 98 pages in 7 days.

Example 9.15 : If 15 men can build a 100 m long wall in some days; how many men shouldbe employed to build 120 m long wall of the same height in the same time.

Solution : Let n denote the number of persons to be employed to build a wall 120 m long.Since the number of men varies directly as the length of the wall, we get the following :

220 Mathematics

Number of men Length of wall (in metres)

15 100

n 120

Equating the two ratios, we get

100 : 120 = 15 : n

⇒100120 =

15n

⇒ n =15 120

100×

= 18

Thus, 18 men will build 120 m long wall in the same time.

Example 9.16 : A jeep travels 50 km in 1 hour. How much distance would it travel in12 minutes ?

Solution : Let d denote the distance (in km) travelled by the jeep in 12 minutes.

Since the distance travelled and time taken vary directly we get the following :

Time (in minutes) Distance (in km)

60 50

12 d

Equating the two ratios we get

60 : 12 = 50 : d

⇒6012 =

50d

⇒ d =50 12

60×

= 10.

Thus, the jeep would travel 10 km in 12 minutes.

Example 9.17 : If a deposit of Rs 3000 carries an interest of Rs 900 in some time, what interestwould a deposit of Rs 5000 earn in the same time ?

Solution : Let I denote the interest that Rs 5000 would earn.

Since the interest earned and the amount deposited vary directly, we get the following :

Principal (in Rs) Interest (in Rs)

3000 900

5000 I


Equating the two ratios, we get

3000 : 5000 = 900 : I

⇒30005000 =

900I

⇒ I =5000 900

3000×

= 1500

Thus, the deposit of Rs 5000 would earn an interest of Rs 1500 in the given time.

9.9 PARTNERSHIP

When two or more individuals enter into business jointly we say that they enter into apartnership. These individuals are called partners.

In partnership, all or some partners invest money in business for same or different periods oftime.

The profits or losses of a partnership business are shared among the partners in the ratio oftheir investments.

The following examples illustrate the above concept.

Example 9.18 : A and B start a business by investing Rs 40000 and Rs 60000 respectively.If at the end of the year they earn a profit of Rs 35000, find the share of each in the profit.

Solution : Let A's share be Rs x. Then B’s share is Rs (35000 – x).

Since share of profit varies directly as the investment, equating the two ratios, we get

40000 : 60000 = x : (35000 – x)

⇒4000060000 =

xx35000−

⇒23 =

xx35000−

⇒ 2(35000 – x) = 3x⇒ 70000 – 2x = 3x⇒ 5x = 70000⇒ x = 14000

Thus A’s share is Rs 14000; and that of B is Rs (35000 – 14000) i.e. Rs. 21000.


1. If 7 pens cost Rs 98, find the cost of 35 pens.

2. If the cost of 12 erasers is Rs 30, how many erasers can be bought for Rs 50 ?

222 Mathematics

3. If 5 litres of petrol cost Rs 140, how much will 8 litres of petrol cost ?

4. A carpenter prepares 18 chairs in 4 days. In how many days would he prepare 27 suchchairs ?

5. A tailor stitches 10 trousers in 4 days. How many trousers can he stitch in 10 days ?

6. A woman can pack 270 bundles in 3 days. How many bundles can she pack in 7 days?

7. If 30 men can weave 40 metres of cloth in a day, how many meters of cloth can be wovenby 240 men in a day ?

8. 4 taps of equal capacity can fill 6 tubs in a given time. How many tubs can be filled by10 taps in the same time ?

9. If a family of 5 members consumes 50 kg of cereals in a month, what quantity of cerealswould be consumed by family of 6 members in a month ?

10. A car travels 160 km in 4 hours. How long will it take to travel 400 km?

11. For every 50 students, 3 teachers are appointed. How many teachers, should be appointed,if a school has 1250 students ?

12. A man earns Rs 1750 per week. How much will he earn in 4 days ?

13. 10 men dig a trench in 9 days. In how many days would 15 men dig the same trench?

14. 20 farmers plough a field in 8 days. How many days will 16 farmers take to plough thefield.

15. If the thickness of a pile of 20 cardboards is 50 mm, find the thickness of a pile of 30similar cardboards ?

16. A deposit earns Rs 120 in 24 years. How much will it earn in 5 years ?

17. If the interest on Rs 3000 is Rs 150 for a certain period, what will be the interest onRs 4500 for the same period, when the rate of interest remains the same ?

18. A and B start, a business with Rs 60000 and Rs 75000 respectively. At the end of theyear, they earned a profit of Rs 40500. Find the share of each partner in the profit.

19. A and B start a business with Rs 70000 and Rs 80000 respectively. They earn Rs 50700as the annual profit. If Rs 1500 per month out of the profit is paid as rent, find the shareof each after paying the rent.

[Hint. Subtract Rs (1500 × 12) i.e. Rs 18000 paid for the rent from the profit of Rs 50700.The balance of profit will be shared in the ratio 70000 : 80000]

20. Two partners together invested Rs 2.12 lakhs in a joint business. After a year, one getsRs 35000 and the other gets Rs 18000 as profit. Find the sum invested by each.


9.10 INVERSE PROPORTION

We will now take some examples to illustrate the application of inverse proportion (variation)in daily life situations.

Example 9.19 : 4 girls can do a work 3 days. How many days will 6 girls take to do the somework ?

Solution : Let x be the number of days in which 6 girls can do the work,

Since the number of girls varies inversely as the number of days, we get the following :

Girls Days4 36 x

The ratios of like terms are 4 : 6 and 3 : xEquating ratio with the inverse of the other, we get

4 : 6 = x : 3

⇒ 46 = x

3

⇒ x = 4 36× = 2

Thus, 6 girls will complete the work in 2 days.

Example 9.20 : A car takes 4 hours to cover some distance at an average speed of 55 kmper hour. How much time will it take to cover the same distance at an average speed of40 km per hour ?

Solution : Let t (in hours) denote the time that the car will take at the speed of 40 km perhour, to cover the same distance.Since speed varies inversely as time, we get the following :

Speed Time55 440 t

The ratios of like terms are 55 : 40 and 4 : t.Equating the ratio with the inverse of the other, we get

55 : 40 = t : 4

or 5540 =

t4

or t = 55 440× = 5.5

Thus, travelling at a speed of 40 km per hour, the car will cover the same distance in 5.5 hours.

224 Mathematics


1. If 3 men can build a wall in 15 days, how long will 5 men take to build the same wall?

2. 400 men have provision for 23 days. If 60 more men join them, how long will theprovisions last ?

3. To cover some distance, a train, running at a speed of 80 km per hour, takes 12.5 hours.If it runs at a speed of 100 km per hour, how much time will it take to cover the samedistance ?

4. A man deposited Rs 5000 in a bank for one year to get a fixed amount of interest. If thesame interest is to be earned in 8 months at the same rate of interest, what sum head tobe deposited ?

5. 6 taps of equal capacity can fill a reservoir in 15 minutes. How many taps can fill it in10 minutes ?

6. A journey takes 21 minutes if I walk at 4.5 km per hour. How long will it take if I walkat 3.5 km per hour ?

7. 120 men had provisions for 200 days. After 5 days, 30 men left. How long will theremaining provision last ?

LET US SUM UP

Ratio is a comparison of numbers or quantities of the same kind.

In a ratio a : b, a is the first term (antecedent) and b the second term (consequent).

The order terms in a ratio is important

a : b is not the same as b : a

A ratio is said to be in the simplest form if the H.C.F. of its two terms is 1.

A statement indicating equality of two ratios is called a proportion.

A proportion has four terms.

In a proportion a : b : : c : d, a and d are called the extremes and b and c are called themeans.

Four numbers a, b, c, d are in proportion if and only if ad = bc.

If a : b = b : c, then a, b, c are said to be in continued proportion.

If a, b, c are in continued proportion, then b2 = ac and b is called mean proportion ofa and c.

Two quantities are said to be in direct proportion if the increase/decrease in one resultsin a corresponding increase/decrease in the other.

Two quantities are said to be in inverse proportion if an increase/decrease in one resultsin a corresponding decrease/increases in the other.


TERMINAL EXERCISE

1. Write each of the following ratios in the simplest form :

(i) 64 : 160 (ii) 216 : 384 (iii) 1.5 : 9

2. Find the ratio of the following in the simplest form :

(i) 450 m, 5 km (ii) 50 minutes, 1 hour 5 minutes

(iii) 16 mm, 3 cm (iv) 5 kg 600 g, 6 kg 400 g.

3. Divide

(i) Rs 154 in the ratio 5 : 9

(ii) Rs 9292 in the ratio 1 : 3.

4. The ratio between two quantities is 2 : 5. If the second quantity is 9.5 kg, find the firstquantity.

5. The angles of a triangle are in the ratio 3 : 7 : 8. Find the angles.

6. Determine the value of p if 20 : p : : 25 : 450.

7. Are 5, 9, 81, 45 in proportion ?

8. A dish for 250 persons uses 15 cups of sugar. How many cups of sugar are needed for150 persons ?

9. If 12 boxes are required to hold 60 dozen apples, how many boxes will be required tohold 105 dozen ?

10. A 12 m length of an angle iron weighs 24.12 kg. What will a piece of length 8 m of thesame weigh ?

11. A watch gains 42 seconds in 3 days 8 hours. In how much time will it gain 2 minutes6 seconds ?

12. A mini bus uses 35 litres of diesel for a journey of 420 km. How many litres of dieselare needed for 600 km ?

13. A and B started a business by investing Rs 50000 and Rs 75000 respectively. Find theshare of each out of a profit of Rs 15000.

14. A and B entered into a partnership investing Rs 35000 and Rs 42000 respectively. A isa working partner and gets Rs 300 per month for the same. Find the share of each in anannual profit of Rs 13500.

15. If the sales tax on a purchase worth Rs 600 is Rs 42, what will be the sales tax on thepurchase worth Rs 1500 ?

226 Mathematics

ANSWERS


1. (i) 3; 7 (ii) different (iii) 3 : 4

(iv) 30 : 1 (v) simplified form

2. (i) 3 : 5 (ii) 4 : 7 (iii) 43 : 132

(iv) 5 : 6 (v) 4 : 15 (vi) 18 : 31

3. (i) 9 : 100 (ii) 1 : 25 (iii) 3 : 10

(iv) 140 : 3 (v) 240 : 1 (vi) 4 : 1

4. (i) 70 : 1 (ii) 1 : 70

5. (i) 1 : 3 (ii) 3 : 4

6. (i) 7 : 10 (ii) 5 : 17.


1. (i) 6 and 9 (ii) 69 and 115 (iii) 2723, 3112 and 1945

2. (i) Rs 40 and Rs 100 (ii) Rs 66 and Rs 88 (iii) 28 mm and 70 mm

(iv) 21 mm and 84 cm

3. 2.8 kg 4. 36°, 60°, 84° 5. 9

6. 51 7. 8 cm, 12 cm and 16 cm


1. (i), (iv) 2. (i), (iv) 3. x = 10

4. x = 30 5. x = 6

6. 15 : 18 : : 35 : 42 ; 18 : 15 : : 42 : 35 7. 25


1. Rs 490 2. 20 3. Rs 224

4. 6 days 5. 25 6. 630

7. 320 metres 8. 15 9. 60 kg of cereals

10. 10 hours 11. 75 12. Rs 1000

13. 6 days 14. 10 days 15. 75 mm

16. Rs 25 17. Rs 225 18. Rs 18000; Rs 22500

19. Rs 15260; Rs 17440 20. Rs 140000; Rs 72000.



1. 9 days 2. 20 days 3. 10 hours

4. Rs 7500 5. 9 6. 27 minutes

7. 260 days

Terminal Exercises

1. (i) 2 : 5 (ii) 9 : 16 (iii) 1 : 6

2. (i) 9 : 100 (ii) 10 : 13 (iii) 8 : 15

(iv) 7 : 8

3. (i) Rs 55 and Rs 99 (ii) Rs 2323 and Rs 6969

4. 3.8 kg 5. 30°, 70°, 80° 6. 360

7. No 8. 9 9. 21

10. 16.08 kg 11. 10 days 12. 50 litres

13. Rs 6000; Rs 9000 14. Rs 8100; Rs 5400 15. Rs 105

228 Mathematics

10

Percentage and its Applications

10.1 INTRODUCTION

In every day life, you come across situations in which the word ‘percent’ is made use of veryfrequently. For example, you see a banner in the market.

“Sale upto 50 percent off”

You read news in the newspaper

“Votes turnout in the poll was over 65 percent”

“Banks have lowered the rate of interest on fixed deposits from 7.5 percent to 6.5 percent”

There are many such situations in different walks of life where the concept of percentage findsits use. In this lesson we shall study percent as a fraction or a decimal and its application insolving problems of profit and loss, discount, sales tax, instalments etc.

10.2 OBJECTIVES

After studying the lesson, the learner will be able to :

write a fraction and a decimal as a percent and vice-versa

calculate specified percent of a given number or a quantity

solve problems based on percentage

solve problems on profit and loss

calculate simple interest and amount when a given sum of money is interested for aspecified time period on a given rate of interest.

state the need for given discount

define discount and discount series (successive discounts, no. more than three)

find a single discount equivalent to a given discount series

calculate the discount and the selling price of an article, given marked price of the article,and the rate of discount

Percentage and its Applications 229

solve inverse problems pertaining to discount

calculate the sales tax on commission and the selling price of an article, given the markedprice of the article and the rate of sales tax or commission

solve inverse problems pertaining to sales tax

solve inverse problems on commission determine the amount of each instalment whengoods are purchased under investment plan (case of equal instalments only)

determine the rate of interest when equal instalments are given


Four fundamental operations on whole numbers, fractions and decimals.’

10.4 PERCENT

You have learnt a lot about fractions. A fraction denotes part of a whole. For instances 34 means

3 out of 4 equal parts. Similarly, 710 means 7 out of ten equal parts and 17

100 means 17 out

of 100 equal parts.

A fraction whose denominator is 100 is read as a percent, for example 5100 is read as five

percent.

In Fig. 10.1, 33 out of 100 small squares are shaded.

This means 33100 of the larger square is shaded.

The word ‘percent’ is abbreviated form of the Latin word “percentum” which means “per hundred” or “hundredths”.

The symbol ‘%’ is used for the term percent.

A ratio whose second term is 100 is also called a percent.

When we say Anita has secured 80% marks in mathematics. This means that she has secured80 marks out of 100 or 40 marks out of 50. Similarly, when we say a man has spent 20% ofhis income on food, it means that out of every hundred rupees of his income, he has spentRs 20 on food.

Suppose, we wish to compare two fractions 34 and 4

5 .

Since these are fractions having different denominators we need to convert them into fractionswith common denominator.

Fig. 10.1

230 Mathematics

34 = 3 5

4 51520

×× = ; 4

5 = 4 45 4

1620

×× =

Since 16 > 15,

∴ 1620

1520> ; and so 4

534>

Here, we have converted each of the two fractions into fractions with least commondenominator.

As a convention, we convert if possible each fraction into a fraction with denominator 100.In the above example,

34 = 3 25

4 2575

100×× = ; 4

5 = 4 205 20

80100

×× =

Since 80 > 75. we get 45

34

> .

10.5 CONVERSION OF A DECIMAL INTO A PERCENT AND VICE VERSA

Let us consider the following examples :

0.37 = 37100 = 37%, 0.7 = 7

1070

100= = 70%; 1.46 = 146100 = 146%

Thus, to write a decimal as a percent, we move the decimal point two places to the right andput the % sign.

0.25 = 25% 0.61 = 61% 0.37 = 37%

0.1 = 10% 0.07 = 7% 1.4 = 140%

Conversely,

To write a percent as a decimal, we drop the % sign and insert or move the decimal point twoplaces to the left. For example,

37% = 0.37 89% = 0.89 35% = 0.35

99% = 0.99 100% = 1.00 3% = 0.03

110% = 1.10 212% = 2.12 0.1% = 0.001

10.6 CONVERSION OF A PERCENT INTO A FRACTION AND VICE VERSA

To write a percent as fraction, we drop the % sign and divide the number by 100. For example,

69% = 69100 13% = 13

100 3% = 3100

4.5% = 4 5100

451000

. = 170% = 170100 216% = 216

100


In general, x% = x

100 .

Conversely,

To write a fraction as a percent, we multiply the fraction by 100, simplify it and suffix the% sign. For example

14 = 1

4 100×FH IK% = 25%

32 = 3

2 100×FH IK% = 150%

35 = 3

5 100×FH IK% = 60%


1. Write each of the following decimals as percent :

(a) 0.56 (b) 0.03 (c) 0.75 (d) 0.02

(e) 0.97 (f) 0.8 (g) 0.04 (h) 1.4

2. Write each of the following percents as decimal :

(a) 75% (b) 14% (c) 3% (d) 115%

(d) 2% (f) 25% (g) 400% (h) 350%

3. Write each of the following percents as fraction :

(a) 35% (b) 40% (c) 70% (d) 85%

4. Convert each of the following fractions into percent :

(a) 34 (b) 1

5 (c) 310 (d) 4

25

5. Aruna obtained 19 marks in a test of 25 marks. What was her percentage of marks ?

6. Gurmeet got half the answers correct. What percent of their answers were correct ?

7. A suit piece consists of cotton and rayon fibre in which cotton is 3 out of 8 parts. Whatis the percentage of cotton in the suit piece ?

8. Kavita read 84 pages of 100-page book. What percent of the book did she read ?

9. A class of a school had 45% girls. What percent of the class were boys ?

10. One-fourth of the shoes in a shop were on sale. What percent of the shoes were thereon normal price ?

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11. In the word PERCENTAGE, what percent of the letters are E’s ?

12. If three fourths of students of a class wear glasses, what per cent of students of the classdo not wear glasses ?

13. There are 20 eggs in a fridge and 6 of them are brown. What percent of eggs arebrown ?

14. 60 candidates appeared in an examination and 45 of them passed. What percent ofcandidate passed ?

15. Mr. X spends Rs 310 out of Rs 500 and Mr. Y spent Rs 500 out of Rs 800 every week.Compute their spending as percentages and state who spends higher percentage.

16. In a class of 40 students, 10 secured first division, 15 secured second division, and 13just qualified. What is the percentage of students that failed ?

10.7 CALCULATION OF PERCENT OF A QUANTITY

To determine the percent of a number or quantity, we first change the percent to a fractionor a decimal and then multiply with the number.

For example, 45% of 90 = 0.45 × 90 = 40.50

or 45% of 90 =45

10090× = 40.50

60% of 120 = 0.60 × 120 = 72.0018% of 215 = 0.18 × 215 = 38.70135% of 80 = 1.35 × 80 = 108

Example 10.1 : A family spends 35% of its monthly budget of Rs 7500 on food. How muchdo they spend on food ?

Solution : Expenditure on food = 35% of Rs 7500

= 0.35 × Rs 7500= Rs (0.35 × 7500)= Rs 2625.00 or Rs 2625.

Example 10.2 : In a garden, there are 500 plants of which 35% are trees, 20% are shrubs and25% are herbs. The rest are creepers. Find out the number of trees, shrubs, herbs and creepers.

Solution : Number of trees = 35% of 500 = 0.35 × 500 = 175Number of shrubs = 20% of 500 = 0.20 × 500 = 100Number of herbs = 25% of 500 = 0.25 × 500 = 125

Since the remaining plants are creepers,Number of Creepers = 500 – (175 + 100 + 125)

= 500 – 400 = 100


Example 10.3 : 35% of students in a school are girls. If the total number of students is 1240,find the number of boys in the school.

Solution : Number of girls in the school = 35% of 1240

= 0.35 × 1240

= 434

∴ No. of boys in the school = 1240 – 434 = 806

Aliter

Since 35% of the students in the school are girls, (100% – 35%) i.e., 65% of the students inthe school are boys.

∴ Number of boys = 65% of 1240

= 0.65 × 1240

= 806

Example 10.4 : What percent of 240 is 96 ?

Solution : Percent = 96240 100× %

= 40%

Example 10.5 : If 27% of ‘a’ is 54, then find a.

Solution : We have 27% of a = 54

⇒ 27100 × a = 54

⇒ a = 54 10027× = 200

Thus, the value of a is 200.

Example 10.6 : 60 is reduced to 45. What is the reduction percent ?

Solution : Let 45 is less than 60 by x%, then

Reduction = 60 – 45 = 15

Reduction percent = 1560 100× % = 25%.

Example 10.7 : If 80 is increased to 125, what is the increase percent ?

Solution : Increase = 125 – 80 = 45

Increase percent = 4580 100× % = 56.25%.

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Example 10.8 : A voluntary organisation was collecting money for a relief camp. Theirtarget was Rs 20000, but they exceeded their target by 45%. How much money did theycollect ?

Solution : The money collected = 20000 + 45% of 20000= Rs 20000 + Rs (0.45 × 20000)= Rs 20000 + Rs (9000.00)= Rs 20000 + Rs 9000= Rs 29000

Hence, they collected Rs 29000.

Example 10.9 : 44% of the students of a class are girls. If the number of girls is 6 less thanthe number of boys, how many students are there in the class ?

Solution : Given that 44% of the students are girls. So, (100 – 44)% i.e., 56% of the studentsare boys.

Thus, there are 12% less girls than boys.

By the given condition,

12100 of the number of students = 6

∴ Number of students = 6 10012× = 50

Thus, there are 50 students in the class.

Example 10.10 : Raman has to secure 40% marks for passing. He gets 178 marks and failsby 22 marks. Find the maximum marks.

Solution : Since Ramesh secures 178 marks and fails by 22 marks, the pass marks are 178+ 22 = 200.

Let x be maximum marks, then

40% of x = 200

or 40100 × x = 200

or x = 200 10040× = 500

Thus, the maximum marks are 500.


1. Find :

(a) 15% of 440 (b) 16% of 1250

(c) 47% of Rs 1200 (d) 39% of 1700 metres.


2. At a school, 40% of the students come on foot to the school. There are 600 students inthe school. How many students come on foot to the school ?

3. There are 36 children in a class 25% of them are boys. How many are boys ? How manyare girls ?

4. An alloy is a combination of zinc and copper with 30% zinc and 70% copper. If a pieceof this alloy weighs 150 kg , how much zinc it contains ?

5. Naresh earns Rs 15400 per month. He keeps 50% for household expenses, 15% for hispersonal expenses, 25% for expenditure on his children and the rest he saves. What amountdoes he save per month ?

6. There are 32 boys in a class. On a particular day 12.5% of them were absent. How manyboys were absent on that day ?

7. It takes me 45 minutes to go to school and I spend 80% of the time travelling by bus.How long does the bus journey last ?

8. During a general election, 70% of the population voted. If 70000 people cast their votes,what is the population of the town ?

9. The cost of a saree was Rs 450. Its cost has increased to Rs 495. By what percent didthe cost increase ?

10. What percent of 160 is 64 ?

11. If 120 is reduced to 96, what is the percentage reduction ?

12. In an election, 25% voters did not cast their votes. A candidate secured 40% of the votespolled and was defeated by 900 votes. Find the total number of voters.

13. A’s income is 25% more than B’s and B’s income is 8% more than C’s. If A’s incomeis Rs 4050, then find the C’s income.

14. A reduction of 10% in the price of tea enables a dealer to purchase 25 kg more tea forRs 22500. What is the reduced price per kg of tea ? Also, find the original price per kg.

15. A rise of 25% in the price of sugar compels a person to buy 1.5 kg of sugar less forRs 240. Find the increased price as well as the original price per kg of the sugar.

16. A number is first increased by 10% and then decreased by 10%. What is the net increaseor decrease percent ?

17. A man donated 5% of his monthly income to a charity and deposited 12% of the restin a bank. If he has Rs 11704 with him now, what was his monthly income ?

10.8 APPLICATIONS OF PERCENTAGE

We come across a number of situations in our day to day life wherein we use the concept ofpercent. In the following, we discuss applications of the concept of percentage in differentfields.

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10.8.1 PROFIT AND LOSS

Let us recall the terms and formulae related to profit and loss that we have learnt earlier.

Cost Price (C.P.) : The price at which an article is purchased, is called its cost price.

Selling Price (S.P.) : The price at which an article is sold, is called its selling price.

Profit (Gain) : When S.P.>C.P., then there is a profit.

Profit = S.P. – C.P.

Loss : When C.P. > S.P, then there is a loss.

Loss = C.P. – S.P.

Note : Gain or Loss is always calculated on C.P.

Formulae Profit% = Pr. .ofit

C P ×FH IK100 % ...(1)

and Loss% = LossC P. . ×FH IK100 % ...(2)

Formula (1) gives

Profit = C P ofit. . Pr %×100

or S.P. – C.P. = C P ofit. . Pr %×100

or S.P. = C P C P ofit. . . . Pr %+ ×100

or S.P. = 100100+FH IK ×Pr % . .ofit C P ...(3)

Similarly, Formula (2) gives

Loss =C P Loss. . %×

100

or C.P. – S.P =C P Loss. . %×

100

or S.P. = C P C P Loss. . . . %− ×100

or S.P. =100

100−F

HGIKJ ×

Loss C P% . . ....(4)

Let us now consider some examples to illustrate the applications of these formulae in solvingproblems related to profit and loss.


Examples 10.11 : A shopkeeper bought an almirah from a wholesale dealer for Rs 4500 andsold it for Rs 6000. Find his profit or loss percent.

Solution : Here C.P. of the almirah = Rs 4500

S.P. of the almirah = Rs 6000

Since S.P. > C.P., there is a profit

Profit = S.P. – C.P.

= Rs 6000 – Rs 4500

= Rs 1500

∴ Profit % = Pr. . %ofit

C P ×FH IK100

=15004500

100×FHG

IKJ%

=100

3% i.e. 33 1

3% .

Example 10.12 : A retailer buys a cooler for Rs 3800 and overhead expenses on it areRs 50. If he sells the cooler for Rs 4400, determine his profit percent.

Solution : Here, C.P. of the cooler = Rs (3800 + 50) = Rs 3850

S.P. of the cooler = Rs 4400

Since S.P. > C.P., there is a profit

Profit = Rs 4400 – Rs 3850

= Rs 550

∴ Profit % = Pr. . %ofit

C P ×FH IK100

= 5503850 100×FH IK%

= 1007 % i.e. 14 2

7 % .

Example 10.13 : By selling a scooter to a customer for Rs 22400 an auto-dealer makes a profitof 12%. Find the cost price of the scooter.

Solution : Here, S.P. of the scooter = Rs 22400, Profit % = 12%

Using formula (3), we have

S.P. =100

100+F

HGIKJ ×

Pr % . .ofit C P

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or C.P. =S P

ofit. .

Pr %×

+100

100

= Rs 22400 100

100 12×+

= Rs 22400 100

112×

= Rs 20000

Thus, the cost price of the scooter is Rs 20000.

Example 10.14 : By selling a cycle for Rs 2024, a cycle merchant loses 12%. If he wishesto make a gain of 12%, what will be the selling price of the cycle ?

Solution : First Part : S.P. = Rs 2024 and Loss % = 12%

From formula (4), we have

C.P. =S P

Loss. .

%×

−100

100

= Rs 2024 100100 12

×−

= Rs 2024 100

88×

= Rs 2300

Second Part : C.P. of the cycle = Rs 2300 and Gain (Profit)% = 12%

Using formula (3), we have

S.P. = 100100+FH IK ×Pr % . .ofit C P

= 100 12100 2300+ ×

= Rs 112 2300100×

= Rs 2576

Thus, the selling price of the cycle will be Rs 2576.

Example 10.15 : If the cost price of 15 articles is the same as the selling price of 12 articles,find the gain or loss percent in the transaction.

Solution : Let C.P. of an article be Rs x, Then

C.P. of 15 articles = Rs 15x ...(i)


or S.P. of 12 articles = Rs 15x

or S.P. of 1 articles = Rs 1512 xFH IK ...(ii)

Since S.P. > C.P., there is a profit (gain) in the transaction

Gain = Rs 1512

x x−FH IK

= Rs 312

xFH IK or Rs x4FH IK

or Gain % = GainC P. . %×FH IK100

=xx4 100×FHG

IKJ%

= 14 100×FH IK% = 25%

Thus, the gain in the transaction is 25%.

Example 10.16 : By selling 45 oranges for Rs 160, a women loses 20%. How many orangesshould she sell for Rs 112 to gain 20% on the whole ?

Solution : First Part : S.P. of 45 oranges = Rs 160

Loss% = 20%

So, by formula (4), we have

C.P. of 45 oranges = S PLoss

. .%

×−

100100

= Rs 160 100100 20

×−

= Rs 160 10080× = Rs 200.

Second Part : C.P. of 45 oranges = Rs 200

Gain% = 20%

∴ S.P. of 45 oranges = 100+ ×Gain%100 C.P.

= 100 20100 200+ × Rs

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= Rs 120 200

100

×

= Rs 240

Now, number of oranges for Rs 240 = 45

Number of oranges for Rs 112 = 45

240112× = 21

Thus, the women should sell 21 oranges for Rs 112.


1. A shopkeeper buys an article for Rs 320 and sells it for Rs 240. Find his gain or losspercent.

2. A dealer buys a wrist watch for Rs 450 and spends Rs 30 on its to repairs. If he sellsthe same for Rs 600, find the profit per cent.

3. A dealer sold two machines at Rs 2400 each. On selling one machine he gained 20% andon selling the other he lost 20%. Find the dealer’s gain or loss percent.

4. A sells an article costing Rs 1000 to B and earns a profit of 6%. B, in turn sells it toC at a loss of 5%. At what price did C purchase the article ?

5. By selling 90 ball pens for Rs 160, a person loses 20%. How many ball pens should hesell for Rs 96, so as to have a gain of 20% ?

6. If the selling price of 20 articles is equal to the cost price of 23 articles, find the lossor gain percent.

7. A watch was sold at a profit of 12%. Had it been sold for Rs 33 more, the profit wouldhave been 14%. Find the cost price of the watch.

8. By selling a book for Rs 258, a publisher gains 20%. For how much should he sell itto gain 30% ?

9. A vendor bought bananas at 6 for 5 rupees and sold them at 4 for 3 rupees. Find his gainor loss percent.

10.8.2 Simple Interest

All the transactions that take place around us involve money. Sometimes, a person has to borrowsome money as a loan from his friends, relatives, bank etc. He promises to return it after aspecified time period. So, he has to give back not only the money he borrows but also someextra money to the lender for using his money.

The money borrowed is called the principal, usually denoted by P. The extra money paid iscalled interest, usually denoted by I.


The sum of principal and the interest is called the amount usually denoted by A.

A = P + I

Interest is calculated on the principal

Interest is mostly expressed as a rate percent per year (per annum).

Interest depends on how much money (P) has been borrowed and the duration of the time (T)for which it is borrowed. Interest is calculated according to an agreement, which specifies acertain percent of the principal for each year’s use, called the rate of interest.

I = P × R × T

Interest as calculated above is called simple interest.

Example 10.17 : A man borrowed Rs 50000 from a finance company for buying a motor bicyclefor a period of 2 years. If the finance company charged simple interest at the rate of 15% perannum, how much interest was paid by the man to the finance company.

Solution : Here, Principal (P) = Rs 50000

Time (T) = 2 years

Rate (R%) = 15% = 0.15

We have I = P × R × T

= Rs (50000 × 0.15 × 2)

= Rs 15000

Hence, the man paid Rs 15000 as the interest to the finance company.

Example 10.18 : A certain sum of money was deposited for 5 years. Simple interest at therate of 12% was paid. Calculated the sum deposited if the simple interest received by thedepositor is Rs 1200.

Solution : Let the sum deposited be Rs P.

Given that I = Rs 1200; T = 5 years; R = 12%

We have I = P × R × T

or P = IR T×

= Rs 1200012 5. × = Rs 2000

Thus, the sum deposited was Rs 2000.

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Example 10.19 : At what rate of simple interest will a sum of Rs 3000 become Rs 4920 atsimple interest in 4 years ?

Solution : Here, P = Rs 3000; A = Rs 4920; T = 4 years

We have I = A – P = Rs 4920 – Rs 3000 = Rs 1920

Now, I = P × R × T

⇒ R = IP T× = Rs 1920

3000 4×

= 16100 = 16%

Thus, at the rate of 16%, Rs 3000 will become Rs 4920 in 4 years.

Example 10.20 : In what time will Rs 8000 amount to Rs 12000, if simple interest is chargedat the rate of 6% per annum ?

Solution : Here, P = Rs 8000; A = Rs 12000; R = 6% = 0.06

We have I = A – P

= Rs 12000 – Rs 8000 = Rs 4000

Now, I = P × R × T

⇒ T = IP R× = 4000

8000 0 06× .

= 10012 = 8 1

3 years

= 8 years 4 months

Thus, in 8 years 4 months, Rs 8000 will amount to Rs 12000 at the rate of interest 6% perannum.


1. Ramesh borrowed Rs 7000 from his friend at 8% per annum simple interest. He returnedthe money after 2 years. How much did he pay back altogether ?

2. Jaya deposited Rs 15600 in a bank. The bank pays interest at 8% per annum. Find theinterest she will receive at the end of 3 years.

3. Subnam lent Rs 25000 to her friend. She gave Rs 10000 at 10% per annum and theremaining at 12% per annum. How much interest did she receive in 2 years ?

4. Nalini borrowed Rs 5000 from her friend at 8% per annum. She returned the money after6 months. How much amount did she pay to her friend ?


5. Find the interest received by Anil if he deposits Rs 16000 for 8 months at the rate of9% per annum. Also, find the amount.

6. Shalini deposited Rs 14500 in a finance company for 3 years and received Rs 4785 assimple interest. What was the rate of interest per annum ?

7. In how many years will Rs 8000 amount to Rs 16000, if simple interest is earned at therate of 12% per annum ?

8. In how much time will simple interest be 14 th of the principal at the rate of 10% per

annum ?

9. In which case, is interest earned more : (a) Rs 5000 deposited for 5 years at 4% per annum(b) Rs 4000 deposited for 6 years at 5% per annum ?

10. At what rate of interest will simple interest be half the principal in 5 years ?

10.9 DISCOUNT

You must have seen all around advertisements of the following types, especially during thefestival seasons.

SALE

Discount upto 50%

A discount is a reduction in the marked (or list) price.

“25% discount” means a reduction of 25% in the marked price of an article. For instance, ifthe marked price of an article is Rs 100, it is sold for Rs 75, i.e., Rs 25 less than the markedprice.

Note. Discount is always calculated on Marked Price.

Marked Price (or list price) : The marked price (M.P) of an article is the price at which itis listed for sale.

Discount : The discount is the reduction from the marked price of the article.

Net Selling price : In case of discount sale, the price of the article obtained by subtractingdiscount from the list price is called the Net selling Price.

Let us consider the following example to illustrate.

Example 10.21 : A shirt with marked price Rs 165 is sold at a discount of 10%. Find its netselling price.

Solution : Here, Marked Price (M.P.) of the shirt = Rs 165, Discount = 10%

∴ Net selling price = Marked price – Discount

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= Rs 165 – 10% of Rs 165

= Rs 165 – Rs 10100 165×FH IK

= Rs 165 – Rs 16.50

= Rs 148.50

Thus, the net selling price of the shirt is Rs 148.50.

Aliter

Since the discount offered is 10%

S.P. = M.P. – 10% of M.P.

= 90% of M.P.

or S.P. = 90100 × Rs 165

= Rs 148.50

Example 10.22 : A pair of socks is marked at Rs 40 and is being offered at Rs 32. Find thediscount per cent being offered.

Solution : Here, M.P. = Rs 40; and S.P. = Rs 32

So, Discount = M.P. – S.P.= Rs 40 – Rs 32 = Rs 8

∴ Discount % = DiscountM P. . ×100

= 840 100×

= 20

Hence, the discount being offered is 20%.

10.9.1 Discount Series

Sometimes a manufacturer, offers a discount of 10% on discounted price in addition to aprevious discount of 20%, because suddenly he gets a supply of cloth at a very low price. Hemay allow another discount of 5% on the discounted price, to some of his customers for promptpayments. In other words, he allows a discount series.

In a discount series, the first figure denotes the discount on the list price, the seconddenotes the discount on the discounted price and so on.

If a shirt is marked for Rs 120 and a discount series 20%, 10% and 5% is offered, thencomputation for calculating net selling price is as under :


Marked price Rs 120 with a discount series 20%, 10% and 5%.

20% discount on Rs 120 = Rs 120 20100× = Rs 24

∴ Discounted price = Rs (120 – 24) = Rs 96

10% discount on Rs 96 = Rs 96 10100× = Rs 9.60

∴ Discounted price = Rs (96 – 9.60) = Rs 86.40

5% discount on Rs 86.40 = Rs 86.40 × 5100 = Rs 4.32

Net selling price = Rs (86.40 – 4.32) = Rs 82.08.

10.9.2 Conversion of Discount Series to a Single Discount

Instead of computing a series of discounts one by one, it is sometimes more convenient to reducethe series to a single discount.


Example 10.23 : Convert the discount series 20%, 10% and 5% to an equivalent singlediscount.

Solution : Let the list price = Rs 100


Discounted price = Rs (100 – 20) = Rs 80


Discounted price = Rs (80 – 8) = Rs 72

and 5% discount on Rs 72 = Rs 72 5100× = Rs 3.60

Discounted price = Rs (72 – 3.60) = Rs 68.40

∴ Single discount on Rs 100 = Rs (100 – 68.40) = 31.60 or 31.6%

Example 10.24 : An old scooter is sold at three successive discounts of 10%, 5% and 2%.If the marked price of the scooter is Rs 18000, find the selling price of the scooter.

Solution : Here, list price = Rs 18000

First discount of 10% = Rs 18000 × 10100 = Rs 1800

Price after first discount = Rs (18000 – 1800) = Rs 16200

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Second discount of 5% = Rs 16200 × 5100 = Rs 810

Price after second discount= Rs (16200 – 810) = Rs 15390

Third discount of 2% = Rs 15390 × 2100 = Rs 307.80

Price after third discount = Rs (15390 – 307.80) = Rs 15082.20Thus the net selling price of the scooter is Rs 15080.20AliterNet selling price of the scooter = (100 – 2)% of (100 – 5)% of (100 – 10)% of Rs 18000

= Rs 18000 98100

95100

90100× × ×FH IK

= Rs 15082.20Example 10.25 : Find the single discount equivalent to the discount series of 20%, 15%and 10%.

Solution : Let the marked price = Rs 100

Price after the given discount series

= (100 – 10)% of (100 – 15)% of (100 – 20)% of Rs 100= 0.90 × 0.85 × 0.80 × Rs 100= Rs 51.20

Hence, the total discount = M.P. – S.P.= Rs 100 – Rs 51.20= Rs 48.80

Hence, the equivalent single discount= Rs 48.80 on M.P. of Rs 100= 48.8%

Example 10.26 : A dealer buys a table listed at Rs 1500 and gets successive discounts of 20%and 10%. He spends Rs 20 on transportation and sells it at a profit of 10%. Find the sellingprice of the table.

Solution : Here, list price of the table = Rs 1500

Price after a discount series of 20% and 10% = (100 – 10)% of (100 – 20)% of Rs 1500

= 90100

80100× × Rs 1500

= Rs 1080


Since the dealer spends Rs 20 on transportationC.P. of the table = Rs 1080 + Rs 20 = Rs 1100

Profit = 10%

∴ S.P. of the table = 100100+FH IK ×Pr % . .ofit C P

= 100 10100 1100+ × Rs

= 110100 1100×Rs

= Rs 1210Thus, the selling price of the table is Rs 1210.


1. A coat is marked at Rs 1200. Find its selling price if a discount of 15% is offered.

2. A man pays Rs 2100 for a machine listed at Rs 2800. Find the rate of discount offered.

3. An article listed at Rs 2650 is sold at a discount of 10%. Due to festival season, theshopkeeper allows a further discount of 5%. Find the selling price of article.

4. Find a single discount equivalent to a discount series given in each of the followingdiscount series :

(a) 25%, 20% and 10%

(b) 20%, 15% and 10%

(c) 20%, 10% and 5%.

5. Which of the following discount series is better for a customer ?

20%, 10% and 5% OR 10%, 5% and 20%.

6. The list price of a table fan is Rs 840 and it is available to a retailer at 25% discount.For how much should the retailer sell it to earn a profit of 15% ?

7. The marked price of a TV set is Rs 25000. A discount series of 20%, 10%, 5% is allowedon it. How much money does one have to pay for the TV set ?

8. If a shopkeeper marks his goods 50% more than their cost price and allows a discountof 40%, find his gain or loss per cent.

9. The list price of a watch is Rs 320. After two successive discounts it is sold for Rs 244.80.If the first discount is 10%, what is the rate of second discount ?

10. A retailer buys shirts from a manufacturer at the rate of Rs 75 per shirt and marked themat Rs 100 each. He allows some discount and gets a profit of 30% on the cost price. Whatpercentage discount does he allow to his customers ?

248 Mathematics

10.9.3 Sales Tax

Government levies some taxes to have earning called revenue. One such tax which is leviedon the sale of goods is called sales tax. The rates of sales tax are different for differentcommodities. Some essential commodities are exempted from sales tax. This tax is chargedon the net selling price of commodities and its rate is expressed as a percentage.

For example, if an article is sold for Rs 750 and the rate of sales tax is 8%, then

Sales tax = Rs 750 8100× = Rs 60

Price inclusive of sales tax = Rs (750 + 60) = Rs 810

The customer will have to pay Rs 810.

Example 10.27 : The marked price of a pair of shoes is Rs 320. If the rate of sales tax is 4%,calculate the amount to be paid by a customer for the purchase of the shoes.

Solution : Marked price of shoes = Rs 320

Rate of sales tax = 4%

∴ Sales tax = 4% of Rs 320

= 4100 × Rs 320

= Rs 12.80

Thus, the customer has to pay (Rs 320 + Rs 12.80) = Rs 332.80 for purchasing the shoes.

Examples 10.28 : Anita purchased a shirt for Rs 594 including sales tax. If the rate of salestax is 8%, find the list price of the shirt.Solution : Let the list price of the shirt be Rs P

Then P + 8% of P = 594or 108% of P = 594

or108100

P= 594

or P = 594108. = 550

Thus, the list price of the shirt is Rs 550.

Example 10.29 : Hari Om bought a radio set for Rs 1870, after getting 15% discount on thelist price and then 10% sales tax on the reduced price. Find the list price of the radio set.

Solution : Let the list price or the radio set be Rs P.

Thus, selling price of the radio after discount= Rs P – 15% of Rs P

= 85% of Rs P


= Rs 85100

P

Sales tax = 10% of Rs 85100FH IK P

= 10100

85100× P

= Rs 851000 P

The net price to be paid for the purchase of the radio set is

Rs 85100

851000P P+FH IK = Rs 935

1000 P

Equating it to Rs 1870, we get

9351000 P = 1870

or P = 1870 1000935× = 2000

Thus, the list price of the radio set is Rs 2000.Example 10.30 : The list price of a washing machine is Rs 9000. The dealer allows a discountof 5% on the cash payment. How much money will a customer pay to the dealer in cash, ifthe rate of sales tax is 10% ?

Solution : Here, list price = Rs 9000 and discount = 5%

∴ Cash price of the washing machine

= Rs 9000 5100 9000− ×FH IK

= Rs (9000 – 450)

= Rs 8550

∴ Sales tax = 10% of Rs 8550

= 10100 × Rs 8550

= Rs 855

Hence, the customer has to pay Rs 8550 + Rs 855 = Rs 9405 for the purchase of the washingmachine.

Example 10.31 : The list price of the air-conditioner is Rs 25630. The rate of sales tax is 10%The customer requests the dealer to allow a discount to such an extent that the price of the

250 Mathematics

air-conditioner amounts to Rs 25630 inclusive of sales tax. Find the discount in the price ofthe air-conditioner.

Solution : Let Rs P be the net price exclusive of sales tax.

Then, price + sales tax = P P+ 10100 = 110

100 P

This is given as Rs 25630

∴ 110100 P = 25630

or P = 23300

∴ Discount allowed = Rs (25630 – 23300) = Rs 2330


1. The marked price of a sewing machine is Rs 3500. If the sales tax on sewing machineis charged at the rate of 6%, find how much a customer has to pay for purchasing themachine.

2. Amita purchases a pair of socks whose list price is Rs 44. The shopkeeper charges salestax at the rate of 5%. Find how much money Amita has to pay for purchasing the socks.

3. Mrs. Mohini purchased a saree for Rs 1100 including sales tax. If the list price of thesaree is Rs 1000, find the rate of sales tax charged.

4. A refrigerator is available for Rs 13915 including sales tax. If the rate of sales tax is 10%,find the selling price of the refrigerator.

5. Radhika purchased a car with a marked price of Rs 2.1 lakhs at a discount of 5%. If thesales tax is charged at the rate of 12%, find the amount Radhika had to pay for purchasingthe car.

6. Dayakant bought a set of cosmetic items for Rs 345 including 15% sales tax and a pursefor Rs 110 including 10% sales tax. What percent is the sales tax charged on the wholetransaction ?

[Hint. C.P. of cosmetic items = Rs (345 ÷ 1.15) ;

and C.P. of purse = Rs (110 ÷ 1.1)]

7. Kamal wants to buy a suitcase whose list price is Rs 504. The rate of sales tax is 5%.He requests the shopkeeper to reduce the list price to such an extent that he has to payRs 504 only. Calculate the discount given in the price of the suit-case.

10.10 COMMISSION

Manufacturers of goods, farmers and owners of properties, frequently uses the services of amiddle man to find a buyer in order to sell their goods or properties. The middle man is calledan agent, who gets some money for the services rendered by him. This money paid is calledcommission. In general, commission is expressed in terms of percentage.


Example 10.32 : A book agent sold 140 books at Rs 20 each. His commission was 25%. Howmuch money did he earn as commission ?

Solution : Total price of books = Rs (20 × 140) = Rs 2800Amount of commission = 25% of Rs 2800

= 25100 2800× Rs

= Rs 700Thus, the book agent earns Rs 700 as commission.Example 10.33 : A salesman earns Rs 300 as commission at the rate of 8%. For what amountdid he sell the goods ?

Solution : Let the amount be Rs x

Then x × 8100 = 300

⇒ x = 300 1008×

= 3750The sales man sold goods worth Rs 3750.

Example 10.34 : A commission merchant charged Rs 427.50 for selling 1500 packets of salt,at 3% commission. At what price per bag did he sell the salt ?

Solution : Let the selling price per packet be Rs x.

Then amount of commission

= 3% of Rs (1500 × x)

= Rs 1500 3100x ×FH IK = Rs 45x

But this is given to be Rs 427.50.Thus, 45x = 427.50

or x = 427 5045

.

or x = 9.5Thus, the selling price per packet of salt was Rs 9.50.


1. A commission merchant sells 1200 tins of oil at Rs 270 a tin on commission of 2 12 %.

Find the amount of the commission. Also find the net proceeds.

252 Mathematics

2. An auctioneer sold a property worth Rs 17.8 lakhs. His commission was 1.25%. Howmuch did the auctioneer earn as commission ?

3. A commission merchant charged Rs 4212 as commission at rate of 3% for selling riceat Rs 450 per bag. How many bags of rice did he sell ?

4. A commission merchant sells a certain amount of goods at a commission of 10% in thefirst two weeks of a month. His commission is then raised to 12%, and he sells an equalamount of goods in the remaining part of the month. If his total earning (commission)were Rs 8000, how much did he sell altogether ?

10.11 INSTALMENT BUYING

With the cost of articles going up day by day it has become difficult for the common man tobuy some articles like scooter, fridge, colour TV etc. which are needed by him and his family.Such articles are available on easy instalments. We shall study about instalment purchasescheme.

10.11.1 What is Instalment Buying ?

Instalment purchase scheme, enables a person to buy costly goods like colour TV sets,refrigerator, video cameras, scooters etc. on convenient terms of payment. In this scheme, thecustomer does not make full payment of the cost of the article at the time of purchase, butmakes a partial payment in the beginning and takes away the article for use. The remainingpayment is made in easy monthly, quarterly or half yearly instalments, as per the agreementsigned between the customer and the shopkeeper.

Cash Price : It is the amount for which the article can be purchased on full payment i.e., theselling price of the article.

Cash down Payment : It is the partial payment made by the customer at the time of signingthe agreement and taking away the article for use. In fact, it is a part of the selling price.

Instalments : It is the amount which is paid by the customer at regular intervals towards theremaining part of the selling price of the article.

It may be noted that in the instalment plan only part payment of the total cost is paid by thecustomer at the time of purchase. The remaining part of the cost is paid on subsequent dates;and therefore the seller charges some extra amount for deferred payments. This extra amountis actually the interest charged on the amount of money which the customer owes to the sellerat different times of instalments paid.

In the following, we solve a few examples to illustrate the process.

Example 10.35 : Bimla buys a sewing machine, which is available for Rs 2600 cash paymentor under an instalment plan for Rs 1000 cash down payment and 3 monthly instalment ofRs 550 each. Find the rate of interest charged under the instalment plan.

Solution : Cash payment price = Rs 2600

Cash down payment = Rs 1000


Balance to be paid in instalments = Rs 1600

Amount paid in 3 instalments = Rs 550 × 3 = Rs 1650

Interest charged in the instalment plan = Rs (1650 – 1600) = Rs 50

The buyer owes to the seller for the 1st month = Rs 1600

The buyer owes to the seller for the 2nd month = Rs (1600 – 550) = Rs 1050

The buyer owes to the seller for the 3rd month = Rs (1050 – 550) = Rs 500

Total = Rs 3150

∴ She has to pay interest on Rs 3150 for 1 month.

But the interest she has to pay = Rs 50

If R% is the rate of interest p.a.

then, 3150 × 112 100× R = 50

∴ R = 50 12 1003150× × %

= 40021 %

= 19 121 %

∴ Rate of interest paid by her in the instalment plan is 19 121 %.

Example 10.36 : A computer is available for Rs 34000 cash or Rs 20000 cash down paymenttogether with 5 equal monthly instalments. If the rate of interest charged under the instalmentplan is 30% per annum, calculate the amount of each instalment.

Solution : Cash price = Rs 34000

Cash down payment = Rs 20000

Balance to paid in 5 equal installments = Rs 14000

Let each instalment be = Rs PInterest charged under instalment plan = Rs (5P – 14000)The buyer owes to the seller

For the 1st month = Rs 14000Fr the 2nd month = Rs (14000 – P)

For the 3rd month = Rs (14000 – 2P)For the 4th month = Rs (14000 – 3P)For the 5th month = Rs (14000 – 4P)

Total = Rs (70000 – 10P)

254 Mathematics

Thus, he had to pay interest on Rs (70000 – 10P) for 1 month at the rate of 30% p.a.

∴ (70000 – 10P) × 112

30100× = 5P – 14000

or (70000 – 10P) × 140 = 5P – 14000

or 70000 – 10P = 40(5P – 14000)

or 70000 – 10P = 200P – 560000

or 210 P = 560000 + 70000

or P = 630000210

= 3000

∴ Amount of each instalment = Rs 3000.


1. A T.V. set is available for 21000 cash or for Rs 4000 cash down payment and 6 equalmonthly instalments of Rs 3000 each. Calculate the rate of interest charged under theinstalment plan.

2. Anil purchased a type writer priced at Rs 6800 cash payment under the instalment planby making a cash down payment of Rs 2000 and 5 monthly instalments of Rs 1000 each.Find the rate of interest charged under the instalment plan.

3. A scooter is available for Rs 30000 cash or for Rs 15000 cash down payment and 4 equal

monthly instalments. If the rate of interest charged under the instalment plan is 33 13 %,

find the amount of each instalment.

4. A microwave oven is available for Rs 9600 cash or for 4000 cash down payment and

3 equal monthly instalment. If the rate of interest charged is 22 29 % per annum, find the

amount of each instalment.

LET US SUM UP

Percent means ‘per hundred’.

Percents can be written as fractions as well as decimals and vice-versa.

To write a percent as a fraction, we drop the % sign and divide the number by 100.

To write a fraction as a percent, we multiply the fraction by 100, simplify it and suffixthe % sign.


To determine the specific percent of a number or quantity, we change the percent to afraction or a decimal and then multiply.

When the selling price is more than the cost price of the goods, there is a profit (or gain)

When the selling price is less than the cost price of the goods, there is a loss

Profit (Gain) = S.P. – C.P. ; Loss = C.P. – S.P.

Gain% = GainC P. . ×100 ; Loss = Loss

C P. . ×100

Further S.P. = 100100+ ×Gain C P% . . ; S.P. = 100

100− ×Loss C P% . .

The simple interest (S.I.) on a principal (P) at the rate of R% for a time T years, iscalculated, using the formula

S.I. = P × R × T

Discount is a reduction in the list price of goods.

Discount is always calculated on the marked price of the goods.

(Marked price – discount), gives the price, which a customer has to pay while buyingan article.

Two or more successive discounts are said to form a discount series.

A discount series can be reduced to a single discount.

Sales tax is charged on the sale price of goods.

Commission is paid to an agent for his services in arranging the sale or purchase of goodsfrom some one else.

An Instalment plan enables a person to buy costlier goods.

TERMINAL EXERCISE

1. Write each of the following as a per cent :

(a) 720 (b) 0.25 (c) 1.4 (d) 0.07

2. Write each of the following as a decimal :

(a) 63% (b) 13% (c) 3% (d) 0.3%

3. Write each of the following as fraction :

(a) 0.13% (b) 1.3% (c) 11.3% (d) 113%

256 Mathematics

4. Find each of the following :

(a) 37% of 400 (b) 3.5% of 800

5. What percent of 700 is 294 ?

6. By what percent is 60 more than 45 ?

7. Find the number whose 15% is 270 ?

8. What number increased by 10% of itself is 352 ?

9. What number decreased by 7% of itself is 16.74 ?

10. Arun at the beginning of a year had a bank balance of its 17500 and at the end of theyear he had a balance of Rs 21350. By what percent did his balance increase ?

11. A man loses 25% by selling a scooter for Rs 8400. For what amount did he buy thescooter ?

12. A commission merchant charged Rs 700 as commission for selling 100 bags of cottonat 5%. Find the price of one bag at which it was sold ?

13. Shalim deposited Rs 14000 in a bank for 2 years and received Rs 4200 as simple interest.At what rate per annum, was interest paid to him ?

14. Simple interest on a sum of money is 13 rd of the sum itself and the number of years is

thrice the rate percent. Find the rate of interest.

15. Ahmad purchased a bicycle by making a cash down payment of Rs 400 and 3 monthlyinstalments of Rs 275 each. The bicycle was also available on cash payment of Rs 1200.Find the rate of interest per annum charged under the instalment plan.

16. Rita purchased a washing machine for Rs 4000 cash down payment and 4 equal monthlyinstalments. The washing machine was also available for Rs 15000 cash payment. If therate of interest charged in the instalment plan is 18% p.a. Find the amount of eachinstalment.


ANSWERS


1. (a) 56% (b) 3% (c) 75% (d) 2%

(e) 97% (f) 80% (g) 4% (h) 140%

2. (a) 0.75 (b) 0.14 (c) 0.03 (d) 1.15

(e) 0.02 (f) 0.25 (g) 4 (h) 3.5

3. (a)35

100 ; 35 : 100 (b)40

100 ; 40 : 100

(c)70

100 ; 70 : 100 (d)85

100 ; 85 : 100

4. (a) 75% (b) 20% (c) 30% (d) 16%

5. 76% 6. 50% 7. 37.5% 8. 84%

9. 55% 10. 75%` 11. 30% 12. 25%

13. 30% 14. 75% 15. 62% ; 62.5 %; Y16. 5%.


1. (a) 66 (b) 200 (c) Rs 564 (d) 663 metres

2. 240 students 3. 9 boys ; 27 girls 4. 45 kg 5. Rs 1540;

6. 4 boys 7. 36 minutes 8. 1 lakh 9. 10%

10. 40% 11. 20% 12. 6000 votes 13. Rs 3000

14. Rs 90 ; Rs100 15. Rs 40; Rs 32 16. 1% decrease 17. Rs 14000


1. Loss 25% 2. Profit 25% 3. 4% loss 4. Rs 1007

5. 36 ball pens 6. 15% gain 7. Rs 1650 8. Rs 279.50

9. 10% loss


1. Rs 8120 2. Rs 3744 3. Rs 5600 4. Rs 5200

5. Rs 960 ; Rs 16960 6. 11% 7. 8 years 4 months

8. 2 years 6 months 9. (b) 10. 10%

258 Mathematics


1. Rs 1020 2. 25% 3. 2265.75

4. (a) 46% (b) 38.8% (c) 31.6%

5. Both same 6. Rs 724.50 7. Rs 17,100 8. Loss 10%

9. 15% 10. 2.5%


1. Rs 3710 2. 46.20 3. 10% 4. Rs 12650

5. Rs 2,23,440 6. 13 34 % 7. Rs 24


1. Rs 8100 2. Rs 22250 3. 312 bags 4. Rs 80,000


1. 21 119 % 2. 17 1

7 % 3. Rs 4000 4. Rs 2000

Terminal Exercise

1. (a) 35% (b) 25% (c) 140% (d) 7%

2. (a) 0.63 (b) 0.13 (c) 0.03 (d) 0.003

3. (a) 1310000 (b) 13

1000 (c) 1131000 (d) 113

100

4. (a) 148 (b) 28 5. 42% 6. 33 13 %

7. 1800 8. 320 9. 18 10. 22%

11. Rs 11200 12. Rs 140

13. 15% 14. 3 13 % 15. 19 1

21%

16. Rs 2850.86

Compound Interest 259

11

Compound Interest

11.1 INTRODUCTION

You have already studied about simple interest in your earlier classes. Recall that interest isthe extra money paid by the user to the bank or money lender for the use of his money forsome specified time period.

If the money is retained by the user for the next time period also, the user will pay intereston the amount borrowed as well as on the interest accrued for the first time period. Thus, theuser will have to pay more interest for the second period of time.

This type of interest is called compound interest. In this lesson, you will learn to calculatecompound interest.

11.2 OBJECTIVES


illustrate the concept of compound interest.

calculate compound interest.

find the difference between simple interest and compound interest.

calculate amount of a sum invested for a given time.

forecast the population of a town when rate of growth of population is given for a specificperiod of time.

calculate the depreciated value of a machine, a vehicle, a building etc.


Knowledge of percentage

Knowledge of simple interest.

11.4 THE CONCEPT OF COMPOUND INTEREST

Recall that interest is the extra money paid by the user to the bank for the use of money. Themoney which has been borrowed is called the principal (P), the time interval for which the

260 Mathematics

money has been taken is called the time duration (T) and the extra money paid by the userto the bank upto the time the principal amount is returned is called the rate of interest (R).This is expressed as percentage of the principal.

When the interest is calculated on the principal for the entire time period of loan, the interestis called simple interest and is given by

S.I = PRT

But if this interest is due (not paid) after the decided time period, then it becomes a part ofthe principal and added to the principal for the next time period. The interest is calculated forthe next time period on the new principal. This way of calculating interest is called compoundinterest.

The time period after which the interest is added to the principal for the next timeperiod is called the conversion period.

The conversion period may be one year, six month, three months or a month and the interestis said to be compounded annually, half yearly, quarterly or monthly respectively.

Let us go through the following example :

Suppose you borrow Rs 2000 from a bank at the rate of interest 10% per annum.

Interest for one year = Rs 2000 10100 1× ×FH IK

= Rs 200

If this interest is not paid at the end of one year, then it becomes a part of principal for thesecond year.

Therefore, the principal for the second year becomes Rs (2000 + 200) = Rs 2200

Now interest for second year

= Rs 2200 10100 1× ×FH IK

= Rs 220

∴ Amount payable at the end of two years

= Rs (2200 + 220) = Rs 2420.

The total interest to be paid at the end of two years

= Rs (200 + 220) = Rs 420.

The interest charged in this way is called the compound interest.

Thus for calculating the compound interest, the interest due after every year is added to theprincipal and then the interest is calculated for the next year on this new principal.


11.5 FORMULA FOR COMPOUND INTEREST

Let a sum P be borrowed for n years at the rate of r% per annum. We shall calculate thecompound interest payable after n years.

Here Principal = PRate = r%Time = n years

Interest on P at r% for the first year = P r× ×100 1 = Pr100

∴ Amount after one year = P P r+ = +FH IKPr100 1 100

Interest on P r1 100+FH IK at r% for the second year = P r r1 100 100+FH IK ×

Amount after two years = P r P r r1 100 1 100 100+FH IK + +FH IK ×

= P r r1 100 1 100+FH IK +FH IK

= P r1 1002

+FH IK(In compound interest, amount after one year becomes principal for second year)

Similarly, amount after 3 years = P r1 1003

+FH IK and so on.

Thus, amount after n years = P r n1 100+FH IK

If A represents amount and R represents r100 , then

A = P (1 + R)n

Also Compound Interest = A – P= P(1 + R)n – P= P[(1 + R)n – 1]

Note : (i) Simple interest and compound interest are equal for the first year.

(ii) The time period is generally taken in years but this is not necessary. The interestcan be compounded yearly, half-yearly (semi-annually), quarterly (after 3 months)or even monthly. The unit of time after which interest is compounded is calledConversion period.

262 Mathematics

Example 11.1 : Calculate the compound interest on Rs 10000 for 2 years at 8% per annum.

Solution : Here P = Rs 10000, R = 8% and n = 2

∴ C.I. = P[(1 + R)n – 1]

= Rs 10000 1 8100 1

2+FH IK −LNM

OQP

= Rs 10000 2725

2725 1× −LNM OQP

= Rs 1664.

Example 11.2 : Calculate the compound interest on Rs 10000 for 2 years at the rate of 8%per annum when the interest is compounded half yearly.

Solution : Here P = Rs. 10000, R = 82 or 4% Half yearly and n = 4 Half years

∴ A = P (1 + R)n

= 10000 1 4100

4+FH IK

= Rs 10000 26 26 26 2625 25 25 25× × × ×

× × ×= Rs 11698.58

∴ C.I. = Rs (11698.58 – 10000)= Rs 1698.58

Example 11.3 : Calculate the compound interest on Rs 10000 for 1 year at the rate of 8%per annum when the interest is compounded quarterly.

Solution : Here P = Rs 10000, R = 84 or 2% quarterly and n = 4 quarters

∴ A = P (1 + R)n

= Rs 10000 1 2100

4+FH IK

= Rs 10000 51 51 51 5150 50 50 50× × × ×

× × ×

= Rs 10824.32

∴ C.I. = Rs (10824.32 – 10000) = Rs 824.32

Example 11.4 : Calculate the compound interest on Rs 12000 for 1½ years at the rate of 10%per annum when the interest is compounded annually.

Solution : Here P = Rs 12000, R = 10% and n = 1½ years


We note that the interest is compounded annually and therefore at the end of one year, theamount will be given by

A = P (1 + R)1

= Rs 12000 1 10100+FH IK

This A becomes the principal for the next 6 months or half year. The rate per half year willbe

R2 = 10

2 % = 5%

∴ The amount say A', after 1½ year will be given by

A' = A 1 5100+FH IK

= Rs 12000 1 10100 1 5

100+FH IK +FH IK

= Rs 12000 1110

2120× × = Rs 13860

∴ C.I. = Rs (13860 – 12000) = Rs 1860Example 11.5 : Find the sum of money which will amount to Rs 26010 in six months at therate of 8% per annum, when the interest is compounded quarterly.

Solution : Here P = ?, A = Rs 26010, R = 84FHGIKJ % or 2% quarterly and

n = 2 (two quarters)

∴ 26010 = P 1 2100

2+FH IK

= P 5150

2FH IK

∴ P = Rs 26010 50 5051 51× ×× = Rs 25000

Hence the required sum = Rs 25000

Example 11.6 : The difference between simple interest and compound interest for a certainsum of money at the rate of 10% per annum for 1½ years when the interest is compoundedsemi-annually, is Rs 183. Find the sum.

Solution : Let the sum be Rs 100

∴ S.I. = Rs 100 10 3100 2× ×× = Rs 15

264 Mathematics

C.I. = Rs 100 1 5100 1

3+FH IK −RST

UVW= Rs 1261

80

∴ Difference = C.I.– S.I. = Rs 126180 15−FH IK = Rs 61

80

If the difference is Rs 6180 , then sum = Rs 100

If the difference is Rs 183, then sum = Rs 100 80 18361

× ×

= Rs 24000Hence the required sum = Rs 24000

Example 11.7 : A sum of money becomes Rs 17640 in two years and Rs 18522 in 3 yearsat the same rate of interest when the interest is compounded annually. Find the sum and therate of interest per annum.

Solution : In compound interest, we know that

A = P (1 + R)n

Here 17640 = P (1 + R)2 ...(i)

and 18522 = P(1 + R)3 ... (ii)

Dividing (ii) by (i), we have

1 + R = 1852217640

∴ R = 1852217640 1− = 18522 17640

17640−

= 88217640

120= or 5

100

∴ Rate = 5%

Substituting the value of R in (i), we have

17640 = P 1 5100

2+FH IK

∴ P = Rs 17640 20 2021 21× ××

= Rs 16000

Hence the required sum is Rs 16000 and the rate of interest per annum is 5%.


ALITER

Interest for 3rd year = Rs (18522 – 17640)

= Rs 882.

Principal at the end of 2nd year = Rs 17640

If R is the rate % then

R = 88217640 1× = 1

20 or 5100 i.e. 5%

Amount after 2 years is given by

17640 = P 1 5100

2+FH IK where P is the principal

∴ P = Rs 17640 2021

2021× ×

= Rs 16000.

Example 11.8 : A sum of money is invested at compound interest for a year at 8% per annumwhen the interest is compounded half yearly. If the interest were compounded quarterly, it wouldhave fetched Rs 5201 more than in the pervious case. Find the sum.

Solution : Let the sum be Rs x

Here, R = 82 % or 4% half yearly and n = 2 half years

∴ Compound interest in Ist case

= Rs x x1 4100

2+FH IK −

LNM

OQP

= Rs 51625 x

Compound interest in 2nd case

= Rs x x1 2100

4+FH IK −

LNM

OQP

= Rs x 51 51 51 5150 50 50 50 1× × ×

× × × −LNM OQP= Rs 515201

6250000 x

∴ Difference = Rs 5152016250000

51625x x−LNM OQP

266 Mathematics

= Rs 52016250000 x

If the difference is Rs 52016250000 x , the sum = Rs x

If the difference is Rs 5201, the sum = Rs xx

××

×62500005201

5201

= Rs 6250000Hence the required sum is Rs 6250000Example 11.9 : A sum of Rs 15625 amounts to Rs 17576 at 8% p.a. compounded semi-annually.Find the time.

Solution : Here P = Rs 15625, A = Rs 17576, R = 82 % or 4% semi annually

Let the conversion periods be n half years

Now, A = P (1 + R)n

∴ 17576 = 15625 1 4100+FH IK

n

∴ 1 4100+FH IK

n= 17576

15625

2625FH IK

n= 26

253FH IK

∴ n = 3

Hence time = 3 Half years = 1 12 years.

Example 11.10 : Find the rate at which Rs 8000 amounts to Rs 9261 in 3 years, if the interestis compounded annually.

Solution : Here P = Rs 8000, A = Rs 9261, n = 3Let the rate of interest be R%

Now A = P (1 + R)n

∴ 9261 = 8000 ( 1 + R)3

i.e. (1 + R)3 = 92618000

= 2120

3FH IK∴ 1 + R = 21

20

i.e. R = 120

5100= = 5%

∴ Rate of interest is 5%.


Example 11.11 : Find the rate at which Rs 4000 will give Rs 630.50 as compound interestin 9 months, interest being compounded quarterly.

Solution : Here P = Rs 4000, A = Rs (4000 + 630.50) = Rs 4630.50 andn = 3 quarters

Let the rate of interest = R

Now A = P(1 + R)n

∴ 4630.50 = 4000 (1 + R)3

i.e. (1 + R)3 = 4630.504000

= 92618000 = FH IK21

203

∴ 1 + R = 2120

⇒ R = 120

5100= = 5% per quarter.

∴ Rate of interest = 20% p.a.


1. Calculate the compound interest on Rs 8000 for 3 years at 5% per annum when the interestis compounded annually.

2. Calculate the compound interest on Rs 62500 for 1 year at 8% per annum when the interestis compounded quarterly.

3. Calculate the compound interest on Rs 390625 for 2 years at 8% per annum when theinterest is compounded half yearly.

4. Find the sum of money which will amount to Rs 27783 in 3 years at 5% per annum, theinterest being compounded annually.

5. How much money will become Rs 194481 after 2 years at 10% per annum when theinterest is compounded semi-annually ?

6. Find the difference between simple interest and compound interest for 2 years at 10%per annum, when the interest is compounded semi-annually, on Rs 8000.

7. The difference between simple interest and compound interest for a certain sum of moneyat 8% per annum for 1½ years when the interest is compounded half-yearly is Rs 228.Find the sum.

268 Mathematics

8. A sum of money becomes Rs 18522 in three years and Rs 19448.10 in 4 years at thesame rate of interest when the interest is compounded annually. Find the sum and the rateof interest per annum.

9. A sum of money is invested at compound interest for 2 years at 10% per annum whenthe interest is compounded yearly. If the interest were compounded half yearly, it wouldhave earned Rs 881 more than in the previous case. Find the sum.

10. At what rate of interest per annum would the compound interest on Rs 12500 be Rs 9100in 1½ years, interest being compounded half-yearly ?

11.5 RATE OF GROWTH AND DEPRECIATION

In our day to day life, we observe about growth of population, plants, viruses etc. anddepreciation of articles like machinery, cars etc. after use. If the rate of growth/depreciationfor period is denoted by r%, V0 is the value of article in the beginning and Vn is the valueafter ‘n’ conversion periods, then Vn is given by :

Vn = V r n0 1 100+FH IK in case of growth and

Vn = V r n0 1 100−FH IK in case of depreciation.

If the rate of growth/depreciation varies for each conversion period, then Vn is given by

Vn = V r r r0

1 2 31100

1100

1100

+FHGIKJ +FHG

IKJ +FHG

IKJ ..... (For growth)

Vn = V r r r0

1 2 31100

1100

1100

−FHGIKJ −FHG

IKJ −FHG

IKJ ..... (For depreciation)

We solve some examples to illustrate the above concepts.

Examples 11.12 : The population of a town is 281250. What will be its population after 3years if the rate of growth of population is 4% per year ?

Solution : Here V0 = 281250, r = 4% and n = 3

∴ Vn = 281250 1 4100

3+FH IK

= 281250 2625

2625

2625× × ×

= 316368

Hence the population of the town after 3 years = 316368.


Example 11.13 : The cost of a car was Rs 35000 in January 2000. Its value depreciates atthe rate of 10% each year. Find the value of the car in January 2003.

Solution : Here V0 = Rs 350000, Vn= ? r = 10%, n = 3

∴ Vn = Rs 350000 1 10100

3−FH IK

= Rs 350000 910

910

910× × ×

= Rs 255150

In January 2003, The cost of car will be Rs 255150

Example 11.14 : The cost of machinery is Rs 1360000 today. In the first year, the costdepreciates by 15% and subsequently, the price depreciates by 10% each year. By how much,the machinery has depreciated after 4 years.

Solution : Here V0 = Rs 1360,000, n = 4, rate = 15%, first year & then 10% for next years.

∴ Vt = Rs1360000 1 15100 1 10

1003

× −FH IK −FH IK

= Rs1360000 1720

910

910

910× × × ×

= Rs 842724

∴ Depreciation = Rs(1360000 – 842724)

= Rs 517276

Example 11.15 : The application of manure increases the output of a crop by 10% in the firstyear, 5% in the second year and 4% in the third year. If the production of a crop in the year2000 was 3.5 tons per hectare, find the production of crop per hectare in 2003.

Solution : We know that Vt = V r n0 1 100+FH IK

∴ Vt = 3.5 1 10100 1 5

100 1 4100+FH IK +FH IK +FH IK tons

= 3.5 1110× × ×21

202625 tons

= 4.2042 tons

So the production of crop in the year 2003 = 4.2042 tons/hectare.

270 Mathematics

Example 11.16 : The virus of a culture decreases at the rate of 4% per hour due to a medicine.If the virus count in the culture at 9.00 A.M. was 3.5 × 108, find the virus count at 11.00 A.M.on the same day.

Solution : Here V0 = 3.5 × 108, R = 4% and n = 2

∴ Vn = 3.5× −FH IK10 1 4100

82

= 3.2256 × 108

Hence virus count at 11.00 A.M. is 3.2256 × 108.


1. The population of a city is 2.7 × 108. If the rate of growth of population is 5% of whatit was in the beginning of the year, find its population after 3 years.

2. The population of a town is 50000. In the first year, the rate of growth of the populationwas 5%. In the second year, because of some epidemics, the population decreased by 10%.In the third year, the population growth rate was noticed as 4%. Find the population ofthe town after 3 years.

3. The cost of a car was Rs 215000 in January 2001. If the rate of depreciation is 15% forthe first year and 10% for the subsequent years, find its cost after 3 years.

4. A tree gains its height at the rate of 2% of what it was in the beginning of the month.It its height was 1.5 m in the beginning of January 2002, find its height at the end ofthe April of the same year (Give your answer correct to 2 decimal places).

5. The value of property increases at the rate of 20% of what it was in the beginning of theyear. In how many years, the value of property will be almost double ?

LET US SUM UP

If Principal = P, Rate = r%, Time = n conversion periods then

Amount = Principal + Compound Interest

= P (1 + R)n

C.I. = P [(1 + R)n – 1]

Compound Interest is greater than simple interest except for the first conversion period.

If V0 is the value of article in the beginning, Vn is the value after ‘n’ conversion periodsand r be the rate of growth/depreciation per period, then

Vn = V r n0 1 100+FH IK in case of growth and

Vn = V r n0 1 100−FH IK in case of depreciation.


If the rate of growth/depreciation varies for each conversion period, then

Vn = V r r r0

1 2 31 100 1 100 1 100+FHGIKJ +FHG

IKJ +FHG

IKJ ... for growth

and Vn = V r r r0

1 2 31 100 1 100 1 100−FHGIKJ −FHG

IKJ −FHG

IKJ ... for depreciation.

TERMINAL EXERCISE

1. Find the rate of interest per annum if Rs 31250 amounts to Rs 35152 in 1½ years, interestbeing compounded half yearly.

2. A sum of Rs 1000 amounts to Rs 1331 after some time. If the rate of interest is 10%per annum compounded annually, find the time period.

3. Find the sum of money which will amount to Rs 26460 in six months at 20% per annum,when the interest is compounded quarterly.

4. Find the difference between simple and compound interests for 2 years at 10% per annum,when the interest is being compounded annually on Rs 2500.

5. The population of a town increases at the rate of 5% per annum, what will be the populationof the town after two years, if the present population is 16000 ?

6. The price of a scooter depreciates at the rate of 20% per annum of its value at the beginningof each year. If the cost of scooter is 15625, what will be its value at the end of threeyears ?

7. The population of a town is 20000. It increases by 10% during first year. During secondyear, the population decreases by 10%. Find the population at the end of 2 years.

272 Mathematics

ANSWERS


1. 1261 2. Rs 5152.01 3. Rs 66351 4. Rs 24000

5. Rs 160000 6. Rs 124.05 7. Rs 46875 8. Rs 16,000, 5%

9. Rs 1,60,000 10. 40% p.a.


1. 3.1256 × 108 2. 49140 3. Rs 148027.50 4. 1.62 m

5. 4 years

Terminal Exercises

1. 8% 2. 3 years 3. 24000 4. Rs 25

5. 17640 6. Rs 8000 7. 19800

Banking 273

12

Banking

12.1 INTRODUCTION

All of us need money to meet our day-to-day expenses. We earn money, spend money and also,save some money. We would like to keep the money in safe custody. Banks are the institutionswhere we keep the money in the form of deposits and those who need money can borrow asloans on payment of interest with certain conditions. Besides this, banks also help the peoplein various kinds of financial transactions. Some of the functions of a bank are :

1. Keeping money of the depositors and pay interest on deposits.

2. Giving loans to the borrowers on interest.

3. Buying and selling security bonds.

4. Receiving payments – Telephone bills, Electricity bills, water bills, school fees etc.

5. Transferring money from one place to another.

6. Collection of taxes – income tax, sales tax, house tax etc.

7. Issuing/Encashing travellers cheques in Local/Foreign currency.

8. Providing All Time Money (ATM) facility to the customers.

9. Issuing credit cards.

10. Exchange of Foreign currency etc.

For the convenience of customers, the banks offer different types of accounts (deposits) someof which are :

(i) Savings Bank Account

(ii) Current Account

(iii) Fixed deposit Account.

(iv) Recurring deposit Account.

274 Mathematics

12.2 OBJECTIVES


state different types of accounts.

calculate interest for a saving bank account, given the rate of interest.

calculate interest/maturity value of fixed/term deposit given the rate of interest compoundedyearly, semi-annually or quarterly (not more than four conversion periods).


(i) Knowledge of simple interest when principal (P), rate of interest (R) and time (T), forwhich money was invested are given i.e. S.I. = PRT.

(ii) Compound interest when P, R, n are given i.e., C.I. = A – P where

A = P(1 + R)n, n is number of conversion periods.

12.4 TYPES OF ACCOUNTS

You can open different types of accounts in a bank depending upon your needs. Some of theseare :

(a) Savings bank account

(b) Fixed or term deposit account

(c) Current account

12.4.1 Savings Bank Account

This is the most popular account offered by the banks. This account encourages people todevelop the habit of saving. Anybody can open this account with a minimum sum of Rs 1000with cheque book facility. In this account, deposits and withdrawals can be made and the recordis maintained by the bank in a ‘Pass-Book’ given to the Account holder (i.e. the person whoholds the account).

The bank pays interest for the money that an account holder keeps in the account. The prevailingrate of interest is 4% per annum compounded half yearly. The rate of interest changes fromtime to time.

On opening a savings bank account, a pass book is issued to the account holder. It containsdate wise entries of deposits withdrawals and the interest earned. It bears savings bank accountnumber also.

The general format of a page of the savings bank passbook is as given below :

Banking 275

Date Particulars Amount withdrawn Amount Deposited Balance Rs P Rs P

— — — — —

— — — — —

— — — — —

Money can be withdrawn from the bank through a withdrawal slip (sample shown) or by acheque (sample shown below).

Fi.g 12.1 Sample of Withdrawal slip

Fig. 12.2 Sample of Cheque

Note : In a savings bank account,

(i) The bank pays interest for the month on the minimum closing balance from the 10th dayof the month to the last day of the month.

(ii) The interest is credited to the account every six months.

(iii) The present rate of interest is 4% per annum compounded semi-annually.

(iv) Usually not more than 25 withdrawals are allowed in a quarter from Savings Bank account.

276 Mathematics

12.4.2 Computation of Interest

(i) Write down the minimum balance between the closing balance on 10th to the last dayof the month.

(ii) Add all the minimum balances for each month as per step (i) to obtain principal for onemonth.

(iii) Calculate the simple interest on this sum for one month using the formula

Interest = P R× × 112

where P is Principal, R is rate of interest per annum and time 1 month i.e. 1/12 year.

(iv) If account is opened after the 10th day of month, no interest is payable for that month.

(v) No interest is paid for the month in which the account is closed.

Let us illustrate the above with some examples.

Example 12.1 : Sidharth opened a savings bank account in State Bank of India on 5thMarch 2003 with a deposit of Rs 5000. He deposited Rs 1500 on 10th March 2003 and withdrewRs 3000 on 29th March, 2003. Find the principal for which he will earn interest forMarch 2003.

Solution : Balance on 5th March, 2003 = Rs 5000

Balance on 10th March, 2003 = Rs 6500

Balance on 29th March, 2003 = Rs 3500

Here minimum balance between 10th and 31st March, 2003 = Rs 3500.

Therefore, the principal for which interest is earned for the month of March 2003 is Rs 3500.

Example 12.2 : Anisha’s savings bank account pass-book has the following entries :

Date Particulars Amount withdrawn Amount Deposited BalanceRs P Rs P Rs P

2002

Jan 6 By Cash — 20000.00 20000.00

Feb 11 By Cash — 10000.00 30000.00

April 5 By Cheque — 5000.00 35000.00

April 27 To Cheque 18000.00 — 17000.00

May 3 By Cash — 13000.00 30000.00

June 15 To Cheque 20000.00 — 10000.00

Find the sum on which Anisha will earn interest from January to June 2002.

Banking 277

Solution : The qualifying amount for the interest is the minimum balance between the 10thand the last day of the month.

Principal for January = Rs 20000

Principal for February = Rs 20000

Principal for March = Rs 30000

Principal for April = Rs 17000

Principal for May = Rs 30000

Principal for June = Rs 10000

Total Rs 127000

Thus, the sum on which Anisha will earn interest for one month = Rs 127000.

Example 12.3 : Ritus’s pass book has the following entries :


2002

Jan. 1 B/F — — 12000.00

Jan. 9 To Cash Rs 7000.00 — 5000.00

Jan. 11 By Cheque — 20000.00 25000.00

Feb. 10 To Cheque Rs 10000.00 — 15000.00

April 3 By Cash — 5000.00 20000.00

June 5 By Cheque — 10000.00 30000.00

June 25 By Cheque — 5000.00 35000.00

If the rate of interest is 4% per annum, find the interest earned by Ritu at the end of June onher savings bank account.

Solution : Details of qualifying amount for the calculation of interest are as below:

Month AmountJan. Rs 5000.00Feb. Rs 15000.00March Rs 15000.00April Rs 20000.00May Rs 20000.00June Rs 30000.00

Total Rs 105000.00

278 Mathematics

Here P = Rs 105000, R = 4% and T = 1

12 year.

∴ Interest = PRT = Rs 105000 4 1

100 12× ×× = Rs 350.

∴ Interest earned by Ritu = Rs 350.Example 12.4 : Ashok has a savings bank account in a bank. His passbook has the followingentries:


2002July 11 By cash — 6000.00 6000.00Aug 12 By Cheque — 4000.00 10000.00Sept 5 By Cheque — 10000.00 20000.00Sept 21 To Cheque 8000.00 — 12000.00Nov 9 By Cheque — 8000.00 20000.00Dec 10 By Cheque — 10000.00 30000.00Dec 29 To cash 26000.00 — 4000.00

The account is closed on 3rd January 2003. Find the amount received by Ashok, if the rateof interest is 4% per annum.

Solution. Details of qualifying amount for the calculation of interest are :

Month AmountJuly Rs —August Rs 6000.00September Rs 12000.00October Rs 12000.00November Rs 20000.00December Rs 4000.00

Total Rs 54000.00

Here P = Rs 54000, R = 4% and T = 1

12 year


100 12× ××

LNM

OQP = Rs 180.00

∴ Amount received = Rs (4000 + 180) = Rs 4180.

Banking 279

Example 12.5 : The dates and respective balances in the pass book of David’s savings bankaccount are given below :

Dates Balances

2002

October 5 Rs 1500.00

October 25 Rs 3500.00

December 8 Rs 5000.00

2003

January 11 Rs 8000.00

February 5 Rs 6000.00



March 15 Rs 20000.00

March 29 Rs 10000.00

Calculate the interest earned upto March 2003 if the rate of interest is 4% per annum.

Solution : Details of qualifying amount for the calculation of interest are :

Month Amount

Oct, 2002 Rs 1500

Nov, 2002 Rs 3500

Dec, 2002 Rs 5000

Jan, 2003 Rs 5000

Feb, 2003 Rs 7500

March 2003 Rs 7500

Total Rs 30000

Here P = Rs 30000, R = 4% and T = 1/12 year

∴ Interest = PRT = Rs 30000 4 1100 12

× ×× = Rs 100

∴ Interest earned by David = Rs 100.

280 Mathematics

Example 12.6 : Salim opened a savings bank account with a bank on 9th January 2002 witha cash deposit of Rs 10000. Subsequently he deposited Rs 2000 on the 8th day of every month.He withdrew Rs 3000 on 25th April and Rs 6000 on 28th June 2002. Write all entries of thepassbook.

If the bank pays interest at the rate of 4% per annum, calculate the interest upto the last dayof 30th June 2002 and make the entry in the passbook alongwith the balance.

Solution : The entries in the passbook are as below :

Date Particulars Amount withdrawn Amount Deposited Balance

Rs P Rs P Rs P

2002

9th Jan. By cash — 10000.00 10000.00

8th Feb. By Cash — 2000.00 12000.00

8th March By Cash — 2000.00 14000.00

8th April By Cash — 2000.00 16000.00

25th April To Cash 3000.00 — 13000.00

8th May By Cash — 2000.00 15000.00

8th June By Cash — 2000.00 17000.00

28th June To Cash 6000.00 — 11000.00

Details of qualifying amount for the calculation of interest are :

Month January February March April May June Total

Amount 10000 12000 14000 13000 15000 11000 75000(in Rs)

∴ Interest upto 30th June, 2002 = Rs 75000 1 4

100 12× ×× = Rs 250

The interest entry is made on 1–7–2002 as below :

1.7.2002 By Interest — 250.00 11250.00


1. Krishna Murthy opened a savings bank account in State Bank of India on 7th July 2002.His pass book has the following entries :

Banking 281


2002

7th July By cash — 1100.00 1100.00

11th July By Cash — 400.00 1500.00

25th August By Cash — 1000.00 2500.00

10th September By Cheque — 1500.00 4000.00

Calculate the principal for which he will earn interest for the months of July, August andSeptember 2002, together.

2. The entries in the savings bank account pass book of Kamlesh are as under:


2002

1st January B/F — — 15000.00

10th January By Cash — 5000.00 20000.00

9th February By Cheque — 5000.00 25000.00

9th March By Cash — 5000.00 30000.00

11th April By Cash — 5000.00 35000.00

25th June By Cheque — 5000.00 40000.00

Calculate the interest at the end of June 2002 at 4% per annum.

3. The entries in the passbook of a savings bank account holder who opened his accountof 11th January 2002 are as follows :


2002

11th January By cash — 5000.00 5000.00

11th Feb. By Cheque — 5000.00 10000.00

9th March By Cheque — 5000.00 15000.00

6th June To Cheque 10000.00 — 5000.00

10th June By Cash — 15000.00 20000.00

26th June To Cash 16000.00 — 4000.00

282 Mathematics

If the rate of interest is 4% per annum, find the interest earned if the account is closedon :

(i) 30th June, 2002

(ii) 3rd July, 2002.

4. Madhu’s Savings Bank Account passbook has the following entries :


2002

July 1 B/F — — 6000.00

July 9 By Cheque — 4000.00 10000.00

Sept. 10 To Cheque 9000.00 — 1000.00

Sept. 14 By Cash — 4000.00 5000.00

December 5 By Cash — 4500.00 9500.00

December 10 By Cash — 1500.00 11000.00

December 23 To Cheque 9000.00 — 2000.00

If the rate of interest is 6% per annum, calculate the interest entry on 1st January 2003in the passbook alongwith the balance.

5. A page from the pass book of savings bank account is given below :


July 1, 2002 B/F — — 2000.00

July 11, 2002 By Cheque — 8000.00 10000.00

August 9, 2002 By Cheque — 10000.00 20000.00

November 25, 2002 To Cheque 15000.00 — 5000.00

December 19, 2000 By Cash — 15000.00 20000.00

The account is closed on 2nd January 2003. Find the amount received if the rate of interestis 4% per annum.

6. Kavita opens a savings bank account with a bank on 8th January, 2002 with a cash depositof Rs 10000. Subsequently she deposited Rs 6000 on the 6th day of every month. Shewithdraws Rs 4000 on 3rd April and Rs 12000 on 10th June 2002. If the bank pays interestat the rate of 5% per annum, payable at the end of June and December, write all the entries,including interest, which are made upto 1st July 2002.

Banking 283

12.4.3 Current Account

In a saving bank account, the account holder is allowed to have a limited number of withdrawalsin a half year. Big business concerns, companies, government organisations etc., have to doa number of transactions every day. For them banks offer a different type of account calledcurrent account. For this account, there is no limit on number of withdrawals or on the amountof withdrawals but the banks do not pay interest. Rather, sometimes, they charge some moneyas service charges. Here, the minimum balance for individuals account is Rs 5000 while forbig concerns, it is Rs 10000.

Depending upon the goodwill of the individual/company, banks allow the current accountholder to get money over and above their deposits called overdraft.

12.4.4 Fixed Deposit Account

Suppose you have some money which is not required for some time. The scheme suitable fordepositing such money is the Fixed Deposit or Term deposit. Here the depositor agrees tokeep the money with the bank for a fixed time. Obviously, the bank can use this money morefreely than the money kept in the savings bank account. Hence the banks offer higher ratesof interest on such deposits depending upon the period of deposits.

The rate of interest per annum on term deposits is as below :

(i) For 46 days and above but less than 179 days. 5%

(ii) For six months and above but less than 1 year. 6%

(iii) For one year and more. 6.5%

For senior citizens, and additional interest of 0.5% is given on deposits.

The total amount receivable after the expiry of the time is called maturity value.

Example 12.7 : Anju deposited Rs 2920 in a fixed deposit scheme in a bank for 60 days. Ifthe bank pays interest at 6% per annum, find the amount she receives at the end of 60 days.

Solution : Here P = Rs 2920, R = 6% per annum and T = 60365 year


100 365× ××

FHG

IKJ

= Rs 28.80

∴ Amount Received by Anju = Rs (2920 + 28.80)

= Rs 2948.80

Example 12.8 : Joginder makes a fixed deposit of Rs 31250 in a bank for 1½ years. If therate of interest is 8% per annum compounded half yearly, find the maturity value of the moneydeposited by him.

284 Mathematics

Solution : Note here that the interest for the first half year also forms a part of principal forthe second half year and likewise for the third half years well. Thus, we shall make use ofthe formula for compound interest.

Here P = Rs. 31250, R = 82 = 4% per half year

and T = 1 12 year = 3 Half years

⇒ n = 3.

∴ Amount = P(1 + R)n = Rs 31250 1 4100

3+FH IK

= Rs 35152

Maturity value of the deposit = Rs 35152

Example 12.9 : Amit makes a fixed deposit of Rs 31250 in a bank for 1½ years. If the rateof interest is 8% per annum compounded yearly, find the maturity value of the money depositedby him.

Solution : Here P = Rs 31250, R = 8%, T = 1 12 years.

Amount after one year = Rs 31250 1 8100+FH IK

and amount after 1 12 year

A = Rs 31250 1 8100 1 4

100+FH IK +FH IK= Rs 31250 × 1.08 × 1.04 = Rs 35100

Hence, the maturity value = Rs 35100.

Note : Note the difference in the maturity value in Example 8 and Example 9. In example 8,the interest is compounded half yearly while in Example 9, the interest is compounded yearly.

Example 12.10 : Kapil makes a fixed deposit of Rs 20000 in a bank in a year. If the rate ofinterest 8% per annum compounded quarterly, find the maturity value of the money depositedby him.

Solution : Here P = Rs. 20000, R = 2% per quarter, Time = 4 quarters

∴ A = P(1 + R)n = Rs 20000 1 2100

4+FH IK

= Rs 20000 × 1.02 × 1.02 × 1.02 × 1.02= Rs 21648.64

Hence the maturity value = Rs 21649 (approx.)

Banking 285

Example 12.11 : How much money should Nirmal deposit in a fixed deposit account in a bankso that she gets Rs 456976 after two years, the rate of interest being 8% per annum compoundedhalf yearly.

Solution : Here P = ?, A = Rs 456976, R = 4% per half year and

T = 2 year = 4 half years i.e. n = 4

We know that A = P(1 + R)n

∴ 456976 = P P1 4100

2625

4 4+FH IK = FH IK

P = Rs 456976 25 25 25 2526 26 26 26

× × × ×× × ×

= Rs 390625

Hence amount to be deposited = Rs 390625.


1. Find the amount at the end of 9 months if Rs 50000 is deposited and the interest 8% perannum is compounded quarterly.

2. Charu makes a fixed deposit of Rs 8000 in a bank for 1½ years. It the rate of interestis 10% per annum and the interest is compounded half yearly, find the maturity value ofthe money deposited by her.

3. Pankaj deposits Rs 75000 in a fixed deposit account for 3 years. If the rate of interestis 10% per annum compounded annually, find the maturity value of the money depositedby him.

4. How much money should Shanta deposit in a fixed deposit account in a bank so as toenable her to receive a sum of Rs 9261 after 1½ years, the rate of interest being 10%per annum compounded half yearly ?

5. How much money should Kamal deposit in a fixed deposit account in a bank so as toenable him to receive a sum of Rs 1061208 after nine months, the rate of interest being8% per annum compounded quarterly ?

LET US SUM UP

There are different types of accounts in a bank. Some of these are :

(i) Saving bank account

(ii) Fixed or term deposit account

(iii) Current account

286 Mathematics

In a saving bank account :

(i) the bank pays interest for the month on the minimum closing balance from the 10thday of the month to the last day of the month.

(ii) the interest is credited to the account every six months.

Step for computing interest are :

(i) Write down the minimum balance between the closing balance on 10th to the lastday of the month.

(ii) Add all the minimum balances for each month as per step (i) to obtain principal forone month.

(iii) Calculate the simple interest on this sum for one month using the formula

Interest = P R× × 112

where P is Principal, R is rate of interest per annum and time 1 month i.e. 1/12 year.

(iv) If account is opened after the 10th day of month, no interest is payable for that month.

(v) No interest is paid for the month in which the account is closed.

TERMINAL EXERCISE

1. A page for the pass book of Mr. Dass’s Saving bank Account in a particular year is givenbelow :


Jan. 9 By Cash — 15000.00 15000.00

Feb. 10 By Cheque — 8000.00 23000.00

April 25 To Cheque 20000.00 — 3000.00

June 7 By Cash — 2000.00 5000.00

June 11 By Cash — 12000.00 17000.00

If the rate of interest is 4% per annum, find the interest earned by Mr. Dass at the endof June on his savings bank account.

Banking 287

2. Sujata has a savings bank account in a bank. Her passbook has the following entries :


2002

July 1 B.F. — — 2500.00

July 9 By Cheque — 2500.00 5000.00

August 10 By Cash — 4000.00 9000.00

October 19 To Cheque 6000.00 — 3000.00

November 2 By Cash — 9600.00 12600.00

Dec. 20 To Cash 9200.00 — 3400.00

Dec. 27 By Cheque — 10600.00 14000.00

The account is closed on 10th January, 2003. Find the amount received if the rate of interestis 4% per annum.

3. Vandana makes a fixed deposit of Rs 62500 in a bank for 1½ years. If the rate of interestis 8% per annum compounded half yearly, find the maturity value of money depositedby her.

4. Smith makes a fixed deposit of Rs 10000 in a bank for a year. If the rate of interest is8% per annum compounded quarterly, find the maturity value of the money deposited byher.

5. How much money should Sarla deposit in a fixed deposit account in a bank so that shegets Rs 194481 after two years, the rate of interest being 10% per annum compoundedhalf-yearly ?

288 Mathematics

ANSWERS


1. Rs 6600 2. Rs 583.33 3. (i) Rs 166.66 (ii) Rs 180

4. Interest Rs 165, Balance Rs 2165 5. Rs 20240

6. Date Particulars Debit Credit BalanceRs P Rs P Rs P

2002

Jan. 5 By cash — 10000.00 10000.00

Feb. 6 By Cash — 6000.00 16000.00

March 6 By Cash — 6000.00 22000.00

April 3 To Cash 4000.00 — 18000.00

April 6 By Cash — 6000.00 24000.00

May 6 By Cash — 6000.00 30000.00

June 6 By Cash — 6000.00 36000.00

June 10 To Cash 12000.00 — 24000.00

July 1 By interest — 525.00 24525.00


1. Rs 53060.40 2. Rs 9261 3. Rs 99825 4. Rs 8000 5. Rs 1000000

Terminal Exercise

1. Rs 240 2. Rs 14140 3. Rs 70304 4. Rs 10824.32 5. Rs 160000

Lines and Angles 1

Module 3Geometry

Geometry is a branch of Mathematics which deals with the study of different types of figuresand their properties. Geometry means measurement of earth. As such geometry originated inancient times when man started measuring land for making his home and boundaries for hisfields. Egyptians and Babylonians discovered many formulae for finding the areas of differentrectilinear figures and used them practically.

The Indian Mathematicians also had contributed a lot towards development of knowledge ofgeometry as is evident from the civilizations of Harappa and Mohenjodaro. Sulbasutras usedduring vedic period, and the work of the great mathematician Baudhayan, who profounded andproved, what is now known as Pythagoras theorem are contributions to geometry worthmentioning from the glorius past of India.

Euclid a Greek mathematician collected all available knowledge in geometry till his times(330 B.C.), arranged it in a systematic way and gave it a logical approach based on deductivereasoning. Since then efforts have been made to give it a perfect logical shape.

To study geometry in its full and complete logical form, is a difficult task. As such we shallbe studying geometry in a very informal way; defining terms with suitable examples, verifyingand stating properties of figures and proving logically only a few important properties knownas theorems.

In this module on geometry, we shall study about lines, angles, triangles, quadrilaterals andcircles along with their properties.

2 Mathematics

13

Lines and Angles

13.1 INTRODUCTION

Observe the top of your desk or table. Now move your hand on the top of your table. It givesan idea of a plane. Its edges give an idea of a line, its corner that of a point and the edgesmeeting at a corner give an idea of an angle.

13.2 OBJECTIVES


illustrate the concepts of point, line, plane, angle, parallel lines, intersecting lines and pairof angles made by them.

illustrate and verify properties of parallel lines

prove that the sum of angles of a triangle is 180°

explain the concept of locus and find the locus of a point under certain conditions.


We assume that the learner is familiar with the geometrical concepts and figures such as :

point, line, plane, intersecting lines, rays and angles.

parallel lines

13.4 POINT, LINE AND ANGLE

In earlier classes, you have studied about a point, a line, a plane and an angle. Let us quicklyrecall these concepts.

Point : If we press the tip of a pen or pencil on a piece of paper, we get a fine dot, whichis called a point.

Fig. 13.1

Lines and Angles 3

A point is used to show the location and is represented by capital letters A, B, C etc.

13.4.1 Line

Now mark two points A and B. Join them with the help of a ruler or a scale and extend iton both sides. This gives us a straight line or simply called a line.

Fig. 13.2

In geometry a line is extended infinitely on both sides and is marked with arrows to give thisidea. A line is named using any two points on it, viz, AB or by a single small letter l, m etc.(See Fig. 13.3)

Fig. 13.3

The part of the line between two points A and B is called a line segment and will be namedAB.

Observe that a line segment is the shortest path between two points A and B. (See Fig. 13.4)

Fig. 13.4

13.4.2 Ray

If we mark a point X and draw a line, starting from it extending infinitely in one directiononly, then we get a ray XY.

Fig. 13.5

X is called the initial point of the ray XY.

13.4.3 Plane

If we move our palm on the top of a table, we get an idea of a plane.

Fig.13.6

4 Mathematics

Similarly, floor of a room also gives the idea of part of a plane.

Plane also extends infinitely lengthwise and breadthwise.

Mark a point A on a sheet of paper.

How many lines can you draw passing though this point ? As many as you wish.

Fig. 13.7

In fact we can draw an infinite number of lines through a point.

Take another point B, at some distance from A. We can again draw an infinite number of linespassing through B.

Fig. 13.8

Out of these lines how many pass through both the points A and B ? Out of all the lines passingthrough A, only one passes through B. Thus, only one line passes through both the points Aand B. We conclude that one and only one line can be drawn passing through two givenpoints.

Now we take three points in plane.

Fig. 13.9

We observe that a line may or may not pass through the three given points.

If a line can pass through three or more points, then these points are said to be collinear. Forexample, points A, B and C in the Fig. 13.9 are collinear points.

If a line can not be drawn passing through all three points (or more points) then they are saidto be non-collinear. For example points P, Q and R in the Fig. 13.9.

Since two points always lie on a line, we talk of collinear points only when their number isthree or more.

Lines and Angles 5

Let us now take two distinct lines AB and CD in a plane.

Fig. 13.10

How many points can they have in common ? We observe that these lines can have.

either (i) one point in common as in Fig. 13.10 (a) and (b). [In such a case they are calledintersecting lines] or (ii) no points in common as in Fig. 13.10(c). In such a case they are calledparallel lines.

Now observe three (or more) distinct lines in plane.

Fig. 13.11

What are the possibilities ?

(i) They may interest in more than one point as in Fig. 13.11(a) and 13.11(b).

or (ii) They may intersect in one point only as in Fig. 13.11(c). In such a case they are calledconcurrent lines.

or (iii) They may be non intersecting lines parallel to each other as in Fig. 13.11 (d).

13.4.4 Angle

Mark a point O and draw two rays OA and OB starting from O. The figure we get is calledan angle. Thus, an angle is a figure consisting of two rays starting from a common point.

6 Mathematics

Fig. 13.11

This angle may be named as angle AOB or angle BOA or simply angle O; and is written as∠AOB or ∠BOA or ∠O. [see Fig. 13.11]

An angle is measured in degrees. If we take any point O and draw two rays starting from itin opposite directions then the measure of this angle is taken to be 180° degrees, written as180°.

Fig. 13.12

This measure divided into 180 equal parts is called one degree (1°).

Angle obtained by two opposite rays is called a straight angle.

An angle of 90° is called a right angle, for example ∠BOA or ∠BOC.

Fig. 13.13

Two lines or rays making a right angle with each other are called perpendicular lines. InFig. 13.13, we can say OA is perpendicular to OB or vice-versa.

An angle less than 90° is called an acute angle. For example ∠POQ is an acute angle inFig. 13.14(a).

An angle greater than 90° but less than 180° is called an obtuse angle. For example, ∠XOYis an obtuse angle in Fig. 13.14(b).

Lines and Angles 7

(a) (b)

Fig. 13.14

13.5 PAIRS OF ANGLES

Fig. 13.15

Observe the two angles ∠1 and ∠2 in each of the figures in Fig. 13.15. Each pair has a commonvertex O and a common side OA in between OB and OC. Such a pair of angles is called a‘pair of adjacent angles’.

(a) (b)Fig. 13.16

Observe the angles in each pair in Fig. 13.16[(a) and (b)]. They add up to make a total of 90°.

A pair of angles, whose sum is 90°, is called a pair of complementary angles. Each angleis called the complement of the other.

8 Mathematics

(a) (b)

Fig. 13.17

Again observe the angles in each pair in Fig. 13.17[(a) and (b)].

These add up to make a total of 180°.

A pair of angles whose sum is 180°, is called a pair of supplementary angles.

Each angle is called the supplement of the other.

Draw a line AB. From a point C on it draw a ray CD making two angles ∠X and ∠Y.

Fig. 13.18

If we measure ∠X and ∠Y and add, we will always find the sum to be 180°, whatever bethe position of the ray CD. We conclude

If a ray stands on a line then the sum of the two adjacent angles so formed is 180°.

The pair of angles so formed as in Fig. 13.18 is called a linear pair of angles.

Note that they also make a pair of supplementary angles.

Draw two intersecting lines AB and CD, intersecting each other at O.

Fig. 13.19

Lines and Angles 9

∠AOC and ∠DOB are angles opposite to each other. These make a pair of vertically oppositeangles. Measure them. You will always find that

∠AOC = ∠DOB.

∠AOD and ∠BOC is another pair of vertically opposite angles. On measuring, you will againfind

∠AOD = ∠BOC

We conclude :

If two lines intersect each other, the pairs of vertically opposite angles are equal.

Activity for you.

Attach two strips with a nail or a pin as shown in the figure.

Fig. 13.20

Rotate one of the strips, keeping the other in position and observe that the pairs of verticallyopposite angles thus formed are always equal.

A line which intersects two or more lines at distinct points is called a transversal. For exampleline l in Fig. 13.21 is a transversal.

Fig. 13.21

When a transversal intersects two lines, eight angles are formed.

Fig. 13.22

10 Mathematics

These angles in pairs are very important in the study of properties of parallel lines. Some ofthe useful pairs are as follows :

(a) ∠1 and ∠5 is a pair of corresponding angles. ∠2 and ∠6, ∠3 and ∠7 and ∠4 and ∠8are other pairs of corresponding angles.

(b) ∠3 and ∠6 is a pair of alternate angles. ∠4 and ∠5 is another pair of alternate angles.(c) ∠3 and ∠5 is a pair of interior angles on the same side of the transversal.

∠4 and ∠6 is another pair of interior angles.In Fig. 13.22 above, lines m and n are not parallel; as such, there may not exist any relationbetween the angles of any of the above pairs. However, when lines are parallel, there are somevery useful relations in these pairs, which we study in the following,When a transversal intersects two parallel lines, eight angles are formed, whatever be theposition of parallel lines or the transversal

Fig. 13.23

If we measure the angles, we shall always find that

∠1 = ∠5, ∠2 = ∠6, ∠3 = ∠7 and ∠4 = ∠8

that is, angles in each pair of corresponding angles are equal.

Also ∠3 = ∠6 and ∠4 = ∠5

that is, angles in each pair of alternate angle are equal.

Also, ∠3 + ∠5 = 180° and ∠4 + ∠6 = 180°

that is, sum of angles in each pair of interior angles is 180°.

Hence we conclude :

When a transversal intersects two parallel lines, then

(i) each pair of corresponding angles, are equal

(ii) each pair of alternate angles are equal

(iii) each pair of interior angles on the same side of the transversal are supplementary.

You may also verify the truth of these results by drawing a pair of parallel lines (using paralleledges of your scale) and a transversal and measuring angles in each of these pairs.

Lines and Angles 11

Converse of each of these results is also true. To verify the truth of the first converse, we drawa line AB and mark two points C and D on it.

Fig. 13.24

At C and D, we construct two angles ACF and CDH equal to each other, say 50°, as shownin Fig. 13.24. On producing EF and GH on either side, we shall find that they do not intersecteach other, that is, they are parallel.

In a similar way, we can verify the truth of the other two converses.

Hence we conclude that

When a transversal intersects two lines in such a way that

(i) any pair of corresponding angles are equal

or (ii) any pair of alternate angles are equal

or (iii) any pair of interior angles on the same side of transversal are supplementary

then the two lines are parallel.

Example 13.1 : Choose the correct answer out of the alternative options in the followingmultiple choice questions.

Fig. 13.25

12 Mathematics

(i) In Fig. 13.25, ∠FOD and ∠BOD are

(a) supplementary angles (b) complementary angles

(c) vertically opposite (d) a linear pair of angles Ans. (b)

(ii) In Fig. 13.25, ∠COE and ∠BOE are

(a) complementary angles (b) supplementary angles

(c) a linear pair (d) adjacent angles Ans. (d)

(iii) In Fig. 13.25, ∠BOD is equal to

(a) x° (b) (90 + x)°

(c) (90 – x)° (d) (180 – x)° Ans. (c)

(iv) An angle is 4 times its supplement; the angle is

(a) 36° (b) 72°

(c) 108° (d) 144° Ans. (d)

(v) What value of x will make AB a straight angle in Fig. 13.26

Fig. 13.26

(a) 30° (b) 40°

(c) 50° (d) 60° Ans. (c)

Fig. 13.27

In the above figure, l is parallel to m and p is parallel to q.

(vi) ∠3 and ∠5 form a pair of

(a) Alternate angles (b) interior angles

(c) vertically opposite (d) corresponding angles Ans. (d)

Lines and Angles 13

(vii) In Fig. 13.27, if ∠1 = 80° then ∠6 is equal to

(a) 80° (b) 90°

(c) 100° (d) 110° Ans. (c)

Fig. 13.28

(viii) In Fig. 13.28 OA bisects ∠LOB, OC bisects ∠MOB and ∠AOC = 90°. Show that pointsL, O and M are collinear.

Solution : ∠BOL = 2 ∠BOA ...(i)

and ∠BOM = 2∠BOC ...(ii)

Adding (i) and (ii), ∠BOL + ∠BOM = 2∠BOA + 2∠BOC

∴ ∠LOM = 2(∠BOA + ∠BOC)

= 2 × 90°

= 180° = straight angle

∴ L, O and M are collinear.


1. Choose the correct answer out of the given alternative in the following multiple choicequestions :

Fig. 13.29

In Fig. 13.29, AB || CD and PQ intersects them at R and S respectively.

14 Mathematics

(i) ∠ARS and ∠BRS form

(a) a pair of alternate angles

(b) a linear pair

(c) a pair of corresponding angles

(d) a pair of vertically opposite angels

(ii) ∠ARS and ∠RSD form a pair of

(a) Alternate angles (b) Vertically opposite angles

(c) Corresponding angles (c) Interior angles

(iii) If ∠PRB = 60° then ∠QSC is

(a) 120° (b) 60°

(c) 30° (d) 90°

Fig. 13.30

(iv) AB and CD intersect at O. ∠COB is equal to

(a) 36° (b) 72°

(c) 108° (d) 144°

Fig. 13.31

2. In fig 13.31 above, AB is a straight line. Find x

3. In Fig 13.32 below, l is parallel to m. Find angles 1 to 7.

Fig. 13.32

Lines and Angles 15

13.6 TRIANGLE, ITS TYPES AND PROPERTIES

Triangle is the simplest of all the closed figures formed in a plane by three line segments.

Fig. 13.33

It is a closed figure formed by three line segments having six elements, namely three angles

(i) ∠ABC or ∠B (ii) ∠ACB or ∠C (iii) ∠CAB or ∠A and three sides : (iv) AB (v) BC (vi)CA

It is named as ∆ ABC or ∆ BAC or ∆ CBA and read as triangle ABC or triangle BAC or triangleCBA.

13.6.1 Types of Triangles

Triangles can be classified into different types in two ways.

(a) On the basis of sides

(i) (ii) (iii)

Fig. 13.34

(i) Equilateral triangle : A triangle in which all the three sides are equal is called anequilateral triangle. [∆ABC in Fig. 13.34(i)]

(ii) Isosceles triangle : A triangle in which two sides are equal is called an isoscelestriangle. [∆DEF in Fig. 13.34(ii)].

(iii) Scalene triangle : A triangle in which all sides are of different lengths, is called ascalene triangle [∆LMN in Fig. 13.34(iii)]

(b) On the basis of angles :

(i) (ii) (iii)

Fig. 13.35

16 Mathematics

(i) Obtuse angled triangle : A triangle in which one of the angles is an obtuse angle iscalled an obtuse angled triangle or simply obtuse triangle [∆PQR is Fig. 13.35(i)]

(ii) Right angled triangle : A triangle in which one of the angles is a right angle is calleda right angled triangle or right triangle. [∆UVW in Fig 13.35(ii)]

(iii) Acute angled triangle : A triangle in which all the three angles are acute is calledan acute angled triangle or acute triangle [∆XYZ in Fig. 13.35(iii)]

Now we shall study some important properties of angles of a triangle.

13.6.2 Angle Sum Property of a Triangle

We draw two triangles and measure their angles.

Fig. 13.36

In Fig. 13.36 (a), ∠A = 80°, ∠B = 40° and ∠C = 60°

∴ ∠A + ∠B + ∠C = 80° + 40° + 60° = 180°

In Fig. 13.36 (b), ∠P = 30°, ∠Q = 40°, ∠ R = 110°

∴ ∠P + ∠Q + ∠R = 30° + 40° + 110° = 180°

What do you observe ? Sum of the angles of triangle in each case in 180°.

We will prove this result in a logical way naming it as a theorem.

Theorem : The sum of the three angles of a triangle is 180°.

Fig. 13.37

Given : A triangle ABC

To prove : ∠A + ∠B + ∠C = 180°

Construction : Through A draw a line DE parallel to BC.

Lines and Angles 17

Proof : Since DE is parallel to BC and AB is a transversal.

∴ ∠B = ∠DAB (pair of alternate angles)

Similarly ∠C = ∠EAC (pair of alternate angles)

∴ ∠B + ∠C = ∠DAB + ∠EAC ... (1)

Now adding ∠A to both sides of (1)

∠A + ∠B + ∠C = ∠A + ∠DAB + ∠EAC

= 180° (Angles making a straight angle)

13.6.3 Exterior Angle of a Triangle

Let us produce a side BC of the ABC to a point D.

Fig. 13.38

In Fig. 13.39, observe that there are six exterior angles of the ∆ABC, namely ∠1, ∠2, ∠3,∠4, ∠5 and ∠6.

Fig. 13.39

In Fig. 13.38, ∠ACD so obtained is called an exterior angle of the ∆ABC. Thus,

The angle formed by a side of the triangle produced and another side of the triangleis called an exterior angle of the triangle.

Corresponding to an exterior angle of a triangle, there are two interior opposite angles.

Interior opposite angles are the angles of the triangle not forming a linear pairwith the given exterior angle.

For example ∠A and ∠B are the two interior opposite angles corresponding to the exteriorangle ACD of ∆ABC. We measure these angles.

18 Mathematics

∠A = 60°

∠B = 50°

and ∠ACD = 110°

We observe that ∠ACD = ∠A + ∠B.

This observation is true in general.

Thus, we may conclude :

An exterior angle of a triangle is equal to the sum of the two interior oppositeangles.

Examples 13.3 : Choose the correct answer out of the given alternatives in the followingmultiple choice questions :

(i) Which of the following can be the angles of a triangle ?

(a) 65°, 45° and 80° (b) 90°, 30° and 61°

(c) 60°, 60° and 59° (d) 60°, 60° and 60°. Ans. (d)

Fig. 13.40

(ii) In Fig. 13.40 ∠A is equal to

(a) 30° (b) 35°

(c) 45° (d) 75° Ans. (c)

(iii) In a triangle, one angle is twice the other and the third angle is 60°. Then the largest angleis

(a) 60° (b) 80°

(c) 100° (d) 120° Ans. (b)

Example 13.4 :

Fig. 13.41

In Fig. 13.41, bisectors of ∠PQR and ∠PRQ intersect each other at O. Prove that ∠QOR =

90° + 12 ∠P.

Lines and Angles 19

Solution : ∠QOR = 180 12

12°− ∠ + ∠FH IKPQR PRQ

= 180 12°− ∠ + ∠PQR PRQb g

= 180 12 180°− °−∠Pb g

= 180 90 12°− °+ ∠P = 90 1

2°+ ∠P .


1. Choose the correct answer out of given alternatives in the following multiple choicequestions:

(i) A triangle can have

(a) Two right angles (b) Two obtuse angles

(c) At the most two acute angles (d) All three acute angles

(ii) In a right triangle, one exterior angle is 120°, The smallest angle of the triangles is

(a) 20° (b) 30°

(c) 40° (d) 60°

(iii)

Fig. 13.42

In Fig. 13.42, CD is parallel to BA. ∠ACB is equal to

(a) 55° (b) 60°

(c) 65° (d) 70°

2. The angles of a triangle are in the ratio 2 : 3 : 5, find the three angles.

3. Prove that the sum of the four angles of a quadrilateral is 360°.

4. In Fig. 13.43 ABCD is a trapezium such that AB || DC. Find ∠D and ∠C and verify thatsum of the for angles is 360°.

20 Mathematics

Fig. 13.43

5. Prove that if one angle of a triangle is equal to the sum of the other two angles, then itis a right triangle.

6. In Fig. 13.44, ABC is a triangle such that ∠ABC = ∠ACB. Find the angles of the triangle.

Fig. 13.44

13.7 LOCUS

During the game of cricket, when a player hits the ball, it describes a path, before being caughtor touching the ground.

Fig. 13.44

The path described is called Locus.

A figure in geometry is a result of the path traced by a point (or a very small particle) movingunder certain conditions. For example :

(1) Given two parallel lines l and m, also a point P between them equidistant from both thelines .

Fig. 13.45

If the particle moves so that it is equidistant from both the lines, what will be its path?

Lines and Angles 21

Fig. 13.46

The path traced by P will be a line parallel to both the lines and exactly in the middle of themas in Fig. 13.46.

(2) Given a fixed point O and a point P at a fixed distance d.

Fig. 13.47

If the point P moves in a plane so that it is always at a constant distance d from the fixedpoint O, what will be its path.

Fig. 13.48

The path of the moving point P will be a circle as shown in Fig. 13.48.

(3) Place a small piece of chalk stick or a pebble on top of a table. Strike it hard with a pencilor a stick so that it leaves the table with a certain speed and observe its path after it leavesthe table.

Fig. 13.49

22 Mathematics

The path traced by the pebble will be a curve (part of what is known as a parabola) asshown in Fig. 13.49.

Thus, locus of a point moving under certain conditions is the path or the geometrical figure,every point of which satisfies the given condition(s).

13.7.1 Locus of a point equidistant from two given points.

Let A and B the two given points.

Fig. 13.50

We have to find the locus of a point P such that PA = PB.

Joint AB. Mark mid point of AB as M. Mark another point P using compasses such thatPA = PB. Join PM and extend it on both sides. Using a pair of divider or a scale, it can easilybe verified that every point on PM is equidistant from A and B. Also

∠AMP = ∠BMP = 90°That is, PM is the perpendicular bisector of AB.

Fig. 13.51

Thus, we may conclude the following :

The locus of a point equidistant from two given points is the perpendicular bisectorof the line segment joining the two points.

Activity for you :Mark two points A and B on a sheet of paper and join them. Fold the paper along mid-pointof AB so that A coincides with B. Make a crease along the line of fold. This crease is a straightline. This is the locus of the point equidistant from the given points A and B. It can be easilychecked that every point on it is equidistant from A and B.

13.7.2 Locus of a point equidistant from two lines intersecting at O

Let AB and CD be two given lines intersecting at O.

Lines and Angles 23

Fig. 13.52

We have to find the locus of a point P which is equidistant from both AB and CD.

Draw bisectors of ∠BOP and ∠BOC.

Fig. 13.53

If we take any point P on any bisector l or m, we will find perpendicular distances PL andPM of P from the lines AB and CD are equal.

that is, PL = PM

Thus, we may conclude :

The locus of a point equidistant from two intersecting lines is the pair of lines,bisecting the angles formed by the given line.

Activity for you :

Draw two lines AB and CD intersecting at O, on a sheet of paper. Fold the paper through Oso that AO falls on CO and OD falls on OB and mark the crease along the fold. Take a pointP on this crease which is the bisector of ∠BOD and check using a set square that

PL = PM

Fig. 13.54

In a similar way find the other bisector by folding again and getting crease 2. Any point onthis crease 2 is also equidistant from both the lines.

24 Mathematics

Example 13.5 : Find the locus of the centre of a circle passing through two given points.

Solution : Let the two given points be A and B. We have to find the position or positionsof centre O of a circle passing through A and B.

Fig. 3.55

Point O must be equidistant from both the points A and B. As we have already learnt, the locusof the point O will be the perpendicular bisector of AB.

Fig. 13.56

CHECK YOU PROGRESS 13.3

1. Find the locus of the centre of a circle passing through three given points A, B and Cwhich are non-collinear.

2. There are two villages certain distance apart. A well is to be dug so that it is equidistantfrom the two villages such that its distance from each village is not more than the distancebetween the two villages. Representing the villages by points A and B and the well bypoint P, show in a diagram the locus of the point P.

3. Two straight roads AB and CD are intersecting at a point O. An observation post is tobe constructed at a distance of 1 km from O and equidistant from the roads AB and CD.Show in a diagram the possible locations of the post.

4. Find the locus of a point which is always at a distance 5 cm from a given line AB.

Lines and Angles 25

LET US SUM UP

A line extends to infinity on both sides and a line segment is only a part of it betweentwo points.

Two distinct lines in a plane may either be intersecting or parallel.If three or more lines intersect in one point only then they are called cocurrent lines.If a line passes through three or more points, then they are called collinear points.Two rays starting from a common point form an angle.A pair of angles, whose sum is 90° is called a pair of complementary angles.A pair of angles whose sum is 180° is called a pair of supplementary angles.If a ray stands on a line then the sum of the two adjacent angles, so formed is 180°.If two lines intersect each other the pairs of vertically opposite angles are equal.When a transversal intersects two parallel lines, then(i) corresponding angles in a pair are equal.

(ii) alternate angles are equal.(iii) interior angles on the same side of the thransversal are supplementary.The sum of the angles of a triangle is 180°.An exterior angle of a triangle is equal to the sum of the two interior opposite angles.

TERMINAL EXERCISE

1. In Fig. 13.57, if x = 42, then determine (a) y (b) ∠AOD

Fig. 13.572.

Fig. 13.58In the above figure p, q and r are parallel lines intersected by a transversal l at A, B andC. Find ∠1 and ∠2.

26 Mathematics

3. The sum of two angles of a triangle is equal to its third angle. Find the third angle. Whattype of triangle is it ?

4.

Fig. 13.59

In Fig. 13.59 sides of ∆ABC have been produced as shown. Find the angles of the triangle.

5.

Fig. 13.60

In Fig. 13.60, sides ABC of the triangle ABC have been produced as shown. Show thatthe sum of the exterior angles so formed is 360°.

6.

Fig. 13.61

In Fig. 13.61 ABC is a triangle in which bisectors of ∠B and ∠C meet at O. Show that∠BOC = 125°.

Lines and Angles 27

7.

Fig. 13.62

In Fig. 13.62 above, find the sum of the angles, ∠A, ∠F, ∠C, ∠D, ∠B and ∠E.

8.

Fig. 13.63

In Fig. 12.63 in ∆ABC, AD is perpendicular to BC and AE is bisector of ∠BAC. Find∠DAE.

9.

Fig. 13.64

In Fig.13.64 above, in ∆PQR, PT is bisector of ∠P and PR is produced to S show that∠PQR + ∠PRS = 2∠PTR.

10. Prove that the sum of the (interior) angles of a pentagon is 540°.

28 Mathematics

ANSWERS

Check Your Progress 13.11. (i) (b) (ii) (a) (iii) (b) (iv) (c)2. x = 17°.3. ∠1 = ∠3 = ∠4 = ∠6 = 110°

and ∠2 = ∠5 = ∠7 = 70°.

Check Your Progress 13.21. (i) (d) (ii) (b) (iii) (b)2. 36°, 54° and 90° 4. ∠D = 140° and ∠C = 110°6. ∠ABC = 45°, ∠ACB = 45° and ∠A = 90°.

Check Your Progress 13.31. Only a point, which is the point of intersection of perpendicular bisectors of AB and BC.2. Let the villages b A and B, then locus will be the line segment PQ, perpendicular bisector

of AB such thatAP = BP = QA = QB = AB

Fig. 13.65

3. Possible locations will be four points two points P and Q on the bisector of ∠AOC andtwo points R and S on the bisector of ∠BOC.

Fig. 13.66

4.Two on either side of AB and lines parallel to AB at a distance of 5 cm from AB.Terminal Exercise

1. (a) y = 27 (b) = 126° 2. ∠1 = 48° and ∠2 = 132°3. Third angle = 90°, Right triangle 4. ∠A = 35°, ∠B = 75° ∠C = 70°7. 360° 8. 12°.

Lines and Angles 29

Congruence of Triangles 29

14

Congruence of Triangles

14.1 INTRODUCTION

You might have observed that leaves of different trees have different shapes, but leaves of thesame tree have almost the same shape. Although they may differ in size. The geometricalfigures which have same shape and same size are called congruent and the property is calledcongruence.

In this lesson you will study congruence of two triangles, some relations between their sidesand angles in details.

14.2 OBJECTIVES


verify and explain whether two given figures are congruent or not.

state the criteria for congruency of two triangles and apply them in solving problems.

prove that angles opposite to equal sides of a triangle are equal.

prove that sides opposite to equal angles of a triangle are equal.

prove that if two sides of triangle are unequal, then the longer side has the greater angleopposite to it.

state and verify inequalities in a triangle.

solve problems based on the above results.


Recognition of plane geometric figures

Equality of lines and angles

Types of angles

Angle sum property of a triangle

Paper cutting and folding.

30 Mathematics

14.4 CONCEPT OF CONGRUENCE

In our daily life you observe various figures or objects. These figures or objects can becategorised in terms of its shape and size in the following manner.

(i) Which have different shapes and sizes as shown in Fig. 14.1

Fig. 14.1

(ii) Which have same shapes but different sizes as shown in Fig. 14.2

Fig. 14.2

(iii) Which have same shape and same size as shown in Figures 14.3, 14.4 and 14.5.

(a) two' one rupee coins

Fig. 14.3

(b) two postage stamps or post cards

Fig. 14.4


(c) two photo prints of same size from the same negative.

Fig. 14.5

We will deal with the figures which have same shapes and same sizes.

Two figures, which have the same shape and same size are called congruent figuresand this property is called congruence.

14.4.1 ACTIVITY FOR YOU

Take a sheet of paper, fold it in the middle and keep a carbon (paper) between the two folds.Now draw a figure of a leaf or a flower or any object which you like, on the upper part ofthe sheet. You will get a carbon copy of it, on the sheet below.

The figure you drew and its carbon copy are of the same shape and same size. Thus, theseare congruent figures. Observe a butterfly folding its two wings. These appear to be one.

14.4.2 CRITERIA FOR CONGRUENCE OF SOME FIGURES

Congruent figures when placed one over the other, exactly coincide with one another or covereach other. In other words, two figures will be congruent, if parts of one figure are equal tothe corresponding parts of the other. For example :

(1) Two line segments are congruent, when they are of equal length.

Fig. 14.6

(2) Two squares are congruent if their sides are equal.

Fig. 14.7

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(3) Two circles are congruent, if their radii are equal, implying their circumferences are alsoequal.

Fig. 14.8

14.5 CONGRUENCE OF TRIANGLES

Triangle is a basic rectilinear figure in geometry, having minimum number of sides. As suchcongruence of triangles plays a very important role in proving many useful results. Hence thisneeds a detailed study.

Two triangles are congruent, if all the sides and all the angles of one are equalto the corresponding sides and angles of other.

For example, in triangles PQR and XYZ in Fig. 14.9

Fig. 14.9

PQ = XY, PR = XZ, QR = YZ

∠P = ∠X, ∠Q = ∠Y and ∠R = ∠Z

Thus we can say

∆PQR is congruent to ∆XYZ and we write

∆PQR ≅ ∆XYZ

Relation of congruence between two triangles is always written with corresponding or matchingparts in proper order.

Here ∆PQR ≅ ∆XYZ

also means P corresponds to X, Q corresponds to Y and R corresponds to Z.


This congruence may also be written as ∆QRP ≅ ∆YZX which means, Q corresponds to Y,R corresponds to Z and P corresponds to X. It also means corresponding parts, (elements) areequal, namely

QR = YZ, RP = ZX, QP = YX, ∠Q = ∠Y, ∠R = ∠Z

and ∠P = ∠X.

This congruence may also be written as

∆RPQ ≅ ∆ZXY

but NOT as ∆PQR ≅ ∆YZX.

Or NOT as ∆PQR ≅ ∆ZXY.

14.6 CRITERIA FOR CONGRUENCE OF TRIANGLES

In order to prove, whether two triangles are congruent or not, we need to know that all thesix parts of one triangle are equal to the corresponding six parts of the other triangle. We shallnow learn that it is possible to prove the congruence of two triangles, even if we are able toknow the equality of three of their corresponding parts.

Consider a triangle ABC in Fig. 14.10

Fig. 14.10

Construct another triangle PQR such that QR = BC, ∠Q = ∠B and PQ = AB.

(See Fig. 14.11).

Fig. 14.11

If we trace or cut out triangle ABC and place it over triangle PQR, we will observe that onecovers the other exactly. Thus, we may say they are congruent.

Alternatively, we can also measure the remaining parts, and observe that

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AC = PR, ∠A = ∠P and ∠C = ∠R

Showing that ∆PQR ≅ ∆ABC.

It should be noted here that in constructing ∆PQR congruent to ∆ABC we used only two pairsof sides PQ = AB, QR = BC and the included angle between them ∠Q = ∠B.

This means that equality of these three corresponding parts results in congruent triangles. Thuswe have

Criterion 1 : If any two sides and the included angle of one triangle are equal tothe corresponding two sides and the included angle of the other triangle, the twotriangles are congruent.

This criterion is referred to as SAS (side angle side).

Again, consider ∆ABC in Fig. 14.12.

Fig. 14.12

Construct another ∆PQR such that, QR = BC, ∠Q = ∠B and ∠R = ∠C. (See Fig. 14.13)

Fig. 14.13

By superimposition or by measuring the remaining corresponding parts, we observe that∠P = ∠A, PQ = AB and PR = AC establishing that ∆PQR ≅ ∆ABC, which again means thatequality of the three corresponding parts (two angles and the included side) of two trianglesresults in congruent triangles.

We also know that the sum of the three angles of a triangle is 180°, as such if two angles ofone triangle are equal to the corresponding angles of another triangle, then the third angleswill also be equal. Thus instead of included side we may have any pair of corresponding sidesequal. Thus we have

Criterion 2 : If any two angles and a side of one triangle are equal to correspondingangles and the side of the other triangle, then the two triangles are congruent.

This criterion is referred to as ASA or AAS.


14.6.1 ACTIVITY FOR YOU

In order to explore another criterion we again take a triangle ABC (See Fig. 14.14)

Fig. 14.14

Now take three thin sticks equal in lengths to sides AB, BC and CA of ∆ABC. Place themin any order to form ∆PQR or ∆P′Q′R′ near the ∆ABC (Fig. 14.15).

Fig. 14.15

By measuring the corresponding angles, we find that, ∠P = ∠P′ = ∠A, ∠Q = ∠Q′ = ∠B and∠R = ∠R′ = ∠C, establishing that

∆PQR ≅ ∆P′Q′R ≅ ∆ABC

which means that equality of the three corresponding sides of two triangles, results in congruenttriangles. Thus we have

Criterion 3 : If the three sides of one triangle are equal to the three correspondingsides of the other triangle, then the two triangles are congruent.

This is referred to as SSS (side, side, side).

Similarly, we can establish one more criterion which will be applicable for two right trianglesonly.

Criterion 4 : If the hypotenuse and a side of one triangle are respectively equalto the hypotenuse and a side of the other triangle, then the two right triangles arecongruent.

36 Mathematics

This criterion is referred to as R.H.S. (Right Hypotenuse side).

Using these criteria we can easily prove, knowing three corresponding parts only, whether twotriangles are congruent and establish the equality of remaining corresponding parts.

Example 14.1 : In which of the following criteria, two given triangles are NOT congruent.

(a) All corresponding sides are equal

(b) All corresponding angles are equal

(c) Two corresponding sides and their included angles are equal

(d) All corresponding angles and any pair of corresponding sides are equal.

Ans. (b)

Example 14.2 : Two rectilinear figures are congruent if they have only

(a) all corresponding sides equal

(b) all corresponding angles equal

(c) the same area

(d) all corresponding angles and all corresponding sides equal.

Ans. (d)

Example 14.3 : In Fig 14.16, PX and QY are perpendicular to PQ and PX = QY. Show thatAX = AY.

Fig. 14.16

Solution :

In ∆PAX and ∆QAY

∠XPA = ∠YQA (Each is 90°)


∠PAX = ∠QAY (Vertically opposite angles)

and PX = QY

∴ ∆PAX ≅ ∆QAY (AAS)

∴ AX = AY.

Example 14.4 : In Fig. 14.17, ∆ABC is a right triangle in which ∠B = 90° and D is the midpoint of AC.

Prove that BD = 12 AC.

Fig. 14.17

Solution : Produce BD to E such that BD = DE

Fig. 14.18

In ∆ADB and ∆CDE,

AD = CD (D being mid point of AC)

DB = DE (By construction)

and ∠ADB = ∠CDE (Vertically opposite angles)

∴ ∆ADB ≅ ∆CDE ...(i)

∴ AB = EC

Also ∠DAB = ∠DCE

But they make a pair of alternate angles

∴ AB is parallel to EC

∴ ∠ABC + ∠ECB = 180° (Pair of interior angles)

∴ ∠90° + ∠ECB = 180°

∴ ∠ECB = 180° – 90° = 90°

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Now in ∆ABC and ∆ECB,

AB = EC (From (i) above)

BC = CB (Common)

and ∠ABC = ∠ECB (Each 90°)

∴ ∆ABC ≅ ∆ECB∴ AC = EB

But BD =12 EB

∴ BD =12 AC


1. In ∆ABC (Fig. 14.19) if ∠B = ∠C and AD⊥ BC, then ∆ABD ≅ ∆ACD by the criterion.

Fig. 14.19

(a) RHS (b) ASA(c) SAS (d) SSS

2. In Fig. 14.20, ∆ABC ≅ ∆PQR. This congruence may also be written as

Fig. 14.20

(a) ∆BAC ≅ ∆RPQ (b) ∆BAC ≅ ∆QPR

(c) ∆BAC ≅ ∆RQP (d) ∆BAC ≅ ∆PRQ.


3. In order that two given triangles are congruent, along with equality of two correspondingangles, we must know the equality of

(a) No corresponding side

(b) Minimum one corresponding side

(c) Minimum two corresponding sides

(d) All the three corresponding sides

4. Two triangles are congruent,. if

(a) All three corresponding angles are equal

(b) Two angles and a side of one are equal to two angles and a side of the other.

(c) Two angles and a side of one are equal to two angles and the corresponding side ofthe other

(d) One angle and two sides of one are equal to one angle and two sides of the other.

5. In Fig.14.21, ∠B = ∠C and AB = AC. Prove that ∆ABE ≅ ∆ACD. Hence show thatCD = BE.

Fig. 14.21

6. In Fig. 14.22, AB is parallel to CD. If O is the mid point of BC, show that it is also themid point of AD.

Fig. 14.22

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7. In ∆ABC (Fig. 14.23), AD is ⊥ BC, BE is ⊥ AC and AD = BE. Prove that AE = BD.

Fig. 14.23

8. From Fig. 14.24, show that the triangles are congruent and make pairs of equal angles.

Fig. 14.24

14.7 ANGLES OPPOSITE TO EQUAL SIDES OF A TRIANGLE AND VICE VERSA

Using the criteria for congruence of triangles, we shall now prove some important theorems.

Theorem : The angles opposite to equal sides of a triangle are equal.

Given : A triangle ABC in which AB = AC.

To prove : ∠B = ∠C.

Construction : Draw bisector of ∠BAC meeting BC at D.

Proof : In ∆ABD and ∆ACD,

AB = AC (Given)

∠BAD = ∠CAD (By construction)

and AD = AD (Common) Fig. 14.25


∴ ∆ABD ≅ ∆ACD (SAS)

Hence ∠B = ∠C

The converse of the above theorem is also true. We prove it as a theorem.

14.7.1 The sides opposite to equal angles of a triangle are equal

Given : A triangle ABC in which ∠B = ∠C

To prove : AB = AC

Construction : Draw bisector of ∠BAC meeting BC at D.

Proof : In ∆ABD and ∆ACD,

∠B = ∠C (Given)

∠BAD = ∠CAD (By construction)

and AD = AD (Common)

∴ ∆ABD ≅ ∆ACD (AAS)Hence AB = AC

Hence the theorem.

Example 14.5 : Prove that the three angles of anequilateral triangle are equal.

Solution :

Given : An equilateral ∆ABC

To prove : ∠A = ∠B = ∠C

Proof : AB = AC (Given)

∴ ∠C = ∠B (Angles opposite equal sides) ...(i)

Also AC = BC (Given)

∴ ∠B = ∠A ...(ii)

From (i) and (ii),

∠A = ∠B = ∠C

Hence the result.

Example 14.6 : ABC is an isosceles triangle in which AB = AC(Fig. 14.28). If BD⊥ AC and CE⊥ AB, prove that BD = CE.

Solution : In ∆BDC and ∆CEB

∠BDC = ∠CEB (Each is 90°)

Fig. 14.28

Fig. 14.27

Fig. 14.26

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∠DCB = ∠EBC (Angles opposite equal sides of a ∆)

and BC = CB (Common)

∴ ∆BDC ≅ ∆CEB (AAS)

Hence BD = CE

This result can be stated in the following manner :

Perpendiculars drawn to equal sides from opposite vertices (or altitudes) of anisosceles triangle are equal.

The result can be extended to an equilateral triangle after which we can say that all the threealtitudes of an equilateral triangle are equal.

Example 14.7 : In ∆ABC (Fig. 14.29), D and E are mid points of AC and AB respectively.If AB = AC, then prove that BD = CE.

Solution : BE =12 AB

and CD =12 AC

∴ BE = CD ...(i)

In ∆BEC and ∆CDB,

BE = CD [By (i)]

BC = CB (Common)

and ∠EBC = ∠DCB (Q AB = AC)

∴ ∆BEC ≅ ∆CDBHence CE = BD

Example 14.8 : In ∆ABC (Fig. 14.30) AB = AC and ∠DAC = 124°;find the angles of the triangle.

Solution. ∠BAC = 180° – 124° = 56°

∠B = ∠C

(Angles opposite to equal sides of a triangle)

Also ∠B + ∠C = 124°

∴ ∠B = ∠C = 124

2°

= 62°

Hence ∠A = 56°, ∠B = 62°, and ∠C = 62°,

Fig. 14.29

Fig. 14.30



1. In Fig. 14.31, PQ = PR and SQ = SR. Prove that ∠PQS = ∠PRS.

Fig. 14.31

2. Prove that ∆ABC is a isosceles triangle, if the altitude AD bisects the base BC (Fig. 14.32).

Fig. 14.32

3. If the line l in Fig. 14.33 is parallel to the base BC of the isosceles ∆ABC find the angles.

Fig. 14.33

4. ∆ABC is an isosceles triangle such that AB = AC. Side BA is produced to a point D suchthat AB = AD. Prove that ∠BCD is a right angle.

Fig. 14.34

44 Mathematics

5. In Fig. 14.35, D is the mid point of BC and perpendiculars DF and DE to sides AB andAC are equal in length. Prove that ∆ABC is an isosceles triangle.

Fig. 14.35

6. In Fig. 14.36, PQ = PR. QS and RT are the angle bisector of ∠Q = ∠R respectively Provethat QS = RT.

Fig. 14.36

7. ∆PQR and ∆SQR are isosceles triangles on the same base QR (Fig. 14.37). Prove that∠PQS = ∠PRS.

Fig. 14.37

8. In ∆ABC, AB = AC (Fig. 14.38). P is any point in the interior of the triangle such that∠ABP = ∠ACP. Prove that AP bisects ∠BAC.

Fig. 14.38


14.8 INEQUALITIES IN A TRIANGLE

We have learnt the relationship between sides and angles of a triangle when they are equal.We shall now study some relations among sides and angles of a triangle, when they are unequal.

Fig. 14.39

In Fig. 14.39, triangle ABC has side AB longer than the side AC. Measure∠B and ∠C. You will find that these angles are not equal and ∠C is greater than ∠B. If yourepeat this experiment, you will always find that this observation is true. This can be provedeasily, as follows.

14.8.1 Theorem

If two sides of a triangle are unequal, then the longer side has the greater angle opposite toit.

Given. A triangle ABC in which AB > AC.

To prove. ∠ACB > ∠ABC

Construction. Mark a point D on the side AB such thatAD = AC and join DC.

Proof : In ∆ADC,

AD = AC

∴ ∠ACD = ∠ADC (Angles opposite equal sides)

But ∠ADC > ∠ABC

(Exterior angle is greater than opposite interior angles)

Again ∠ACB > ∠ACD(Point D lies in the interior of the ∠ACB).

∴ ∠ACB > ∠ABC

What can we say about the converse of this theorem. Let us examine.

In ∆ABC, (Fig. 14.41) compare ∠C and ∠B. It is clear that ∠Cis greater than ∠B. Now compare sides AB and AC opposite tothese angles by measuring them. We observe that AB is longerthan AC.

Again compare ∠C and ∠A and measure sides AB and BCopposite to these angles. We observe that ∠C > ∠A andAB > BC; i.e. side opposite to greater angle is longer. Fig. 14.41

Fig. 14.40

46 Mathematics

Comparing ∠A and ∠B, we observe a similar result. ∠A > ∠B and BC > AC; i.e. side oppositeto greater angle is longer.

You can also verify this property by drawing any type of triangle, a right triangle or an obtusetriangle.

Measure any pair of angles in a triangle. Compare them and then compare the sides oppositeto them by measurement. You will find the above result always true, which we state as aproperty.

In a triangle, the greater angle has longer side opposite to it.

Observe that in a triangle if one angle is right or an obtuse angle, then the side opposite tothat angle is the longest.

You have already learnt the relationship among the three angles of a triangle i.e., the sum ofthe three angles of a triangle is 180°, we shall now study whether the three sides of a triangleare related in some way.

Draw a triangle ABC.

Fig. 14.42

Measure its three sides AB, BC and CA.

Now find the sum of different pairs AB + BC, BC + CA, and CA + AB separately and compareeach sum of a pair with the third side, we observe that

(i) AB + BC > CA

(ii) BC + CA > AB and

(ii) CA + AB > BC.

Thus we conclude that

Sum of any two sides of a triangle is greater than the third side.

ACTIVITY FOR YOU

Fix three nails P, Q and R on a wooden board or any surface.


Fig. 14.43

Take a piece of thread equal in length to QR and another piece of thread equal inlength QP + PR. Compare the two lengths, you will find that the length corresponding toQP + PR > the length corresponding to QR confirming the above property.

Example 14.9 : In which of the following four cases, is construction of a triangle possiblefrom the given measurements :

(a) 5 cm, 8 cm and 3 cm

(b) 14 cm, 6 cm and 7 cm

(c) 3.5 cm, 2.5 cm and 5.2 cm

(d) 20 cm, 25 cm and 48 cm.

Solution. In (a) 5 + 3 > 8, in (b) 6 + 7 > 14

in (c) 3.5 + 2.5 > 5.2, 3.5 + 5.2 > 2.5 and 2.5 + 5.2 > 3.5 and

in (d) 20 + 25 > 48.

Ans. (c)

Example 14.10 : In Fig. 14.44, AD is a median of ∆ABC. Prove that AB + AC > 2AD.

Fig. 14.44 Fig. 14.45

Solution : Produce AD to E such that AD = DE and join C to E.

Consider ∆ABD and ∆ECD

Here, BD = CD

48 Mathematics

∠ADB = ∠EDCAD = ED

∴ ∆ABD ≅ ∆ECD∴ AB = EC

Now in ∆ACE,EC + AC > AE

or AB + AC > 2AD (Q AD = ED ⇒ AE = 2AD)


1. PQRS is a quadrilateral in which diagonals PR and QS intersect at O. Prove thatPQ + QR + RS + SP > PR + QS

2. In a triangle ABC, AB = 5.7 cm, BC = 6.2 cm and CA = 4.8 cm. Name the greatest andthe smallest angle.

3. In Fig. 14.46, if ∠CBD > ∠BCE then prove that AB > AC.

Fig. 14.46

4. In Fig. 14.47, D is any point on the base BC of a ∆ABC. If AB > AC then prove thatAB > AD.

Fig. 14.47

5. Prove that the sum of the three sides of triangle is greater than the sum of its three medians.(Use Example 14.10)


6. In Fig. 14.48, if AB = AD then prove that BC > CD.

[Hint : ∠ADB = ∠ABD].

Fig. 14.48

7. In Fig. 14.49, AB is parallel to CD. If ∠A > ∠B then prove that BC > AD.

Fig. 14.49

LET US SUM UP

Figures which have the same shape and same size are called congruent figures.

Congruent figures, when placed one over the other completely cover each other. All partsof one figure are equal to the corresponding parts of the other figure.

To prove that two triangles are congruent we need to know the equality of only threecorresponding parts. These corresponding parts must satisfy one of the four criteria

(i) SAS. (ii) ASA or AAS

(iii) SSS (iv) RHS

Angles opposite to equal sides of a triangle are equal.

Sides opposite to equal angles of a triangle are equal.

If two sides of a triangle are unequal, then the longer side has the greater angle oppositeto it.

In a triangle, the greater angle has the longer side opposite to it.

Sum of any two sides of a triangle is greater than the third side.

50 Mathematics

TERMINAL EXERCISE

1. Two lines AB and CD bisect each other at O. Prove that CA = BD (Fig. 14.50)

2. In a ∆ABC, if the median AD is perpendicular to the base BC then prove that the triangleis an isosceles triangle.

Fig. 14.50 Fig. 14.51

3. In Fig. 14.51, ∆ABC and ∆CDE are such that BC = CE and AB = DE. If ∠B = 60°,∠ACE = 30° and ∠D = 90°, then prove that the two triangles are congruent.

4. In Fig. 14.52, two sides AB and BC and the altitude AD of ∆ABC are respectively equalto the sides PQ and QR and the altitudes PS. Prove that ∆ABC ≅ ∆PQR.

Fig. 14.52

5. In a right triangle, one of the acute angles is 30°. Prove that the hypotenuse is twice theside opposite to the angle of 30°.

6. Line segments AB and CD intersect each other at O suchthat O is the midpoint of AB. If AC is parallel to DB thenprove that O is also the mid point of CD.

7. In Fig. 14.53, AB is the longest side and DC is the shortestside of a quadrilateral ABCD. Prove that ∠C > ∠A and∠D > ∠B.[Hint : Join AC and BD]. Fig. 14.53


8. ABC is an isosceles triangle in which AB = AC and AD is the altitude from A to thebase BC. Prove that BD = DC.

Fig. 14.54

9. Prove that the medians bisecting the equal sides of an isosceles triangle are also equal.

[Hint : Show that ∆DBC ≅ ∆ECB)

Fig. 14.55

52 Mathematics

ANSWERS


1. (a) 2 (b)

3. (b) 4. (c)

8. ∠P = ∠C ∠Q = ∠A and ∠R = ∠B.


3. ∠B = ∠C = 65°; ∠A = 50°


2. Greatest angle is A and smallest angle is B.

Concurrent Lines 53

15

Concurrent Lines

15.1 INTRODUCTION

You have already learnt about concurrent lines, in the lesson on lines and angles. You havealso studied about triangles and some special lines; i.e, medians, right bisectors of sides, anglebisectors and altitudes, which can be drawn in a triangle. In this lesson, we shall study theconcurrency property of these lines, which are quite useful.

15.2 OBJECTIVES


distinguish between an angle bisector and perpendicular bisector of a side, an altitude anda median of a triangle.

state and apply the concurrency property of angle bisectors, perpendicular bisectors ofsides, altitudes and medians of a triangle.


Properties of intersecting lines, such as :

Two lines in a plane can either be parallel [See Fig. 15.1(a)] or intersecting [See Fig.15.1(b) and (c)].

(a) (b) (c)

Fig. 15.1

54 Mathematics

Three lines in a plane may :

(i) be parallel to each other, i.e., intersect in no point [See Fig 15.2(a). or

(ii) intersect each other in exactly one point [Fig. 15.2(b), or

(iii) intersect each other in two points [Fig. 15.2(c), or

(iv) intersect each other at the most in three points [Fig. 15.2(d)]

(a) (b) (c) (d)

Fig. 15.2

15.4 CONCURRENT LINES

Three or more lines in a plane which intersect each other in exactly one point or which passthrough the same point are called concurrent lines and the common point is called the pointof concurrency (See Fig. 15.3).

(a) (b) (c)

Fig. 15.3

15.4.1 ANGLE BISECTORS OF A TRIANGLE

In triangle ABC, the line AD bisects ∠A of the triangle. (See Fig. 15.4)

Fig.15.4

Concurrent Lines 55

A line which bisects an angle of a triangle is called an angle bisector of the triangle.

How many angle bisectors can a triangle have ? Since a triangle has three angles, we can drawthree angle bisectors in it. AD is one of the three angle bisectors of ∆ABC. Let us draw secondangle bisector BE of ∠B (See Fig 15.5).

Fig. 15.5 Fig. 15.6

The two angle bisectors of the ∆ABC intersect each other as I. Let us draw the third anglebisector CF of ∠C (See Fig. 15.6). We observe that this angle bisector of the triangle also passesthrough I. In other words they are concurrent and the point of concurrency is I.

We may take any type of triangle— acute, right or obtuse triangle, and draw its angle bisectors,we will always find that the three angle bisectors of a triangle are concurrent. (see Fig. 15.7)

Fig. 15.7

Thus we conclude the following :

Angle bisectors of a triangle pass through the same point, that is they areconcurrent.

The point of concurrency I is called the ‘Incentre’ of the triangle.

Can you reason out, why the name incentre for this point ?

Recall that the locus of a point equidistant from two intersecting lines is the pair of anglebisectors of the angles formed by the lines. Since I is a point on bisector of ∠BAC, it mustbe equidistant from them. Also I is a point on angle bisector of ∠ABC, (See Fig. 15.8), it must

56 Mathematics

also be equidistant from them. Thus this point of concurrency I is at the same distance fromthe three sides of the triangle.

Fig. 15.8

Thus, we have IL = IM = IN (Fig. 15.8). Taking I as the centre and IL as the radius, we candraw a circle touching all the three sides, of the triangle, called ‘Incircle’ of the triangle. Ibeing the centre of the incircle is called the Incentre and IL the radius of the incircle is calledthe inradius of the triangle.

Note : That incentre always lies in the interior of the triangle.

15.4.2 Perpendicular bisectors of the sides of a triangle

ABC is a triangle, line DP bisects side BC at right angle. A line which bisects a side of a triangleat right angle is called the perpendicular bisector of the side. Since a triangle has three sides,so we can draw three perpendicular bisectors in a triangle. DP is one of the three perpendicularbisector of ∆ABC (Fig. 15.9). We draw the second perpendicular bisector EQ, intersecting DPat O (Fig. 15.10). Now if we also draw the third perpendicular bisector FR, then we observethat it also passes through the point O (Fig. 15.11). In other words, we can say that the threeperpendicular bisectors of the sides are concurrent at O.

Fig. 15.9 Fig. 15.10

Fig. 15.11

Concurrent Lines 57

We may repeat this experiment with any type of triangle, but we will always find that the threeperpendicular bisectors of the sides of a triangle pass through the same point.

(a) (b)

Fig. 15.12

Thus we conclude that :

The three perpendicular bisectors of the sides of a triangle pass through the samepoint, that is, they are concurrent.

The point of concurrency O is called the ‘circumcentre’ of the triangle

Can you reason out; why the name circumcentre for this point ?

Recall that the locus of a point equidistant from two given points is the perpendicular bisectorof the line joining the two points. Since O lies on the perpendicular bisector of BC, so it mustbe equidistant from both the point B and C, so that BO = CO (Fig. 15.13).

Fig. 15.13

The point O also lies on the perpendicular bisector of AC, so it must be equidistant from bothA and C, that is, AO = CO. Thus, we have AO = BO = CO.

58 Mathematics

If we take O as the centre and AO as the radius, we can draw a circle passing through thethree vertices, A, B and C of the triangle, called ‘circumcircle’ of the triangle. O being thecentre of this circle is called the circumcentre and AO the radius of the circumcircle is calledcircumradius of the triangle.

Note that the circumcentre will be

1. in the interior of the triangle for an acute triangle (Fig. 15.11)

2. on the hypotenuse for a right triangle [(Fig. 15.12(a)]

3. in the exterior of the triangle for an obtuse triangle [(Fig. 15.12(b)].

15.4.3 ALTITUDES OF A TRIANGLE

In ∆ABC, the line AL is the perpendicular drawn from vertex A to the opposite side BC. (Fig.15.14)

Fig. 15.14

Perpendicular drawn from a vertex of a triangle to the opposite side is called its altitude. Howmany altitudes can be drawn in a triangle ? There are three vertices in a triangle, so we candraw three of its altitudes. AL is one of these altitudes. Now we draw the second altitude BM,which intersects the first altitude at a point H (see Fig. 15.15). We also draw the third altitudeCN and observe that it also passes through the point H (Fig. 15.16). This shows that the threealtitudes of the triangle pass through the same point.

Fig. 1515 Fig. 15.16

Concurrent Lines 59

We may take any type of triangle and draw its three altitudes. We always find that the threealtitudes of a triangle are concurrent.

Fig. 15.17 Fig. 15.18


In a triangle, the three altitudes pass through the same point, that is, they areconcurrent.

The point of concurrency is called the ‘Orthocentre’ of the triangle.

Again observe that the orthocentre will be

1. in the interior of the triangle for an acute triangle (Fig. 15.16)

2. in the exterior of the triangle for an obtuse triangle (Fig. 15.17)

3. at the vertex containing the right angle for a right triangle (Fig. 15.18)

15.4.4 MEDIANS OF A TRIANGLE

In ∆ABC, AD joins the vertex A to the mid point D of the opposite side BC (Fig. 15.19)

(a) (b)Fig. 15.19

60 Mathematics

A line joining a vertex to the mid point of the opposite side of a triangle is called its median.Clearly, three medians can be drawn in a triangle, AD is one of the medians. If we draw allthe three medians in any triangle, we always find that the three medians pass through the samepoint [Fig. 15.20 (a), (b), (c)]

(a) (b) (c)

Fig. 15.20

Here in each of the triangles ABC given above (Fig. 15.20) the three medians AD, BE andCF are concurrent at G. In each triangle we measure the parts into which G divides each median.On measurement, we observe that

AG = 2GD, BG = 2GE

and CG = 2GF

that is, the point of concurrency G divides each of the medians in the ratio 2 : 1


Medians of a triangle pass through the same point, which divides each of themedians in the ratio 2 : 1.

The point of concurrency G is called the ‘centroid’ of the triangle.

ACTIVITY FOR YOU

Cut out a triangle from a piece of cardboard. Draw its three medians and mark the centroidG of the triangle. Try to balance the triangle by placing the tip of a pointed stick or a needleof compasses below the point G or at G. If the position of G is correctly marked then the weightof the triangle will balance at G (Fig. 15.21).

Fig. 15.21

Concurrent Lines 61

Can you reason out, why the point of concurrency of the medians of a triangle is called itscentroid. It is the point where the weight of the triangle is centred or it is the point throughwhich the weight of the triangle acts.

We consider some examples using these concepts.

Example 15.1 : In an isosceles triangle, show that the bisector of the angle formed by the equalsides is also a perpendicular bisector, an altitude and a median of the triangle.

Solution : In ∆ABD and ∆ACD

AB = AC (Given)

∠BAD = ∠CAD [Q AD is bisector of ∠A]

AD = AD

∴ ∆ABD ≅ ∆ACD

∴ BD = CD

⇒ AD is also a median

⇒ Also ∠ADB = ∠ADC = 90°

⇒ AD is perpendicular bisector of side BC

[Q ∠ADB = 90° ⇒ AD is an altitude also]

Example 15.2 : In an equilateral triangle, show that the three angle bisectors are also the threeperpendicular bisectors of sides, three altitudes and the three medians of the triangle.

Solution : Since AB = AC

∴ AD, the angle bisector of ∠A is also a perpendicularbisector of BC, an altitude and a median of the ∆ABC

(Refer Example 1 above)

Similarly, since AB = BC and BC = AC

∴ BE and CF, angle bisectors of ∠B and ∠C respectively,are also perpendicular bisectors, altitudes and medians of the∆ABC.

Example 15.3 : Find the circumradius of circumcircle and inradius of incircle of an equilateraltriangle of side a.

Solution : We draw perpendicular from the vertex A to the side BC.

AD is also the angle bisector of ∠A, perpendicular bisector of side BC and a median joiningvertex to the midpoint of BC

Fig. 15.22

Fig. 15.23

62 Mathematics

Fig. 15.24

∴ AD =3

2a as BC = a.

⇒ AG = circumradius in this case = 23

32

× a = 3

3a

and GD = inradius in this case = 13

32

× a = 3

6a .


1. In the given figure if BF = FC, ∠BAE = ∠CAE and ∠ADE = ∠GFC = 90° then namea median, an angle bisector, and altitude and a perpendicular bisector of the triangle.

Fig. 15.25

2. In an equilateral triangle show that the incentre, the circumcentre, the orthocentre and thecentroid are the same point.

3. In an equilateral ∆ABC (Fig. 15.26), G is the centroid of the triangle. If AG is 4.8 cm,find BE.

Concurrent Lines 63

Fig. 15.26

4. If H is the orthocentre of ∆ABC, then show that A is the orthocentre of the ∆HBC.

5. Choose the correct answers out of the given alternatives in the following questions :

(i) In a plane, the point equidistant from vertices of a triangle is called its

(a) centroid (b) incentre

(c) circumcentre (d) orthocentre

(ii) In the plane of a triangle, the point equidistant from the sides of the triangle is calledits

(a) entroid (b) incentre

(c) circumcentre (d) orthocentre

LET US SUM UP

Three or more lines in a plane which intersect each other in exactly one point are calledconcurrent lines.

A line which bisects an angle of a triangle is called an angle bisector of the triangle.

A line which bisects a side of a triangle at right angle is called a perpendicular bisectorof the triangle.

A line drawn perpendicular from a vertex of a triangle to its opposite side is called analtitude of the triangle.

A line which joins a vertex of a triangle to the mid-point of the opposite side is calleda median.

In a triangle

(i) Angle bisectors are concurrent and the point of concurrency is called incentre.

(ii) Perpendicular bisectors of the sides are concurrent and the point of concurrency iscalled circumcentre.

(iii) Altitudes are concurrent and the point of concurrency is called orthocentre.

(iv) Medians are concurrent and the point of concurrency is called centroid, which divideseach of the medians in the ratio 2 : 1.

64 Mathematics

TERMINAL EXERCISE

1. In the given Fig. 15.27 D, E and F are the mid points of the sides of ∆ABC. Show that

BE + CF > 32 BC.

Fig. 15.27

2. ABC is an isosceles triangle such that AB = AC and D is the midpoint of BC. Show thatthe centroid, the incentre, the circumcentre and the orthocentre, all lie on AD.

Fig. 15.28

3. ABC is an isosceles triangle such that AB = AC = 17 cm and base BC = 16 cm. If Gis the centroid of ∆ABC, find AG.

4. ABC is an equilateral triangle of side 12 cm. If G be its centroid, find AG.

ACTIVITIES FOR YOU :

1. Draw a triangle ABC and find its circumcentre. Also draw the circumcircle of the triangle.

2. Draw an equilateral triangle. Find its incentre and circumcentre. Draw its incircle andcircumcircle.

3. Draw the circumcircle and the incircle for an equilateral triangle of side 5 cm.

Concurrent Lines 65

ANSWERS


1. Median–AF, Angle bisector AE

Altitude – AD and perpendicular bisector – GF

3. AD = 7.2 cm also BE = 7.2 cm

5. (i) (c) (ii) (b)

Terminal Exercise

3. AG = 10 cm

4. AG = 4 3 cm.

66 Mathematics

16

Quadrilaterals

16.1 INTRODUCTION

If you look around, you will find many objects bounded by four lines. A book, window door,some parts of window-grill, slice of bread, the floor of your room are all examples of a closedfigure bounded by four line segments, such a figure is called a quadrilateral.

The word quadrilateral has its origin from the two words “quadric” meaning four and “lateral”meaning sides. Thus, a quadrilateral is that geometrical figure which has four sides, enclosinga part of the plane.

In this lesson, we shall study about terms and concepts related to quadrilateral with theirproperties.

16.2 OBJECTIVES


describe various types of quadrilaterals viz. trapeziums, parallelograms, rectangles,rhombuses and squares.

verify properties of different types of quadrilaterals.

verify that in a triangle the line segment joining the mid-points of any two sides is parallelto the third side and is half of it.

verify that the line drawn through the mid-point of a side of a triangle parallel to anotherside bisects the third side.

verify that if there are three or more parallel lines and the intercepts made by them ona transversal are equal, the corresponding intercepts on any other transversal are also equal.

verify that a diagonal of a parallelogram divides it into two triangles of equal area.

prove that parallelograms on equal (or same) bases and between the same parallels areequal in area.

verify that triangles on the same or equal bases and between the same parallels are equalin area and its converse.

Quadrilaterals 67


Drawing line-segments and angles of given measure.

Drawing circles/arcs of given radius.

Drawing parallel and perpendicular lines.

Four fundamental operations on numbers.

16.4 QUADRILATERAL

Recall that if A, B, C and D are four points in a plane such that no three of them are collinearand the line segment AB, BC, CD and DA do not intersect except at their end points, thenthe closed figure made up of four line segments is called a quadrilateral with vertices A, B,C and D. A quadrilateral with vertices A, B, C and D is generally denoted by quad. ABCD.In Fig. 16.1 (i) and (ii), both the quadrilaterals can be named as quad. ABCD.

In quadrilateral ABCD,

(i) (ii)

Fig. 16.1

(i) AB and DC ; BC and AD are two pairs of opposite sides.

(ii) ∠A and ∠C ; ∠B and ∠D are two pairs of opposite angles.

(iii) AB and BC ; BC and CD are two pairs of consecutive sides or adjacent sides. Can youname the other pairs of consecutive sides ?

(iv) ∠A and ∠B ; ∠B and ∠C are two pairs of consecutive angles or adjacent angles. Canyou name the other pairs of consecutive angles ?

(v) AC and BD are the two diagonals.

In Fig. 16.2, angles denoted by 1, 2, 3 and 4 are the interior angles or the angles of thequad. ABCD. Angles denoted by 5, 6, 7 and 8 are the exterior angles of the quad. ABCD.

Measure ∠1, ∠2, ∠3 and ∠4

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(i) (ii)

Fig. 16.2

What is the sum of these angles ? You will find that ∠1 + ∠2 + ∠3 + ∠4 = 360°.

i.e. sum of interior angles of a quadrilateral equals 360°.

Also what is the sum of exterior angles of the quadrilateral ABCD ?

You will again find that ∠5 + ∠6 + 7 + ∠8 = 360°

i.e., sum of exterior angels of a quadrilateral is also 360°.

16.5 TYPES OF QUADRILATERALS

You are familiar with quadrilaterals and their different shapes. You also know how to namethem. However, we will now study different types of quadrilaterals in a systematic way. Afamily tree of quadrilaterals is given in Fig. 16.3 below :

Quadrilateral

Parallelogram Trapezium Kite

Rectangle Rhombus

SquareFig. 16.3

Let us describe them one by one.

16.5.1 Trapezium

A quadrilateral which has one pair of opposite sides parallel is called a trapezium. In Fig. 16.4[(i) and (ii)] ABCD and PQRS are trapeziums with AB || DC and PQ || SR respectively.

Quadrilaterals 69

(i) (ii)

Fig. 16.4

16.5.2 Parallelogram

A quadrilateral which has both pairs of opposite sides parallel, is called a parallelogram. InFig. 16.5 [(i) and (ii)] ABCD and PQRS are parallelograms with AB||DC and AD||BC. Theseare denoted by ||gm ABCD and ||gm PQRS.

(i) (ii)

Fig. 16.5

16.5.3 Rhombus

A rhombus is a parallelogram in which any pair of adjacent sides is equal.

In Fig. 16.6 ABCD is a rhombus.

Fig. 16.6

70 Mathematics

You may note that ABCD is a parallelogram with AB = BC = CD = DA i.e., each pair of adjacentsides is equal.

16.5.4 Rectangle

A parallelogram one of whose angles is a right angle is called a rectangle.

In Fig. 16.7, ABCD is a rectangle in which AB||DC, AD||BC

and ∠A = ∠B = ∠C = ∠D = 90°.

Fig. 16.7

16.5.5 Square

A square is a rectangle, with a pair of adjacent sides equal.

In other words, a parallelogram having all sides equal and each angle equal to a right angleis called a square.

Fig. 16.8

In Fig. 16.8, ABCD is a square in which AB||DC, AD||BC, and AB = BC = CD = DA and∠A = ∠B = ∠C = ∠D = 90°.


Quadrilaterals 71

Example 16.1 : In Fig. 16.9, PQR is a triangle. S and T are two points on the sides PQ andPR such that ST||QR. Name the type of quadrilateral STRQ so formed.

Fig. 16.9

Solution : Quadrilateral STRQ is a trapezium, because ST||QR.Example 16.2 : The three angles of a quadrilateral are 100°, 50° and 70°. Find the measureof the fourth angle.Solution : We know that the sum of the angles of a quadrilateral is 360°.Let the fourth angle be x.

Then 100° + 50° + 70° + x° = 360°220° + x° = 360°

x = 140Hence, the measure of fourth angle is 140°.


1. Name each of the following quadrilaterals.

(i) (ii) (iii)

(iv) (v) (vi)Fig. 16.10

72 Mathematics

2. States which of the following statements are correct ?

(i) Sum of interior angles of a quadrilateral is 360°.

(ii) All rectangles are squares.

(iii) A rectangle is a parallelogram.

(iv) A square is a rhombus.

(v) A rhombus is a parallelogram.

(vi) A square is a parallelogram.

(vii) A parallelogram is a rhombus.

(viii) A trapezium is a parallelogram.

(ix) A trapezium is a rectangle.

(x) A parallelogram is a trapezium.

3. In a quadrilateral, all its angles are equal. Find the measure of each angle.

4. The angles of a quadrilateral are in the ratio 5 : 7 : 7 : 11. Find the measure of each angle.

5. If a pair of opposite angles of a quadrilateral are supplementary, what can you say aboutthe other pair of angles ?

16.6. PROPERTIES OF DIFFERENT TYPES OF QUADRILATERALS

16.6.1 Properties of a Parallelogram

We have learnt that a parallelogram is a quadrilateral with both pairs of opposite sides parallel.Now let us establish some relationship between sides, angles and diagonals of a parallelogram.

Draw a pair of parallel lines l and m as shown in Fig. 16.11. Draw another pair of parallellines p and q such that they intersect l and m. You observe that a parallelogram ABCD is formed.Join AC and BD. They intersect each other at O.

(i) (ii)

Fig. 16.11

Quadrilaterals 73

Now measure the sides AB, BC, CD and DA. What do you find ?

You will find that AB = DC and BC = AD.

Also measure ∠ABC, ∠BCD, ∠CDA and ∠DAB.

What do you find ?

You will find that ∠DAB = ∠DCB and ∠ABC = ∠CDA

Again, Measure OA, OC, OB and OD.

What do you find ?

You will find that OA = OC and OB = OD

Draw another parallelogram and repeat the activity you will find that

(i) The opposite sides of a parallelogram are equal.

(ii) The opposite angles of a parallelogram are equal.

(iii) The diagonals of a parallelogram bisect each other.

The above said properties of a parallelogram can also be verified by Cardboard model whichis as follows :

Let us take a card board. Draw any parallelogram ABCD on it. Draw its diagonal AC as shownin Fig 16.12. Cut the parallelogram ABCD from the cardboard. Now cut this parallelogramalong the diagonal AC. Thus, the parallelogram has been divided into two parts and each partis a triangle.

(i) (ii)

Fig. 16.12

In other words, you get two triangles, ∆ABC and ∆ADC. Now place ∆ADC on ∆ABC in sucha way that the vertex D falls on the vertex B and the side CD falls along the side AB.

Where does the point C fall ?

Where does the point A fall ?

You will observe that ∆ADC will coincide with ∆ABC. In other words ∆ABC≅ ∆ADC. AlsoAB = CD and BC = AD and ∠B = ∠D.

74 Mathematics

You may repeat this activity by taking some other parallelograms, you will always get the sameresults as verified earlier, thus, proving the above two properties of the parallelogram.

Now you can prove the third property of the parallelogram, i.e., the diagonals of a parallelogrambisect each other.

Again take a thin cardboard. Draw any parallelogram PQRS on it. Draw its diagonals

PR and QS which intersect each other at O as shown in Fig. 16.13. Now cut the parallelogramPQRS.

Fig. 16.13

Also cut ∆POQ and ∆ROS.

Now place ∆ROS and ∆POQ in such a way that the vertex R coincides with the vertex P andRO coincides with the side PO.

Where does the point S fall ?

Where does the side OS fall ?

Is ∆ROS ≅ ∆POQ ? Yes, it is.

So, what do you observe ?

We find that RO = PO and OS = OQ

You may also verify this property by taking another pair of triangles i.e. ∆POS and ∆ROQ.You will again arrive at the same result.

You may also verify the following properties which are the converse of the properties of aparallelogram proved earlier.

(i) A quadrilateral is a parallelogram if its opposite sides are equal.

(ii) A quadrilateral is a parallelogram if its opposite angles are equal.

(iii) A quadrilateral is a parallelogram if its diagonals bisect each other.

16.6.2 Properties of a Rhombus

In the previous section we have defined a rhombus. We know that a rhombus is a parallelogramin which a pair of adjacent sides are equal. In Fig. 16.14, ABCD is a rhombus.

Quadrilaterals 75

Fig. 16.14

Thus, ABCD is a parallelogram with AB = BC. Since every rhombus is a parallelogram,therefore all the properties of a parallelogram are also true for rhombus, i.e.

(i) Opposite sides are equal,

i.e., AB = DC and AD = BC

(ii) Opposite angles are equal,

i.e., ∠A = ∠C and ∠B = ∠D

(iii) Diagonals bisect each other

i.e., AO = OC and DO = OB

Since adjacent sides of a rhombus are equal and by the property of a parallelogram

AB = BC = CD = DA

Thus, all the sides of a rhombus are equal.

Measure ∠AOD and ∠BOC.

What is the measures of these angles ?

You will find that each of them equals 90°

Also ∠AOB = ∠COD (Each pair is a vertically opposite angles)

and ∠BOC = ∠DOA

∠AOB = ∠COD = ∠BOC = ∠DOA = 90°

Thus, the diagonals of a rhombus bisect each other at right angles.

You may repeat this experiment by taking different rhombuses, you will find in each case, thediagonals of a rhombus bisect each other.

Thus, we have the following properties of a rhombus.

(i) All sides of a rhombus are equal

(ii) The opposite angles of a rhombus are equal

(iii) The diagonals of a rhombus bisect each other at right angles.

76 Mathematics

16.6.3 Properties of a Rectangle

We know that a rectangle is a parallelogram one of whose angles is a right angle. Can yousay whether a rectangle possesses all the properties of a parallelogram or not ?

Yes it is. Let us study some more properties of a rectangle

Draw a parallelogram ABCD in which ∠B = 90°.

Join AC and BD as shown in the Fig. 16.15

Fig. 16.15

Measure ∠ΒAD, ∠BCD and ∠ADC, what do you find ?

What are the measures of these angles ?

The measure of each angle is 90°. Thus, we can conclude that ∠A = ∠B = ∠C = ∠D = 90°i.e., each angle of a rectangle measures 90°. Now measure the diagonals AC and BD. Alsomeasure AO, OC, BO and DO. Do you find that AC = BD ? Yes, it is

You will also find that AO = OC and BO = DO.

Draw some more rectangles of different dimensions. Label them again by ABCD. Join ACand BD in each case. Let them intersect each other at O. Also measure AO, OC and BO, ODfor each rectangle. In each case you will find that

The diagonals of a rectangle are equal and they bisect each other.

Thus, we have the following properties of a rectangle :

(i) The opposite sides of a rectangle are equal

(ii) Each angle of a rectangle is a right-angle.

(iii) The diagonals of a rectangle are equal.

(iv) The diagonals of a rectangle bisect each other.

16.6.4 Properties of a Square

You know that a square is a rectangle, with a pair of adjacent sides equal. Now, can you concludefrom definition of a square that a square is rectangle and possesses all the properties of arectangle ? Yes it is, Let us now study some more properties of a square.

Quadrilaterals 77

Draw a square ABCD as shown in Fig. 16.16.

Fig. 16.16

Since ABCD is a rectangle, therefore we have

(i) AB = DC, AD = BC

(ii) ∠A = ∠B = ∠C = ∠D = 90°

(iii) AC = BD and AO = OC, BO = OD

But in a square we have AB = AD

∴ By property (i) we have

AB = AD = CD = BC.

Now measure ∠AOB, ∠BOC, ∠COD and ∠AOD.

What do you find ?

Does each angle measure 90° ? Yes

Thus, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

Thus, we conclude that the diagonals AC and BD of a square bisect each other at right angles.You may also observe that since a square is a parallelogram also with AB = AD, thereforea square ABCD is a rhombus also.

Thus, we have the following properties of a square

(i) All the sides of a square are equal.

(ii) Each of the angles measure 90°.

(iii) The diagonals of a square are equal.

(iv) The diagonals of a square bisect each other at right angles.

Let us study some examples to illustrate :

78 Mathematics

Example 16.3 : In Fig. 16.17, ABCD is a parallelogram. If ∠A = 80°, find the measures ofthe remaining angles.

Solution : As ABCD is a parallelogram.

∴ ∠A = ∠C and ∠B = ∠D.

It is given that

∠A = 80°

∴ ∠C = 80°.

Q AB || DC

∴ ∠A + ∠D = 180°

∴ ∠D = (180 – 80)° = 100°

∴ ∠B = ∠D = 100°

Hence, ∠C = 80°, ∠B = 100° and ∠D = 100°.

Example 16.4 : Two adjacent angles of a rhombus are in the ratio 4 : 5. Find the measureof all its angles.

Solution : We know that the sum of two adjacent angles of a rhombus is 180°.

Let the two angels be 4x and 5x,

then 4x + 5x = 180

i.e. 9x = 180

x = 20

∴ The two angles are 80° and 100°

i.e. ∠A = 80° and ∠B = 100°

Since ∠A = ∠C ⇒ ∠C = 80°

Also, ∠B = ∠D ⇒ ∠D = 100°

Hence, the angles of the rhombus are 80°, 100°, 80° and 100°.

Example 16.5 : One of the diagonals of a rhombus is equal to one of its sides. Find the anglesof the rhombus.

Solution : In rhombus, ABCD,

AB = AD = BD

∴ ∆ABD is an equilateral triangle.

∴ ∠DAB = ∠1 = ∠2 = 60° ...(i)

Similarly, ∠BCD = ∠3 = ∠4 = 60° ...(ii)

Fig. 16.17

Fig. 16.18

Fig. 16.19

Quadrilaterals 79

Also from (i) and (ii)∠ABC = ∠B = ∠1 + ∠3 = 60° + 60° = 120°∠ADC = ∠D = ∠2 + ∠4 = 60° + 60° = 120°

Hence, ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.

Example 16.6 : The diagonals of a rhombus ABCD intersect at O. If ∠ADC = 120° andOD = 6 cm, find

(i) ∠OAD

(ii) side AB

(iii) perimeter of the rhombus ABCD.

Solution : Given that

∠ADC = 120°

i.e., ∠ADO + ∠ODC = 120°

But ∠ADO = ∠ODC (∆AOD ≅ ∆COD)

∴ 2∠ADO = 120°

i.e. ∠ADO = 60° ...(i)

Also, we know that the diagonals of a rhombus bisect each that at 90°.

∴ ∠DOA = 90° ...(ii)

Now, in ∆DOA

∠ADO + ∠DOA + ∠OAD = 180°

From (i) and (ii), we have

60° + 90° + ∠OAD = 180°

⇒ ∠OAD = 30°

∴ ∠DAB = 60°

∴ ∆DAB is an equilateral triangle

(ii) Now OD = 6 cm

⇒ OD + OB = BD

6 cm + 6 cm = BD

⇒ BD = 12 cm

Since, AB = BD = AD = 12 cm

⇒ AB = 12 m.

(iii) Now Perimeter = 4 × side

= (4 × 12) cm

= 48 cm

Hence, the perimeter of the rhombus = 48 cm.

Fig. 16.20

80 Mathematics


1. In a parallelogram ABCD, ∠A = 62°. Find the measures of the other angles.

2. The sum of the two opposite angles of a parallelogram is 150°. Find all the angles ofthe parallelogram.

3. In a parallelogram ABCD, ∠A = (2x + 10)° and ∠C = (3x – 20)°.

Find the value of x.

4. ABCD is a parallelogram in which ∠DAB = 70° and ∠CBD = 55°. Find ∠CDB and∠ADB.

5. ABCD is a rhombus in which ∠ABC = 58°. Findthe measure of ∠ACD.

6. In Fig. 16.21, the diagonals of a rectangle PQRSintersect each other at O. If ∠ROQ = 40°, find themeasure of ∠OPS.

7. AC is one diagonal of a square ABCD. Find themeasure of ∠CAB.

16.7 MID POINT THEOREM

Draw any triangle ABC. Find the mid points of the side AB and AC. Mark them as D andE respectively. Join DE, as shown in Fig. 16.22.

Measure BC and DE.

What relation do you find between the length of BC and DE ?

Of course, it is, DE =12 BC

Again, measure ∠ADE and ∠ABC.

Are these angles equal ?

Yes, they are equal, you know that these angles make a pair of corresponding angles. You knowthat when a pair of corresponding angles are equal, the lines are parallel.

∴ DE || BC

You may repeat this experiment with another two or three triangles and naming each of themas triangle ABC and the mid points as D and E of sides AB and AC.

You will always find that DE = 12 BC and DE || BC.

Thus, we conclude that

Fig. 16.21

Fig. 16.22

Quadrilaterals 81

In a triangle the line-segment joining the mid points of any two sides is parallelto the third side and is half of it.

We can also verify the converse of the above stated result.

Draw any ∆PQR. Find the mid point of side RQ, and markit as L. From L, draw a line LX || PQ, which intersects,PR at M.

Measure PM and MR. Are they equal ? Yes, they areequal.

You may repeat with different triangles and by namingeach of them as PQR and taking each time L as the mid-point RQ and drawing a line LM || PQ, you will find ineach case that RM = MP. Thus, we conclude that

“The line drawn through the mid point of one side of a triangle parallel to theanother side bisects the third side.

Let us consider some examples.

Example 16.7 : In Fig. 16.24, D is the mid-point of the side AB of ∆ABC and DE || BC. IfAC = 8 cm, find AE.

Solution : In ∆ABC, DE || BC and D is the mid point of AB

∴ E is also the mid point of AC

i.e. AE =12 AC

=12

8×FHGIKJ cm [Q AC = 8 cm]

= 4 cm

Hence AE = 4 cm.

Example 16.8 : In Fig. 16.25, ABCD is a trapezium in whichAD and BC are its non-parallel sides and E is the mid-point ofAD. EF||AB. Show that F is the mid-point of BC.

Solution : Since EG||AB and E is the mid-point of AD

∴ G is the mid point of DB

In ∆DBC, GF||DC and G is the mid-point of DB,

∴ F is the mid-point of BC.

Fig. 16.23

Fig. 16.24

Fig. 16.25

82 Mathematics

Example 16.9 : ABC is a triangle, in which P, Q and R are mid-points of the sides AB, BCand CA respectively. If AB = 8 cm, BC = 7 cm, and CA = 6 cm, find the sides of the trianglePQR.

Solution : P is the mid-point of AB and R is the mid-point of AC

∴ PR||BC and PR =12 BC

=12 × 7 [Q BC = 7 cm]

= 3.5 cm

Similarly, PQ =12 AC

=12 × 6 cm [Q AC = 6 cm]

= 3 cm

and QR =12 AB

=12 × 8 cm [Q AB = 8 cm]

= 4 cm.

Hence, the sides of ∆PQR are PQ = 3 cm, QR = 4 cm and PR = 3.5 cm.


1. In Fig. 16.27, ABC is an equilateral triangle D, E and F are the mid-points of the sidesAB, BC and CA respectively. Prove that DEF is also an equilateral triangle.

Fig. 16.27

Fig. 16.25

Quadrilaterals 83

2. In Fig. 16.28, D and E are the mid-points of the sides AB and AC respectively of a ∆ABC.If BC = 10 cm; find DE.

Fig. 16.28

3. In Fig. 16.29, AD is a median of a ∆ABC and E is the mid-point of AD. BE is producedto meet AC at F. DG||EF, meets AC at G. If AC = 9 cm, find AF.

Fig. 16.29

4. In Fig. 16.30, A and C divide the side PQ of ∆PQR into three equal parts. AB||CD||QR.Prove that B and D also divide PR into three equal parts.

Fig. 16.30

5. In Fig. 16.31, ABC is a isosceles triangle in which AB = AC. M is the mid-point of ABand MN||BC. Show that ∆AMN is also an isosceles.

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Fig. 16.31

16.8 INTERCEPT THEOREM

Recall that a line which intersects two or more lines is called a transversal. The segment cutoff from the transversal by a pair of lines is called an intercept. Thus, in Fig. 16.32, XY isan intercept made by line l and m on transversal n.

Fig. 16.32

The intercepts made by parallel lines on a transversal have some special properties which weshall study here.

Let l and m be two parallel lines and XY is an intercept made on the transversal “n”. If thereare three parallel lines and they are intersected by a transversal, there will be two interceptsAB and BC as shown in Fig. 16.33 (ii).

(i) (ii)

Fig. 16.33

Quadrilaterals 85

Now let us learn an important property of intercepts made on the transversals by the parallellines.

On a page of your note-book, draw any two transversals l and m intersecting the parallel linesp, q, r and s as shown in Fig. 16.34. These transversal make different intercepts. Measure theintercept AB, BC and CD. Are they equal ? Yes, they are equal.

Fig. 16.34

Also, measure LM, MN and NX. Do you find that they are also equal ? Yes, they are.

Repeat this experiment by taking another set of two or more equidistant parallel lines andmeasure their intercepts as done earlier. You will find in each case that the intercepts madeare equal.

Thus, we conclude the following :

If there are three or more parallel lines and the intercepts made by them on atransversal are equal, the corresponding intercepts made on any other transversalare also equal.

Let us illustrate it by some examples :

Example 16.10 : In Fig 16.35, p || q || r. The transversals l, m andn cut them at L, M, N; A, B, C and X, Y, Z respectively suchthat XY = YZ. Name the other pairs of equal intercepts.

Solution. Given that XY = YZ

∴ AB = BC (Intercept theorem)

and LM = MN

Thus, the other pairs of equal intercepts are

AB = BC and LM = MN. Fig. 16.35

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Example 16.11. In Fig. 16.36, l || m || n and PQ = QR.

If XZ = 20 cm, find YZ.

Solution. We have PQ = QR

∴ By intercept theorem,

XY = YZ

Also XZ = XY + YZ

= YZ + YZ

∴ 20 = 2YZ ⇒ YZ = 10 cm

Hence, YZ =10 cm.


1. In Fig 16.37, l, m and n are three equidistant parallel lines. AD, PQ and GH are threetransversals. If BC = 2 cm and LM = 2.5 cm and AD ||PQ, find MS and MN.

Fig. 16.37

2. From Fig 16.38, when can you say that AB = BC and XY = YZ ?

Fig. 16.38

Fig. 16.36

Quadrilaterals 87

3. In Fig. 16.39, LM = MZ = 3 cm, find XY, XP and BZ. Given that l || m || n andPQ = 3.2 cm, AB = 3.5 cm and YZ = 3.4 cm.

Fig. 16.39

16.9 THE DIAGONAL OF A PARALLELOGRAM AND DIVISION OF AREA

Draw a parallelogram ABCD. Join its diagonal AC. Draw DP ⊥ DC and QC ⊥ DC.

Consider the two triangles ADC and ACB in which theparallelogram ABCD has been divided by the diagonalAC. Because AB || DC, therefore PD = QC.

Now, Area ∆ADC =12 DC × PD ...(i)

Area of ∆ACB =12 AB × QC ...(ii)

As AB = DC and PD = QC

∴ Area (∆ADC) = Area (∆ACB)


A diagonal of a parallelogram divides it into two triangles of equal area.

16.10 PARALLELOGRAMS AND TRIANGLES BETWEEN THE SAME PARALLELS

Two parallelograms or triangles, having equal or same bases and having their other verticeson a line parallel to their bases, are said to be on the same or equal bases and between thesame parallels.

We will prove an important theorem on parallelogram and their area

Theorm : Parallelograms on the same base (or equal bases) and between the sameparallels are equal in area.

Let us prove it logically.

Fig. 16.40

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Given : Parallelograms ABCD and PBCQ standon the same base BC and between the sameparallels BC and AQ.

To prove : Area (ABCD) = Area (BCQP)

Proof : Consider two triangles ABP and DCQ,

we have AB = DC

(Opposite sides of a parallelogram)

and BP = CQ (Opposite sides of a parallelogram)

∠1 = ∠2 (Corresponding Angles)

∴ ∆ABP ≅ ∆DCQ

∴ Area (∆ABP) = Area (∆DCQ) ...(i)

Now, Area (||gm ABCD) = Area(∆ABP) + Area(Trap. BCDP) ...(ii)

Area (||gm BCQP) = Area (∆CQD) + Area(Trap. BCDP) ...(iii)

From (i), (ii) and (iii), we get

Area (||gm ABCD) = Area (||gm BCQP)

Note : ||gm and Trap. stands for parallelogram and trapezium respectively.

Result : Triangles, on the same base and between the same parallels,are equal in area

Consider Fig. 16.41. Join the diagonals BQ and AC of the two parallelograms BCQP and ABCDrespectively. We know that a diagonal of a ||gm divides it in two triangles of equal area.

∴ Area (∆BCQ) = Area (∆PBQ)

and Area(∆ABC) = Area (∆CAD)

∴ Area (∆ABC) = Area (∆BCQ)

Parallelogram on the same base (or equal bases) and between the same parallelsare equal in area.

Thus, we also conclude the following :

Triangles on the same base (or equal bases) and between the same parallels areequal in area.

16.11 TRIANGLES ON THE SAME OR EQUAL BASES HAVING EQUAL AREASHAVE THEIR CORRESPONDING ALTITUDES EQUAL

Recall that the area of triangle = 12 (Base) × Altitude

Fig. 16.41

Quadrilaterals 89

Fig. 16.42

Here BC = QR

and Area (∆ABC) = Area (∆DBC) = Area (∆PQR) [Given] ...(i)

Draw perpendiculars DE and PS from D and P to the line m meeting it in E and S respectively.

Now Area (∆ABC) =12 BC × DE

Area (∆DBC) =12 BC × DE ...(ii)

and Area (∆PQR) =12 QR × PS

Also, BC = QR (Given) ...(iii)From (i), (ii) and (iii), we get

12 BC × DE =

12 QR × PS

or,12 BC × DE =

12 BC × PS

∴ DE = PSi.e., Altitudes of ∆ABC, ∆DBC and ∆PQR are equal in length.


Triangles on the same or equal bases, having equal areas have their correspondingaltitudes equal.

Let us consider some examples :

Example 16.12 : In Fig. 16.43, the area of parallelogram ABCD is 40 sq cm. If BC = 8m,find the altitude of parallelogram BCEF.

Solution : Area of ||gm BCEF = Area of ||gm ABCD = 40 sq cm ...(i)

we know that Area (||gm BCEF) = EF × Altitude

or 40 = BC × Altitude of ||gm BCEF

or, 40 = 8 × Altitude of ||gm BCEF

∴ Altitude of ||gm BCEF =408 cm or 5 cm.

Fig. 16.43

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Example 16.13 : In Fig. 16.44, the area of ∆ABC is given to be 18 cm2. If the altitude DLequals 4.5 cm, find the base of the ∆BCD.

Solution : Area (∆BCD) = Area (∆ABC) = 18cm2

Let the base of ∆BCD be x cm

∴ Area of ∆BCD =12 x × DL

=12

4 5x ×FHGIKJ. cm2

or 18 =94

xFHGIKJ

∴ x = 18 49

×FHGIKJ cm = 8 cm.

Example 16.14 : In Fig. 16.45, ABCD and ACED are two parallelograms. If area of ∆ABCequals 12 cm2, and the length of CE and BC are equal, find the area of the trapezium ABED.

Solution : Area(||gm ABCD) = Area(||gm ACED)

The diagonal AC divides the ||gm ABCD into twotriangles of equal area.

∴ Area (∆ABC) =12 Area (||gm ABCD)

∴ Area (||gm ABCD)=Area (||gm ACED) = 2 × 12 cm2

= 24 cm2

∴ Area of Trapezium ABED

= Area (∆ABC) + Area (||gm ACED)

= (12 + 24) cm2

= 36 cm2.


1. When are two parallelograms on the same base (or equal bases) of equal areas ?

2. The area of a triangle formed by joining the diagonal AC of a ||gm ABCD is 16 cm2. Findthe area of the ||gm ABCD.

Fig. 16.44

Fig. 16.45

Quadrilaterals 91

3. The area of ∆ACD in Fig 16.46 is 8 cm2. If EF = 4 cm, find the altitude of ||gm BCFE.

Fig. 16.46

LET US SUM UP

A quadrilateral is a four sided closed figure, enclosing some area of the plane.The sum of the interior or exterior angles of a quadrilateral is each equal to 360°.A quadrilateral is a trapezium if its one pair of opposite sides is parallel.A quadrilateral is a parallelogram if both pair of sides are parallel.In a parallelogram :(i) opposite sides and angles are equal.

(ii) diagonals bisect each other.A parallelogram is a rhombus if its adjacent sides are equal.The diagonals of a rhombus bisect each other at right angle.A parallelogram is a rectangle if its one angle is 90°.The diagonals of a rectangle are equal.A rectangle is a square if its adjacent sides are equal.The diagonals of a square intersect at right angles.The diagonal of a parallelogram divides it into two triangles of equal area.Parallelogram on the same base (or equal bases) and between the same parallels are equalin area.The triangles on the same base (or equal bases) and between the same parallels are equalin area.Triangles on equal bases having equal areas have their corresponding altitudes equal.

TERMINAL EXERCISE

1. Which of the following are trapeziums ?

(i) (ii) (iii)

Fig. 16.47

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2. In Fig. 16.48, PQ || FG || DE || BC. Name all the trapeziums in the figure.

Fig. 16.48

3. In Fig. 16.49, ABCD is a parallelogram with an area of 48 cm2.Find the area of (i) shaded region (ii) unshaded region.

Fig. 16.49

4. Fill in the blanks in each of the following to make them true statements :

(i) A quadrilateral is a trapezium if ...

(ii) A quadrilateral is a parallelogram if ...

(iii) A rectangle is a square if ...

(iv) The diagonals of a quadrilateral bisect each other at right angle. If none of the anglesof the quadrilateral is a right angle, it is a ...

(v) The sum of the exterior angles of a quadrilateral is ...

5. If the angles of a quadrilateral are (x – 20)°, (x + 20)°, (x – 15)° and (x + 15)°, find xand the angles of the quadrilateral.

6. The sum of the opposite angles of a parallelograms is 180°. What special type of aparallelogram is it ?

Quadrilaterals 93

7. The area of a ∆ABD in Fig. 16.50 is 24 cm2. If DE = 6 cm, and AB || CD, BD || CE,AE || BC, find

Fig. 16.50

(i) Altitude of the parallelogram BCED.

(ii) Area of the parallelogram BCED

8. In Fig. 16.51, the area of parallelogram ABCD is 40 cm2. If EF = 8 cm, find the altitudeof ∆DCE.

Fig. 16.51

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ANSWERS


1. (i) Rectangle (ii) trapezium (iii) Rectangle (iv) Parallelogram(v) Rhombus (vi) Square

2. (i) true (ii) False (iii) True (iv) True(v) True (vi) True (vii) False (viii) False

(ix) False (x) False3. 90° 4. 60°, 84° 84° and 132°5. Other pair of opposite angles will also be supplementary


1. ∠B = 118°, ∠C = 62° and ∠D = 118°2. ∠A = 105°, ∠B = 75°, ∠C = 105° and ∠D = 75°3. 30 4. ∠CDB = 55° and ∠ADB = 55°5. ∠ACD = 61° 6. ∠OPS = 70° 7. ∠CAB = 45°


2. 5 cm 3. 3 cm


1. MS = 2 cm and MN = 2.5 cm2. l, m and n are three equidistant parallel lines3. XY = 3.4 cm, XP = 3.2 cm and BZ = 3.5 cm


1. When they are lying between the same parallel lines2. 32 cm2 3. 4 cm

Terminal Exercise

1. (i) and (iii)2. PFGQ, FDEG, DBCE, PDEQ, FBCG and PBCQ3. (i) 24 cm2 (ii) 24 cm2

4. (i) any one pair of opposite sides is parallel(ii) both pairs of opposite sides are parallel

(iii) pair of adjacent sides is equal (iv) rhombus (v) 360°5. x = 90°, angles are 70°, 110°, 75° and 105° respectively.6. It is a rectangle7. (i) 8 cm (ii) 48 cm2 8. 5 cm

Similarity of Triangle 95

17

Similarity of Triangle

17.1 INTRODUCTION

Looking around you will see many objects which are of the same shape but of same or differentsizes. For examples, leaves of a tree have almost the same shape but same or different sizes.Similarly, photographs of different sizes developed from the same negative are of same shapebut different sizes, the miniature model of a building and the building itself are of same shapebut different sizes. All those objects which have the same shape but different sizes are calledsimilar objects.

Let us examine the similarity of plane figures :

(i) Two line-segments of the same length are congruent but of different lengths are similar.

(ii) Two circles of the same radius are congurent but circles of different radii are similar.

(iii) Two equilateral triangles of different sides are similar.

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(iv) Two squares of different sides are similar.

In this lesson, we shall study about the concept of similarity, especially similarity of trianglesand the conditions thereof. We shall also study about various results related to them.

17.2 OBJECTIVES


identify similar figures

distinguish between congurent and similar plane figures

cite the criteria for similarity of triangles viz. AAA, SSS and SAS.

verify and use unstarred results given in the curriculum based on similarity experimentally

prove the Baudhayan/Pythagoras Theorem

apply these results in verifying experimentally (or proving logically) problems based onsimilar triangles.


Knowledge of

plane figures like triangles, quadrilaterals, circles, rectangles, squares, etc.

criteria of congruency of triangles

finding squares and square-roots of numbers

ratio and proportion

internal and external bisectors of angles of a triangle.

17.4 SIMILAR PLANE FIGURES

Fig. 17.2


In Fig. 17.2, the two pentagon seem to be of the same shape.We can see that ∠A = ∠A′, ∠B = ∠B′, ∠C = ∠C′, ∠D = ∠D′ and ∠E = ∠E′ and

ABA B

BCB C

CDC D

DED E

EAE A' ' ' ' ' ' ' ' ' '= = = = . We say that the two pentagons are similar. Thus we say

thatAny two polygons, with corresponding angles equal and corresponding sidesproportional, are similar

Thus, two polygons are similar, if they satisfy the following two conditions :

(i) Corresponding angles are equal

(ii) The corresponding sides are proportional.

Even if one of the conditions does not hold, the polygons are not similar as in the case of arectangle and square given in Fig. 17.3. Here all the corresponding angles are equal but thecorresponding sides are not proportional.

Fig. 17.3

17.5 SIMILARITY OF TRIANGLES

Triangles are special type of polygons and therefore the conditions of similarity of polygonsalso hold for triangles. Thus,Two triangles are similar if

(i) their corresponding angles are equal, and(ii) their corresponding sides are proportional

Fig. 17.4

We say that ∆ABC is similar to ∆DEF and denote it by writing

∆ABC ~ ∆DEF

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The symbol ‘~’ stands for the phrase “ is similar to”

If ∆ABC ~ ∆DEF, then by definition

∠A = ∠D, ∠B = ∠E, ∠D = ∠F and ABDE

BCEF

CAFD= =

17.5.1 AAA criterion for similarity

We shall show that if either of the above two conditions is satisfied then the other automaticallyholds in the case of triangles.

Let us perform the following experiment.

Construct two ∆’s ABC and PQR in which ∠P = ∠A, ∠Q = ∠B and ∠R = ∠C as shownin Fig. 17.5.

Fig. 17.5

Measure the sides AB, BC and CA of ∆ABC and also measure the sides PQ, QR and RP of∆PQR.

Now find the ratio ABPQ

BCQR, and CA

RP .

What do you find ? You will find that all the three ratios are equal and therefore the trianglesare similar.

Try this with different triangles with equal corresponding angles. You will find the same result.

Thus, we can say that

If in two triangles, the corresponding angles are equal the triangles are similar.

This is called AAA similarity criterion.

17.5.2 SSS criterion for similarity.

Let us now perform the following experiment :

Draw a triangle ABC with AB = 3 cm, BC = 4.5 cm and CA = 3.5 cm


(i) (ii)

Fig. 17.6

Draw another ∆PQR as shown in Fig. 17.6 (ii)

We can see that ABPQ

BCQR

ACPR= =

i.e., the sides of the two triangles are proportional

Now measure ∠A, ∠B and ∠C of ∆ABC and ∠P, ∠Q and ∠R of ∆PQR.

You will find that ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R.

Repeat the experiment with another two triangles having corresponding sides proportional, youwill find that the corresponding angles are equal and so the triangle are similar.


If the corresponding sides of two triangles are proportional the triangles aresimilar.

17.5.3 SAS Criterian for Similarity

Let us conduct the following experiment.

Take a line AB = 3 cm and at A construct an angle of 60°. Cut off AC = 4.5 cm. Join BC

Fig. 17.7

Now take PQ = 6 cm. At P, draw an angle of 60° and cut off PR = 9 cm.

Measure ∠B, ∠C, ∠Q and ∠R. We shall find that ∠B = ∠Q and ∠C = ∠R

Thus, ∆ABC ~ ∆PQR

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Thus, we conclude that

If one angle of a triangle is equal to one angle of the other triangle and the sidescontaining these angles are proportional, the triangles are similar.

Thus, we have three important criteria for the similarity of triangles. They are given below:

(i) If in two triangles, the corresponding angles are equal, the triangles are similar.

(ii) If the corresponding sides of two triangles are proportional, the triangles aresimilar.

(iii) If one angle of a triangle is equal to one angle of the other triangle and the sidescontaining these angles are proportional, the triangle are similar.

Example 17.1 : In Fig. 17.8 are given two triangles ABC and PQR

Fig. 17.8

Is ∆ABC ~ ∆PQR ?

Solution : We are given that

∠A = ∠P and ∠B = ∠Q

We also know that

∠A + ∠B + ∠C = ∠P + ∠Q + ∠R = 180°

Thus, according to first criterion of similarity

∆ABC ~ ∆PQR.

Example 17.2 :

Fig. 17.9

In Fig. 17.9, ∆ABC ~ ∆PQR. If AC = 4.8 cm, AB = 4 cm and PQ = 9 cm, find PR.


Solution : It is given that ∆ABC ~ ∆PQR

∴ ABPQ = AC

PRLet PR = x cm

∴ 49 = 4 8.

x⇒ 4x = 9 × 4.8⇒ x = 10.8i.e., PR = 10.8 cm.


Find the values of x and y if ∆ABC ~ ∆PQR

(i)

Fig. 17.10

(ii)

Fig. 17.11

(iii)

Fig. 17.12

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17.6 BASIC PROPORTIONALITY THEOREM

We state below the Basic Proportionality Theorem :

If a line is drawn parallel to one side of a triangle, the other two sides of the triangleare divided proportionally.

Thus, in Fig. 17.13, DE || BC, According to the above result

ADDB

AEEC=

We can easily verify this by measuring AD, DB, AE and EC. Youwill find that

ADDB

AEEC=

We state the converse of the above result as follows :

If a line divides any two sides of a triangle in the same ratio, the line is parallelto third side of the triangle.

Thus, in Fig. 17.13, if DE divides sides AB and AC of ∆ABC such that ADDB

AEEC= , then

DE || BC.

We can verify this by measuring ∠ADE and ∠ABC and finding that

∠ADE = ∠ABC

These being alternate angles, the lines DE and BC are parallel.

We can verify the above two results by taking different triangles.

Let us solve some examples based on these.

Example 17.3 : In Fig. 17.14, DE || BC. If AD = 3 cm, DB = 5 cm and AE = 6 cm, find AC.

Solution : DE || BC (Given). Let EC = x

∴ ADDB = AE

EC

∴ 35 = 6

x⇒ 3x = 30

⇒ x = 10

∴ EC = 10 cm

∴ AC = AE + EC = 16 cm.

Fig. 17.13

Fig. 17.14


Example 17.4 : In Fig. 17.15, AD = 4 cm, DB = 5 cm, AE = 4.5 cm and EC = 5 58 cm. Is

DE || BC ? Given reasons for your answer.

Solution : We are given that AD = 4 cm and DB = 5 cm.

∴ ADDB = 4

5

Similarly, AEEC = 4.5

458

= 92

845

45× =

∴ According to converse of Basic Proportionality TheoremDE || BC.


1. In Fig. 17.16 (i), (ii) and (iii), PQ || BC. Find the value of x in each case.

Fig. 17.16

2. In Fig. 17.17 [(i), (ii) and (iii)], find whether DE is parallel to BC or not ? Give reasonsfor your answer.

Fig. 17.17

17.7 BISECTOR OF AN ANGLE OF A TRIANGLE

We now state an important result as given below :

The internal bisector of an angle of a triangle divides the opposite side in the ratioof sides containing the angle

Fig. 17.15

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Thus, according to the above result, if AD is the internalbisector of ∠A of ∆ABC, then

BDDC = AB

AC

We can easily verify this by measuring BD, DC, AB andAC and finding the ratios. We will find that

BDDC = AB

AC

Repeating the same activity with other triangles, we may verify the result.

Let us solve some examples to illustrate this.

Example 17.5 : The sides AB and AC of a triangle are 6 cm and 8 cm. The bisector AD of∠A intersects the opposite side BC in D such that BD = 4.5 cm. Find the length of segmentCD.

Solution : According to the above result, we have

BDDC = AB

AC

(Q AD is the internal bisector of ∠A of ∆ABC)

or 4 5.x = 6

8

⇒ 6x = 4.5 × 8

x = 6

i.e., the length of line-segment CD = 6 cm.

Example 17.6 : The sides of a triangle are 28 cm, 36 cm and 48 cm. Find the lengths of theline-segments into which the smallest side is divided by the bisector of the angle opposite toit.

Solution : The smallest side is of length 28 cm and thesides forming the angle. A opposite to it are 36 cm and48 cm. Let the angle bisector AD meet BC in D.

∴ BDDC = 36

4834=

⇒ 4BD = 3DC or BD = 3648

34DC = DC

BC = BD + DC = 28 cm

∴ DC + 34 DC = 28

Fig. 17.19

Fig. 17.18

Fig. 17.20


∴ DC = 28 47×FH IK cm = 16

∴ BD = 12 cm and DC = 16 cm


1. In Fig. 17.21, AD is the bisector of ∠A, meeting BC in D. If AB = 4.5 cm, BD = 3 cm,DC = 5 cm, find x.

Fig. 17.21

2. In Fig. 17.22, PS is the internal bisector of ∠P of ∆PQR. The dimensions of some ofthe sides are given in Fig. 17.22. Find x.

Fig. 17.22

3. In Fig. 17.23, RS is the internal bisector of ∠R of ∆PQR. For the given dimensions, expressp, the length of QS in terms of x, y and z.

Fig. 17.23

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17.8 SOME MORE IMPORTANT RESULTS

Let us study another important result on similarity in connection with a right triangle and theperpendicular from the vertex of the right angle to the opposite side. We state the result belowand try to verify the same.

If a perpendicular is drawn from the vertex of the right angle of a right triangleto the hypotenuse, the triangles on each side of the perpendicular are similar toeach other and to the triangle.

Let us try of verify this by an activity.

Draw a ∆ABC, right angled at A. Draw AD ⊥ to thehypotenuse BC, meeting it in D.

Let ∠DBA = α , As ∠ADB = 90°,

∴ ∠BAD = 90° – αAs ∠BAC = 90° and ∠BAD = 90° – α ,

Therefore ∠DAC = α

Similarly, ∠DCA = 90° – α

∴ ∆ADB and ∆CDA are similar, as it has all the corresponding angles equal.

Also, the angles of ∆BAC are α , 90° and 90° – α

∴ ∆ADB ~ ∆CDA ~ ∆CAB

Another important result is about relation between sides and areas of similar triangles.

It states that

The ratio of the areas of similar triangles is equal to the ratio of the squares ontheir corresponding sides

Let us verify this result by the following activity. Draw two triangles ABC and PQR whichare similar i.e., their sides are proportional.

Fig. 17.25

Fig. 17.24


Draw AD ⊥ BC and PS ⊥ QR

Measure the lengths of AD and PS.

Find the product AD × BC and PS × QR

You will find that AD × BC = BC2 and PS × QR = QR2

Now AD × BC = 2. Area of ∆ABC

PS × QR = 2. Area of ∆PQR

∴ Area of ABCArea of PQR

∆∆ = AD BC

PS QR×× =

BCQR

22 ...(i)

As BCQR = AB

PQACPR=

∴ Area of ABCArea of PQR

∆∆ =

BCQR

22 =

ABPQ

ACPR

22

22=

The activity may be repeated by taking different pairs of similar triangles.

Let us illustrate these results with the help of examples.

Example 17.7 : Find the ratio of the area of two similar triangles if one pair of theircorresponding sides are 2.5 cm and 5.0 cm.

Solution : Let the two triangles be ABC and PQR

Let BC = 2.5 cm and QR = 5.0

Area ABCArea PQR

∆∆b gb g =

BCQR

2

2 = 2 5

5 014

2

2.

.b gb g

=

Example 17.8 : In a ∆ABC, PQ || BC and intersects AB and AC at P and Q respectively. If

APBP = 2

3 , find the ratio of areas of ∆APQ and ∆ABC.

Solution : In Fig. 17.26,

PQ || BC

∴ APBP =

AQQC = 2

3

∴ BPAP = 3

2

∴ 1+ BPAP = 1 3

252+ =

Fig. 17.26

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⇒ ABAP = 5

2 ⇒ APAB = 2

5

∴Area APQArea ABC

∆∆b gb g =

APAB

2

2 = APABFHGIKJ = FHGIKJ

2 225 =

425 .


1. In Fig. 17.27, ABC is a right triangle with ∠A = 90° and ∠C = 30°. Show that∆DAB ~ ∆DCA ~ ∆ACB.

Fig. 17.27

2. Find the ratio of the areas of two similar triangles if the corresponding sides are of lengths3 cm and 5 cm.

3. In Fig. 17.28, ABC is a triangle in which DE || BC. If AB = 6 cm and AD = 2 cm, findthe ratio of the area of ∆ADE and trapezium DBCE.

Fig. 17.28

4. P, Q and R are the mid-points of the sides AB, BC and CA of the ∆ABC respectively.Show that the area of ∆PQR is one-fourth the area of ∆ABC.

5. In two similar triangles ABC and PQR, if the corresponding altitudes AD and PS are inthe ratio of 4 : 9, find the ratio of the areas of ∆ABC and ∆PQR.

Hint : Use ABPQ

= = =LNM

OQP

ADPS

BCQR

CAPR

6. If the ratio of the areas of two similar triangles is 16 : 25, find the ratio of theircorresponding sides.


17.9 BAUDHAYAN/PYTHAGORAS THEOREM

We know prove an important theorem, called Baudhayan/Phythagorus Theorem using theconcept of similarity.

Theorem : In a right triangle, the square on the hypotenuse is equal to the sumof the squares on the other two sides.

Given. A right triangle ABC, in which ∠B = 90°.

To Prove : AC2 = AB2 + BC2

Construction. From B, draw BD ⊥ AC (See Fig. 17.29)

Proof : BD ⊥ AC

∴ ∆ADB ~ ∆ABC ...(i)

and ∆BDC ~ ∆ABC ...(ii)

From (i), we get ABAC = AD

AB

⇒ AB2 = AC.AD ...(A)

From (ii), we get BCAC = DC

BC

⇒ BC2 = AC.DC ...(B)

Adding (A) and (B), we get

AB2 + BC2 = AC (AD + DC)

= AC. AC = AC2

The theorem is known after the name of famous Greek Mathematician Pythagoras. This wasoriginally stated by the Indian Mathematician. Baudhayan about 200 years before Pythagoras.

17.9.1 Converse of Pythagoras Theorem

The conserve of the above theorem states :

In a triangle, if the square on one side is equal to sum of the squares on the othertwo sides, the angle opposite to first side is a right angle.

This result can be verified by the following activity.

Draw a triangle ABC with side 3 cm, 4 cm and 5 cm.

i.e., AB = 3 cm, BC = 4 cm

and AC = 5 cm.

You can see that AB2 + BC2 = (3)2 + (4)2

= 9 + 16 = 25Fig. 17.30

Fig. 17.29

110 Mathematics

AC2 = (5)2 = 25

∴ AB2 + BC2 = AC2

The triangle in Fig. 17.30 satisfies the condition of the above result.

Measure ∠ABC, you will find that ∠ABC = 90°. Construct triangles of sides 5 cm, 12 cmand 13 cm, and of sides 7 cm, 24 cm, 25 cm. You will again find that the angles oppositeto side of length 13 cm and 25 cm are 90° each.

Let us solve some examples using above results.

Example 17.9 : In a right triangle, the sides containing the right angle are of length 5 cm and12 cm. Find the length of the hypotenuse.

Solution : Let ABC be the right triangle, right angled at B

∴ AB = 5 cm, BC = 12 cm

Also, AC2 = BC2 + AB2

= (12)2 + (5)2

= 144 + 125

= 169

∴ AC = 13

i.e., the length of the hypotenuse is 13 cm.

Example 17.10 : Find the length of diagonal of a rectangle the lengths of whose sides are3 cm and 4 cm.

Solution : In Fig. 17.31, is a rectangle ABCD. Join thediagonal BD. Now DCB is a right triangle.

∴ BD2 = BC2 + CD2

= 42 + 32

= 16 + 9 = 25

BD = 5

i.e., the length of diagonal of rectangle ABCD is 5 cm.

Example 17.11 : In an equilateral triangle, verify that three timesthe square on one side is equal to four times the square on itsaltitude.

Solution : The altitude AD ⊥ BC

and BD = CD

Let AB = BC = CA = 2a

and BD = CD = a

Let AD = x

Fig. 17.31

Fig. 17.32


∴ x2 = (2a)2 – (a)2 = 3a2

3.(Side)2 = 3. (2a)2 = 12 a2

4.(Altitude) = 4. 3a2 = 12a2

Hence the result.

Example 17.12 : ABC is a right triangle, right angled at C. If CD, the length of perpendicularfrom C on AB is p, BC = a, AC = b and AB = c, show that

(i) pc = ab

(ii)12p = 1 1

2 2a b+

Solution : (i) CD ⊥ AB

∴ ∆ABC ~ ∆ACD

∴cb =

ap

⇒ pc = ab.

(ii) AB2 = AC2 + BC2

or c2 = b2 + a2

abpFHIK

2

= b2 + a2

or12p

=a ba b

2 2

2 2+

= 1 12 2a b

+


1. The sides of certain triangles are given below. Determine which of them are right triangles: [AB = c, BC = a, CA = b]

(i) a = 4 cm, b = 5 cm, c = 3 cm

(ii) a = 1.6 cm, b = 3.8 cm, c = 4 cm

(iii) a = 9 cm, b = 16 cm, c = 18 cm

(iv) a = 7 cm, b = 24 cm, c = 25 cm

2. Two poles of height 6 m and 11 m, stand on a plane ground. If the distance between theirfeet is 12 m, find the distance between their tops.

Fig. 17.33

112 Mathematics

3. Find the length of the diagonal of a square of side 10 cm.

4. In Fig. 17.34, ∠C is acute and AD ⊥ BC. Show that AB2 = AC2 + BC2 – 2BC. DC

Fig. 17.34

5. L and M are the mid-points of the sides AB and AC of ∆ABC, right angled at B. Showthat

4LC2 = AB2 + 4BC2

6. P and Q are points on the sides CA and CB respectively of ∆ABC, right angled at C.Prove that

AQ2 + BP2 = AB2 + PQ2

7. PQR is an isosceles right triangle with ∠Q = 90°. Prove that PR2 = 2PQ2.

8. A ladder is placed against a wall such that its top reaches upto a height of 4 m of thewall. If the foot of the ladder is 3 m away from the wall, find the length of the ladder.

LET US SUM UP

Objects which have the same shape but different sizes are called similar objects.

Any two polygons, with corresponding angles equal and corresponding sides proportional,are similar.

Two triangles are said to be similar, if

(a) their corresponding angles are equal and

(b) their corresponding sides are proportional

Criteria of similarity

– AAA criterion

– SSS criterion

– SAS criterion

If a line is drawn parallel to one-side of a triangle, it divides the other two sides in thesame ratio and its converse.


The internal bisector of an angle of a triangle divides the opposite side in the ratio ofsides containing the angle.

If a perpendicular is drawn from the vertex of the right angle of a right angled triangleto the hypotenuse, the triangles so formed are similar to each other and to the giventriangle.

The ratio of the areas of two similar triangles is equal to the ratio of squares of theircorresponding sides.

In a right triangle, the square on the hypotenuse is equal to sum of the squares on theremaining two sides – (Baudhyan) Pythagoras Theorem

In a triangle, if the square on one side is equal to the sum of the squares on the remainingtwo sides, then the angle opposite to the first side is a right angle – converse of(Baudhayan) Pythagoras Theorem.

TERMINAL EXERCISE

1. Write the criteria for the similarity of two polygons.

2. Enumerate different criteria for the similarity of the two triangles.

3. In which of the following cases, ∆’s ABC and PQR are similar

(i) ∠A = 40°, ∠B = 60°, ∠C = 80°, ∠P = 40°, ∠Q = 60° and ∠R = 80°

(ii) ∠A = 50°, ∠B = 70°, ∠C = 60°, ∠P = 50°, ∠Q = 60° and ∠R = 70°

(iii) AB = 2.5 cm, BC = 4.5 cm, CA = 3.5 cm

PQ = 5.0 cm, QR = 9.0 cm, RP = 7.0 cm

(iv) AB = 3 cm, BC = 4 cm, CA = 5.0 cm

PQ = 4.5 cm, QR = 7.5 cm, RP = 6.0 cm.

4. In Fig. 17.35, AD = 3 cm, AE = 4.5 cm, DB = 4.0 cm, find CE, given that DE || BC.

Fig. 17.35 Fig. 17.36

5. In Fig. 15.36, DE || AC. From the dimension given in the figure, find the value of x.

114 Mathematics

6. In Fig. 17.37 is shown a ∆ABC in which AD = 5 cm, DB = 3 cm, AE = 2.50 cm andEC = 1.5 cm. Is DE || BC ? Give reasons for your answer.

Fig. 17.37 Fig. 17.38

7. In Fig. 17.38, AD is the internal bisector of ∠A of ∆ABC. From the given dimension,find x.

8. The perimeter of two similar ∆’s ABC and DEF are 12 cm and 18 cm. Find the ratio ofthe area of ∆ABC to that of ∆DEF.

9. The altitudes AD and PS of two similar ∆’s ABC and PQR are of length 2.5 cm and 3.5cm. Find the ratio of area of ∆ABC to that of ∆PQR.

10. Which of the following are right triangles ?

(i) AB = 5 cm, BC = 12 cm, CA = 13 cm

(ii) AB = 8 cm, BC = 6 cm, CA = 10 cm

(iii) AB = 10 cm, BC = 5 cm, CA = 6 cm

(iv) AB = 25 cm, BC = 24 cm, CA = 7 cm

(v) AB = a2 + b2, BC = 2ab, CA = a2 – b2

11. Find the area of an equilateral triangle of side 2a.

12. Two poles of height 12 m and 17 m, stand on a planeground and the distance between their feet is 12 m.Find the distance between their tops.

13. In Fig. 17.39, show that

AB2 = AC2 + BC2 + 2BC.CD

14. A ladder is placed against a wall and its top reaches a point at a height of 8 m from theground. If the distance between the wall and foot of the ladder is 6 m, find the lengthof the ladder.

15. In an equilateral triangle, show that three times the square of a side equals four timesthe square on medians.

Fig. 17.39


ANSWERS


1. (i) x = 4.5, y = 3.5 (ii) x = 70, y = 50 (iii) x = 2 cm, y = 7 cm


1. (i) 6 (ii) 6 (iii) 10 cm

2. (i) No (ii) Yes (iii) Yes


1. 7.5 cm 2. 4 cm

3.yzx (x = –1 is not possible)


2. 9 : 25 3. 1 : 8 5. 16 : 81

6. 4 : 5


1. (i) Yes (ii) No (iii) No

(iv) Yes

2. 13 m 3. 10 2 cm 8. 5 m

Terminal Exercise

3. (i) and (iii) 4. 6 cm 5. 4.5 cm

6. Yes : ADDB

AEEC

= 7. 4.5 cm

8. 4 : 9 9. 25 : 49

10. (i), (ii), (iv) and (v)

11. 3 2a 12. 13 m 14. 10 m

116 Mathematics

18

Circles

18.1 INTRODUCTION

You are already familiar with geometrical figures such as a line segment, an angle, a triangle,a quadrilateral and a circle. Examples, of a circle are a wheel, a bangle, alphabet O, etc. Inthis lesson we shall study in some details about the circle and related concepts.

18.2 OBJECTIVES


define a circle

give examples of various terms related to a circle

illustrate congruent circles and concentric circles

identify and illustrate terms connected with circles like chord, arc, sector, segment, etc

verify experimentally results based on arcs and chords of a circle

use the results in solving problems


Line segment and its lengthAngle and its measureParallel and perpendicular linesClosed figures such as triangles, quadrilaterals polygons, etc.Perimeter of a closed figureRegion bounded by a closed figureCongruence of closed figures

18.4 CIRCLE AND RELATED TERMS

18.4.1 Circle : A circle is a collection of all points in a plane which are at a constant distancefrom a fixed point in the same plane.

Circles 117

Radius : A line segment joining the centre of the circle to a pointon the circle is called its radius.

In Fig. 18.1, there is a circle with centre O and one of its radiusis OA. OB is another radius of the same circle.

Activity for you : Measure the length OA and OB and observethat they are equal. Thus

All radii (plural of radius) of a circle are equal

The length of the radius of a circle is generally denoted by theletter ‘r’. It is customary to write radius instead of the length ofthe radius.

A closed geometric figure in the plane divides the plane into threeparts namely, the inner part of the figure, the figure and the outerpart. In Figure 18.2 the shaded portion is the inner part of thecircle, the boundary is the circle and the unshaded portion is theouter part of the circle.

Activity for you

(a) Take a point Q in the inner part of the circle (See Figure 18.3).Measure OQ and find that OQ < r. The inner part of the circleis called the interior of the circle.

(b) Now take a point P in the outer part of the circle (Figure 18.3).Measure OP and find that OP > r. The outer part of the circleis called the exterior of the circle.

18.4.2 Chord

A line segment joining any two points of a circle is called a chord.In Figure 18.4 AB and PQ and CD are three chords of a circlewith centre O and radius r. The chord PQ passes through thecentre O of the circle. Such a chord is called a diameter of thecircle. Diameter is usually denoted by ‘d’.

A chord passing though the centre of circle is called itsdiameter.

Activity for you :

Measure the length d of PQ, the radius r and find that d is the same as 2r. Thus we have

d = 2r

i.e. the diameter of a circle = twice the radius of the circle.

Fig. 18.1

Fig. 18.2

Fig. 18.3

Fig. 18.4

118 Mathematics

Measure the length PQ, AB and CD and find that PQ > AB and PQ > CD, we may conclude

Diameter is the longest chord of a circle.

18.4.3 Arc

A part of a circle is called an arc. In Figure 18.5(a) ABC is an arc and is denoted by arc ABCor .

(a) (b)

Fig. 18.5

18.4.4 Semicircle

A diameter of a circle divides a circle into two equal arcs, each known as a semicircle.

In Figure 18.5(b), PQ is a diameter and is a semicircle and so is .

18.4.5 Sector

The region bounded by an arc of a circle and two radii at its endpoints is called a sector.

In Figure 18.6, the shaded portion is a sector formed by the arcPRQ and the unshaded portion is a sector formed by the arc PTQ.

18.4.6 Segment

A chord divides the interior of a circle into two parts, each calleda segment. In Figure 18.7, the shaded region PAQP and theunshaded region PBQP are both segments of the circle. PAQPis called a minor segment and PBQP is called a major segment.

18.4.7 Circumference

Choose a point P on a circle. If this point moves along the circle once and comes back to itsoriginal position then the distance covered by P is called the circumference of the circle

Fig. 18.6

Fig. 18.7

Circles 119

Fig. 18.8

Activity for you :

Take a wheel and mark a point P on the wheel where it touches the ground. Rotate the wheelalong a line till the point P comes back on the ground. Measure the distance between the Istand last position of P along the line. This distance is equal to the circumference of the circle.Thus,

The length of the boundary of a circle is the circumference of the circle.

Activity for you

Consider different circles and measure their circumference(s) and diameters. Observe that ineach case the ratio of the circumference to diameter turns out to be the same.

The ratio of the circumference of a circle to its diameter is always a constant. Thisconstant is universally denoted by Greek letter π .

Therefore, cd

cr= =2 π , where c is the circumference of the circle, d its diameter and r is its

radius.

An approximate value of π is 227 . Aryabhata–I (476 AD), a famous Indian Mathematician

gave a more accurate value of π which is 3.1416. In fact this number π is an irrational number.

18.5. MEASUREMENT OF AN ARC OF A CIRCLE

Consider an arc PAQ of a circle (Fig 18.9). To measure its length we put a thread along PAQand then measure the length of the thread with the help of a scale.

Similarly, you may measure the length of the arc PBQ.

18.5.1 Minor arc

An arc of a circle whose length is less than that or a semicircleof the same circle is called a minor arc. PAQ a minor arc (SeeFig. 18.9)

Fig. 18.9

120 Mathematics

18.5.2 Major arc

An arc of a circle whose length is greater than that of a semicircle of the same circle is calleda major arc. In Figure 18.9, arc PBQ is a major arc.

18.6 CONCENTRIC CIRCLES

Circles having the same centre but different radii are calledconcentric circles (See Fig. 18.10).

18.7 CONGRUENT CIRCLES OR ARCS

Two circles (or arcs) are said to be congruent if we can superpose (place) one over the othersuch that they cover each other completely.

18.8 SOME IMPORTANT RULES

Activity for you :

(i) Draw two circles with centre O1 and O2 and radius r and s respectively (See Fig. 18.11)

(i) (ii)

Fig. 18.11

(ii) Supeimpose the circle (i) on the circle (ii) so that O1coincides with O2

(iii) We observe that circle (i) will cover circle (ii) if and onlyif r = s

Two circles are congurent if and only if they have equal radii.

In Figure 18.12 if arc PAQ = arc RBS then ∠POQ = ∠ROS andconversely if ∠POQ = ∠ROS then arc PAQ = arc RBS.

Two arcs of a circle are congurent if and only if the anglessubtended by them at the centre are equal.

In Figure 18.13, if arc PAQ = arc RBS

then PQ = RS

and conversely if PQ = RS then

Fig. 18.10

Fig. 18.12

Circles 121

arc PAQ = arc RBS.

Two arcs of a circle are congurent if and only if theircorresponding chords are equal.

Activity for you :

(i) Draw a circle with centre O

(ii) Draw equal chords PQ and RS (See Fig. 18.14)

(iii) Join OP, OQ, OR and OS

(iv) Measure ∠POQ and ∠ROS

we observe that ∠POQ = ∠ROS

Conversely if ∠POQ = ∠ROS

then PQ = RS

Equal chords of a circle subtend equal angles at the centre and conversely if theangles subtended by the chords at the centre of a circle are equal, then the chordsare equal.

Note : The above results also hold good in case of congruent circles.

We take some examples using the above properties :

Example 18.1 : In Fig. 18.15, chord PQ = chord RS. Show that chordPR = chord QS.

Solution : The arcs corresponding to equal chords PQ and RS areequal.

Add to each arc, the arc QR,

yielding arc PQR = arc QRS

∴ chord PR = chord QS

Example 18.2 : In Figure 18.16 arc AB = arc BC, ∠AOB = 30°and ∠AOD = 70°. Find ∠COD.

Solution : Since arc AB = arc BC

∴ ∠AOB = ∠BOC

(Equal arcs subtend equal angles at the centre)

∴ ∠BOC = 30°

Since ∠COD = ∠COB + ∠BOA + ∠AOD

= 30° + 30° + 70°

= 130°.

Fig. 18.16

Fig. 18.15

Fig. 18.13

Fig. 18.14

122 Mathematics

Activity for you :

(i) Draw a circle with centre O (See Fig. 18.17).

(ii) Draw a chord PQ.

(iii) Draw ⊥ ON from O on the chord PQ.

(iv) Measure PN and NQ

You will observe that .

PN = NQ.

The perpendicular drawn from the centre of a circle to a chord bisects the chord.

Activity for you :

(i) Draw a circle with centre O (See Fig. 18.18).

(ii) Draw a chord PQ.

(iii) Find the mid point M of PQ.

(iv) Join O and M.

(v) Measure ∠OMP or ∠ OMQ with set square or protractor

We observe that ∠OMP = ∠ OMQ = 90°.

The line joining the centre of a circle to the mid point of a chord is perpendicularto the chord.

Activity for you :

Take three non collinear points A, B and C. Join AB and BC.Draw perpendicular bisectors MN and RS of AB and BCrespectively.

Since A, B, C are not collinear, MN is not parallel to RS. Theywill intersect only at one point O. Join OA, OB and OC andmeasure them.

We observe that OA = OB = OC

Now taking O as the centre and OA as radius draw a circle which passes through A, B andC.

Repeat the above procedure with another three non-collinear points and observe that there isonly one circle passing through there given non-collinear points. This gives us a method todraw a circle passing through three non-collinear points.

There is one and only one circle passing through three non-collinear points.

Note. It is important to note that a circle can not be drawn to pass through three collinearpoints.

Fig. 18.18

Fig. 18.17

Fig. 18.19

Circles 123

Activity for you :

(i) Draw a circle with centre O [Fig. 18.20(a)].

(ii) Draw two equal chords AB and PQ of the circle.

(iii) Draw OM⊥ AB and ON⊥ PQ.

(iv) Measure OM and ON and observe that they are equal.

Equal chords of a circle are equidistant from the centre.

In Fig. 18.20(b) OM = ON.

Measure and observe that AB = PQ. Thus,

Chords that are equidistant from the centre of a circle areequal.

The above results hold good in case of congurent circles also.

We now take a few examples using these properties of circles.

Examples 18.3 : In Figure 18.21, O is the centre of the circleand ON⊥ PQ. If PQ = 8 cm and ON = 3 cm, find OP.

Solution : ON⊥ PQ (given) and since perpendicular drawn fromthe centre of a circle to a chord bisects the chord.

∴ PN = NQ = 4 cmIn a right triangle OPN,

∴ OP2 = PN2 + ON2

OP2 = 42 + 32 = 25∴ OP = 5 cm.

Examples 18.4 : In Figure 18.22, OD is perpendicular to thechord AB of a circle whose centre is O and BC is a diameter.Prove that CA = 2OD.

Solution : Since OD⊥ AB (Given)∴ D is the mid point of AB (Perpendicular through the centre

bisects the chord)Also O is the mid point of CB (Since CB is a diameter)Now in ∆ABC, O and D are mid points of the two sides BC and BA of the triangle ABC.Since the line segment joining the mid points of any two sides of a triangle is parallel andhalf of the third side.

∴ OD = 12 CA

i.e. CA = 2OD.

Fig. 18.20b

Fig. 18.20a

Fig. 18.21

Fig. 18.22

124 Mathematics

Example.18.5 : A regular hexagon is inscribed in a circle. What angle does each side of thehexagon subtend at the centre ?

Solution : A regular hexagon has six sides which are equal.Therefore each side subtends the same angle at the centre.

Let us suppose that a side of the hexagon subtends an angle x°at the centre.

Then, we have

6x° = 360° ⇒ x = 60

Hence, each side of the hexagon subtends an angle of 60° at the centre.

Example 18.6 : In Fig. 18.24, two parallel chords PQ and ABof a circle are of lengths 7 cm and 13 cm respectively. If thedistance between PQ and AB is 3 cm, find the radius of the circle.

Solution : Let O be the centre of the circle. Draw perpendicularbisector OL of PQ which also bisects AB at M. Join OQ and OB(Fig. 18.24).

Let OM = x cm and radius of the circle be r cm

Then OB2 = OM2 + MB2 and OQ2 = OL2 + LQ2

∴ r2 = x2213

2+ FH IK ...(i)

and r2 = x + + FH IK3 72

2 2b g ...(ii)

Therefore from (i) and (ii) ,

x2213

2+ FH IK = x + + FH IK3 72

2 2b g

∴ 6x = 1694 9 49

4− −

∴ x = 72

∴ r2 = 72

132

2 2FH IK + FH IK = 494

1694

2184+ =

∴ r = 2182

Hence the radius of the circle is 2182 cm.

Fig. 18.24

Fig. 18.23

Circles 125


In questions 1 to 5, fill in the blanks to make each of thestatements true.

1. In Figure 18.25,

(i) AB is a ... of the circle.

(ii) Minor arc corresponding to AB is ... .

2. A ... is the longest chord of a circle.

3. The ratio of the circumference to the diameter of a circle is always ... .

4. The value of π as 3.1416 was given by great Indian Mathematician ... .

5. Circles having the same centre are called ... circles.

6. Diameter of a circle is 30 cm. If the length of a chord is 20 cm, find the distance of thechord from the centre.

7. Find the circumference of a circle whose radius is

(i) 7 cm (ii) 11 cm. Takeπ=FH IK227

8. In the Figure 18.26, RS is a diameter which bisects the chords PQ and AB at the pointsM and N respectively. Is PQ || AB ? Give reasons.

Fig. 18.26 Fig. 18.27

9. In the Figures 18.27, a line l intersects the two concentric circles with centre O at pointsA, B, C and D. Is AB = CD ? Give reasons.

LET US SUM UP

The circumference of a circle of radius r is equal to 2πr .

Two arcs of a circle are congurent if and only if either the angles subtended by them atthe centre are equal or their corresponding chords are equal.

Equal chords of a circle subtend equal angles at the centre and vice versa.

Fig. 18.25

126 Mathematics

Perpendicular drawn from the centre of a circle to a chord bisects the chord.

The line joining the centre of a circle to the mid point of a chord is perpendicular to thechord.

There is one and only one circle passing through three non-collinear points.

Equal chords of a circle are equidistant from the centre and the converse.

TERMINAL EXERCISE

1. If the length of a chord of a circle is 16 cm and the distance of the chord from the centreis 6 cm, find the radius of the circle.

2. Two circles with centre O and O' (See Fig. 18.28) are congurent. Find the length of thearc CD.

Fig. 18.28

3. A regular pentagon is inscribed in a circle. Find the angle which each side of the pentagonsubtend at the centre.

4. In Figure 18.29, AB = 8 cm and CD = 6 cm are two parallel chords of a circle with centreO. Find the distance between the chords.

Fig. 18.29

5. In Figure 18.30, arc PQ = arc QR. ∠POQ = 15° and ∠SOR = 110°. Find ∠SOP.

Fig. 18.30

Circles 127

6. In Figure 18.31, AB and CD are two equal chords of a circle with centre O. Ischord BD = chord CA ? Give reasons.

Fig. 18.31

7. If AB and CD are two equal chords of a circle with centre O (Fig. 18.32) and OM AB⊥ ,ON CD⊥ . Is OM = ON ? Give reasons.

Fig. 18.32

8. In Figure 18.33, AB = 14 cm and CD = 6 cm are two parallel chords of a circle withcentre O. Find the distance between the chords AB and CD.

Fig. 18.33

9. In Figure 18.34, AB and CD are two chords of a circle with centre O, intersecting at apoint P inside the circle.

Fig. 18.34

OM CD⊥ , ON AB⊥ and ∠OPM = ∠OPN

Is (i) OM = ON, (ii) AB = CD ? Give reasons.

128 Mathematics

10. C1 and C2 are concentric circles with centre O (See Fig 18.35), l is a line intersectingC1 at points P and Q and C2 at points A and B respectively. If ON⊥ l, is PA = BQ ?Give reasons.

Fig. 18.35

Circles 129

ANSWERS


1. (i) Chord (ii) APB 2. Diameter 3. Constant 4. Aryabhata–I

5. Concentric 6. 5 5 cm. 7. (i) 44 cm (ii) 69.14 cm.

8. Yes 9. Yes

Terminal Exercise 18.2

1. 10 cm 2. 2a cm 3. 72°

4. 4 – 3 = 1 cm. 5. 80°

6. Yes (Equal arcs have corresponding equal chords of a circle)

7. Yes (equal chords are equidistant from the centre of the circle)

8. 10 2 cm 9. (i) Yes (ii) Yes (∆OMP ≅ ∆ONP)

10. Yes (N is the middle point of chords PQ and AB).

130 Mathematics

19

Angles in a Circle and Cyclic Quadrilateral

19.1 INTRODUCTION

You must have measured the angles between two straight lines, let us now study the anglesmade by arcs and chords in a circle and a cyclic quadrilateral.

19.2 OBJECTIVES


prove that angles in the same segment of a circle are equal

cite examples of concyclic points

define cyclic quadrilaterals

prove that sum of the opposite angles of a cyclic quadrilateral is 180°

use properties of a cyclic quadrilateral

solve problems based on Theorems (proved) and solve other numerical problems basedon verified properties.


Angles of a triangle

Arc, chord and circumference of a circle

Quadrilateral and its types

19.4 ANGLES IN A CIRCLE

Central Angle. The angle made at the centre of a circle by theradii at the end points of an arc (or a chord) is called the centralangle or angle subtended by an arc (or chord) at the centre.

In Figure 19.1, ∠POQ is the central angle made by arc PRQ.

The length of an arc is closely associated with the central angle subtended by the arc. Let usdefine the “degree measure” of an arc in terms of the central angle.

Fig.19.1

Angles in a Circle and Cyclic Quadrilateral 131

The degree measure of a minor arc of a circle is the measure ofits corresponding central angle.

In Figure 19.2, Degree measure of PQR = x°

The degree measure of a semicircle in 180° and that of a majorarc is 360° minus the degree measure of the corresponding minorarc.

Relationship between length of an arc and its degree measure.

Length of an arc = circumference degree measure of the arc360× °

If the degree measure of an arc is 40°

then length of the arc PQR = 2 40360

29π πr r. °

° =

Inscribed angle : The angle subtended by an arc (or chord) onany point on the remaining part of the circle is called an inscribedangle.

In Figure 19.3, ∠PAQ is the angle inscribed by arc PRQ at pointA of the remaining part of the circle or by the chord PQ at thepoint A.

19.5. SOME IMPORTANT PROPERTIES

ACTIVITY FOR YOU :

Draw a circle with centre O. Let PAQ be an arc and B any pointon the circle.

Measure the central angle POQ and an inscribed angle PBQ bythe arc at remaining part of the circle. We observe that

∠POQ = 2∠PBQ

Repeat this activity taking different circles and different arcs. Weobserve that

The angle subtended at the centre of a circle by an arcis double the angle subtended by it on any point on theremaining part of the circle.

Let O be the centre of a circle. Consider a semicircle PAQ andits inscribed angle PBQ

∴ 2 ∠PBQ = ∠POQ

Fig.19.2

Fig.19.3

Fig.19.4

Fig.19.5

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(Since the angle subtended by an arc at the centre is double the angle subtended by it at anypoint on the remaining part of the circle)

But ∠POQ = 180° (Since PQ is a diameter of the circle)

2∠PBQ = 180°

∴ ∠PBQ = 90°


Angle in a semicircle is a right angle.

Theorem : Angles in the same segment of a circle are equal.

Given : A circle with centre O and the angles ∠PRQ and ∠PSQ in the same segment formedby the chord PQ (or arc PAQ)

To prove : ∠PRQ = ∠PSQ

Construction : Join OP and OQ.

Proof : As the angle subtended by an arc at the centre is doublethe angle subtended by it at any point on the remaining part ofthe circle, therefore we have

∠POQ = 2 ∠PRQ ...(i)

and ∠POQ = 2∠PSQ ...(ii)

From (i) and (ii), we get

2∠PRQ = 2∠PSQ

∴ ∠PRQ = ∠PSQ

We take some examples using the above results

Example 19.1 : In Figure 19.7, O is the centre of the circle and∠AOC = 120°. Find ∠ABC.

Solution : It is obvious that ∠x is the central angle subtendedby the arc APC and ∠ABC is the inscribed angle.

∴ ∠x = 2∠ABC

But ∠x = 360° – 120°

∴ 2∠ABC = 240°

∴ ∠ABC = 120°

Example 19.2 : In Figure 19.8 O is the centre of the circle and ∠PAQ = 35°. Find ∠OPQ.

Solution : ∠POQ = 2∠PAQ = 70° ...(i)

(Angle at the centre is double the angle on the remaining part of the circle)

Fig.19.6

Fig.19.7


Since OP = OQ (Radii of the same circle)

∴ ∠OPQ = ∠OQP ...(ii)(Angles opposite to equal sides are equal)

But ∠OPQ + ∠OQP + ∠POQ = 180°

∴ 2∠OPQ = 180° – 70° = 110°

∴ ∠OPQ = 55°.

Example 19.3 : In Figure 19.9, O is the centre of the circle and AD bisects ∠BAC. Find ∠BCD.

Solution : Since BC is a diameter

∠BAC = 90°

(Angle in the semicircle is a right angle)

As AD bisects ∠BAC

∴ ∠BAD = 45°

But ∠BCD = ∠BAD

(Angles in the same segment of a circle are equal)

∴ ∠BCD = 45°.

Example 19.4 : In Figure 19.10, O is the centre of the circle, ∠POQ = 70° and PS OQ⊥ . Find∠MQS.

Solution :

2∠PSQ = ∠POQ = 70°

(Angle subtended at the centre of a circle is twice the anglesubtended by it on the remaining part of the circle)

∴ ∠PSQ = 35°Since ∠MSQ + ∠SMQ + ∠MQS = 180°

(Sum of the angles of a triangle)∴ 35° + 90° + ∠MQS = 180°∴ ∠MQS = 180° – 125° = 55°.


1. In Figure 19.11, ADB is an arc of a circle with centre O, if ∠ACB = 35°, find ∠AOB.

Fig. 19.11

Fig.19.9

Fig.19.10

Fig.19.8

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2. In Figure 19.12, AOB is a diameter of a circle with centre O. Is ∠APB = ∠AQB = 90°?Give reasons.

Fig. 19.12

3. In Figure 19.13, PQR is an arc of a circle with centre O. If ∠PTR = 35°, find ∠PSR.

Fig. 19.13

4. In Figure 19.14, O is the centre of a circle and ∠AOB = 60°. Find ∠ADB.

Fig. 19.14

19.6 CONCYCLIC POINTS

Definition : Points which lie on a circle are calledconcyclic points.

Let us now find certain conditions under whichpoints are concyclic.

If you take a point P, you can draw not only onebut many circles passing through it as inFig. 19.15.

Now take two points P and Q on a sheet of a paper. You can draw as many circles as youwish, passing through the points. (Fig. 19.16).

Fig.19.15


Fig. 19.16

Let us now take three points P, Q and R which do not lie on the same straight line. In thiscase you can draw only one circle passing through these three non-colinear points(Figure 19.17).

Fig. 19.17

Further let us now take four points P, Q, R, and S which do not lie on the same line. You willsee that it is not always possible to draw a circle passing through four non-collinear points.

In Fig 19.18 (a) and (b) points are noncyclic but concyclic in Fig 19.18(c).

(a) (b) (b)

Fig. 19.18

Note. If the points P, Q and R are collinear then it is not possible to draw a circle passingthrough them.

Thus we conclude

1. Given one or two points there are infinitely many circles passing through them.

2. Three non-collinear points are always concyclic and there is only one circle passingthrough all of them.

3. Three collinear points are not concyclic (or noncyclic).

4. Four non-collinear points may or may not be concyclic.

136 Mathematics

19.6.1 CYCLIC QUADRILATERAL

A quadrilateral is said to be a cyclic quadrilateral if there is acircle passing through all its four vertices.

For example, Fig. 19.19 shows a cyclic quadrilateral PQRS.

Theorem. Sum of the opposite angles of a cyclic quadrilateral is 180°.

Given : A cyclic quadrilateral ABCDTo prove : ∠BAD + ∠BCD = ∠ABC + ∠ADC = 180°Construction : Draw AC and DBProof : ∠ACB = ∠ADB

and ∠BAC = ∠BDC[Angles in the same segment]

∴ ∠ACB + ∠BAC = ∠ADB + ∠BDC = ∠ADCAdding ∠ABC on both the sides, we get

∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABCBut ∠ACB + ∠BAC + ∠ABC = 180° [Sum of the angles of a triangle]∴ ∠ADC + ∠ABC = 180°∴ ∠BAD + ∠BCD = 360° – (∠ADC + ∠ABC) = 180°.Hence proved.Converse of this theorem is also true.If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral is cyclic.

Verification :

Draw a quadrilateral PQRS

Since in quadrilateral PQRS,

∠P + ∠R = 180°

and ∠S + ∠Q = 180°

Therefore draw a circle passing through the point P, Q and R and observe that it also passesthrough the point S. So we conclude that quadrilateral PQRS is a cyclic quadrilateral.

We solve some examples using the above results.Example 19.5 : ABCD is a cyclic parallelogram. Show that itis a rectangle.Solution : ∠A + ∠C = 180°

(ABCD is a cyclic quadrilateral)Since ∠A = ∠C

[Opposite angles of a parallelogram]

Fig.19.19

Fig.19.20

Fig.19.21

Fig.19.22


or ∠A + ∠A = 180°

∴ 2∠A = 180°

∴ ∠A = 90°

Thus ABCD is a rectangle.

Example 19.6 : A pair of opposite sides of a cyclic quadrilateral is equal. Prove that itsdiagonals are also equal (See Figure 19.23).

Solution : Let ABCD be a cyclic quadrilateral and AB = CD.

⇒ arc AB = arc CD (Corresponding arcs)

Adding arc AD to both the sides;

arc AB + arc AD = arc CD + arc AD

∴ arc BAD = arc CDA

⇒ Chord BD = Chord CA

⇒ BD = CA

Example 19.7 : In Figure 19.24, PQRS is a cyclic quadrilateral whose diagonals intersect atA. If ∠SQR = 80° and ∠QPR = 30°, find ∠SRQ.

Solution : Given ∠SQR = 80°

Since ∠SQR = ∠SPR

[Angles in the same segment]

∴ ∠SPR = 80°

∴ ∠SPQ = ∠SPR + ∠RPQ

= 80° + 30°.

or ∠SPQ = 110°.

But ∠SPQ + ∠SRQ = 180° (Sum of the opposite angles of a cyclicquadrilateral is 180°)

∴ ∠SRQ = 180° – ∠SPQ

= 180° – 110° = 70°

Example 19.8 : PQRS is a cyclic quadrilateral.

If ∠Q = ∠R = 65°, find ∠P and ∠S.

Solution : ∠P + ∠R = 180°

∴ ∠P = 180° – ∠R = 180° – 65°

∴ ∠P = 115°

Similarly, ∠Q + ∠S = 180°

∴ ∠S = 180º – ∠Q = 180° – 65°

∴ ∠S = 115°.

Fig.19.23

Fig.19.25

Fig.19.24

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1. In Figure 19.26, AB and CD are two equal chords of a circle with centre O. If∠AOB = 55°, find ∠COD.

Fig. 19.26

2. In Figure 19.27, PQRS is a cyclic quadrilateral, and the side PS is extended to the pointA. If ∠PQR = 80°, find ∠ASR.

Fig. 19.27

3. In Figure 19.28, ABCD is a cyclic quadrilateral whose diagonals intersect at O. If∠ACB = 50° and ∠ABC = 110°, find ∠BDC.

Fig. 19.28

4. In Figure 19.29, ABCD is a quadrilateral. If ∠A = ∠BCE, is the quadrilateral a cyclicquadrilateral ? Give reasons.

Fig. 19.29


LET US SUM UP

The angle subtended by an arc (or chord) at the centre of a circle is called central angleand an angle subtended by it at any point on the remaining part of the circle is calledinscribed angle.

Points lying on the same circle are called concyclic points.

The angle subtended by an arc at the centre of a circle is double the angle subtended byit at any point on the remaining part of the circle.

Angle in a semicircle is a right angle.

Angles in the same segment of a circle are equal.

Sum of the opposite angles of a cyclic quadrilateral is 180°.

If a pair of opposite angles of a quadrilateral is supplementary, then the quadrilateral iscyclic.

TERMINAL EXERCISES

1. A square PQRS is inscribed in a circle with centre O. What angle does each side subtendat the centre O ?

2. In Figure 19.30, C1 and C2 are two circles with centre O1 and O2 and intersect each otherat points A and B. If O1O2 intersect AB at M then show that

(i) ∆O1AO2 ≅ ∆O1BO2

(ii) M is the mid point of AB

(iii) AB O O⊥ 1 2

Fig. 19.30

(Hint. From (i) conclude that ∠1 = ∠2 and then prove that ∆AO1M ≅ ∆BO1M(by SAS rule)).

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3. Two circle intersect in A and B. AC and AD are the diameters of the circles. Prove thatC, B and D are collinear.

Fig. 19.31

[Hint. Join CB, BD and AB, Since ∠ABC = 90° and ∠ABD = 90°]

4. In Figure 19.32, AB is a chord of a circle with centre O. If ∠ACB = 40°, find ∠OAB.

Fig. 19.32

5. In Figure 19.33, O is the centre of a circle and ∠PQR = 115°. Find ∠POR.

Fig. 19.33

6. In Figure 19.34, O is the centre of a circle, ∠AOB = 80° and ∠PQB = 70°. Find ∠PBQ.

Fig. 19.34


ANSWERS


1. 70° 2. Yes 3. 35° 4. 30°


1. 55° 2. 80° 3. 20° 4. Yes

Terminal Exercise

1. 90° 4. 50° 5. 130° 6. 70°.

Secants, Tangents and Properties 143

20

Secants, Tangents and Properties

8.1 INTRODUCTION

Look at a moving cycle. You will observe that at any instant of time, the wheels of the movingcycle touch the road at a very limited area, more correctly a point.

If you roll a coin on a smooth surface, say a table or floor, you will find that at any instantof time, only one point of the coin comes in contact with the surface it is rolled upon.

What do you observe from the above situations ?

Fig. 20.1

If you consider a wheel or a coin as a circle and the touching surface (road or table) as a line,the above illustrations show that a line touches a circle. In this lesson, we shall study aboutthe possible contacts that a line and a circle have and try to study their properties.

144 Mathematics

20.2 OBJECTIVES


define a secant and a tangent to the circle

differentiate between a secant and a tangent

verify and use important results (given in the curriculum) related to tangents and secantsto circles.


Measurement of angles and line segments

Drawing circles of given radii

Drawing lines perpendicular and parallel to given lines

Knowledge of previous results about lines and angles, congruence and circles.

Knowledge of Pythagoras Theorem.

20.4 SECANTS AND TANGENTS—AN INTRODUCTION

You have read about lines and circles in your earlier lessons. Recall that a circle is the locusof a point in a plane which moves in such a way that its distance from a fixed point in theplane always remains constant. The fixed point is called the centre of the circle and the constantdistance is called the radius of the circle. You also know that a line is a collection of points,extending indefinitely to both sides, whereas a line segment is a portion of a line bounded bytwo points.

Fig. 20.2

Now consider the case when a line and a circle co-exist in the same plane. There can be threedistinct possibilities as shown in Fig. 20.2.

You can see that in Fig. 20.2(i), the line XY does not intersect the circle, with centre O. Inother words, we say that the line XY and the circle have no common point. In Fig. 20.2 (ii),the line XY intersects the circle in two distinct points A and B, and in Fig. 20.2 (iii), the lineXY intersects the circle in only one point and is said to touch the circle at the point P.


Thus, we can say that in case of intersection of a line and a circle, the following threepossibilities are there :

(i) The line does not intersect the circle at all, i.e., the line lies in the exterior of the circle.

(ii) The line intersects the circle at two distinct points. In that case, a part of the line lies inthe interior of the circle, the two points of intersection lie on the circle and the remainingportion lies in the exterior of the circle.

(iii) The line touches the circle in exactly one point. We therefore define the following :

Tangent

A line which touches a circle at exactly one point is called a tangent line and thepoint where it touches the circle is called the point of contact.

Thus, in Fig. 20.2 (iii), XY is a tangent to the circle at P, which is called the point of contact.

Secant

A line which intersects the circle in two distinct points is called a secant line(usually referred to as a secant).

In Fig. 20.2 (ii), XY is a secant line to the circle and A and B are called the points of intersectionof the line XY and the circle with centre O.

20.5 TANGENT AS A LIMITING CASE OF A SECANT

Consider the secant XY of the circle with centreO, intersecting the circle in the points A and B.Imagine that one point A, which lies on the circle,of the secant XY is fixed and the secant rotatesabout A, intersecting the circle at B', B'', B''', B''''as shown in Fig. 20.3 and ultimately attains theposition of the line XAY, when it becomes tangentto the circle at A.

Thus, we say that

A tangent is the limiting position of a secantwhen the two points of intersection coincide.

20.6 TANGENT AND RADIUS THROUGH THE POINT OF CONTACT

Let XY be a tangent to the circle, with centre O, at the point P. Join OP.

Take points Q, R, S and T on the tangent XY and join OQ, OR, OS and OT.

As Q, R, S and T are points in the exterior of the circle and P is on the circle.

Fig. 20.3

146 Mathematics

∴ OP is less than each of OQ, OR, OS and OT.

From our, “previous study of Geometry, we know that ofall the segments that can be drawn from a point (not onthe line) to the line, the perpendicular segment is theshortest” :

As OP is the shortest distance from O to the line XY

∴ OP ⊥ XY

Thus, we can state that

A radius, through the point of contact of tangent to a circle, is perpendicular tothe tangent at that point.

The above result can also be verified by measuring angles OPX and OPY and finding eachof them equal to 90°.

20.7 TANGENTS FROM A POINT OUTSIDE THE CIRCLE

Take any point P in the exterior of the circle with centreO. Draw lines through P. Some of these are shown as PT,PB, PA, PC, PD and PT' in Fig.20.5.

How many of these touch the circle ? Only two.

Repeat the activity with another point and a circle. Youwill again find the same result.


From an external point, two tangents can bedrawn to a circle.

If the point P lies on the circle, can there still be two tangents to the circle from thatpoint ? You can see that only one tangent can be drawn to the circle in that case. What aboutthe case when P lies in the interior of the circle ? Note that any line through P in that casewill intersect the circle in two points and hence no tangents can be drawn from an interiorpoint to the circle.

(A) Now, measure the lengths of PT and PT′. You will find that

PT = PT′ ...(i)

(B) Let us see this logically also.

Consider two ∆s OPT and OPT′

∠OTP = ∠OT'P (Each being a right angle)

OT = OT′ (Radii of the same circle)

OP = OP (Common)

Fig. 20.5

Fig. 20.4


∴ ∆OPT ≅ ∆OPT′

∴ PT = PT′ ...(ii)

From (A) and (B), we can say that

The lengths of two tangents from an externalpoint are equal.

Also, in Fig. 20.6 as ∆OPT ≅ ∆OPT′

∴ ∠OPT = ∠OPT′


The tangents drawn from an external point to a circle are equally inclined to theline joining the point to the centre of the circle.

Let us now take some examples to illustrate.

Example 20.1 : In Fig. 20.7, OP = 5 cm and radius of the circle is 3 cm. Find the length ofthe tangent PT from P to the circle, with centre O.

Solution : ∠OTP = 90°, Let PT = x

∴ In right triangle OTP, we have

OP2 = OT2 + PT2

or 52 = 32 + x2

or x2 = 25 – 9 = 16

∴ x = 4

i.e. the length of tangent PT = 4 cm

Example 20.2 : In Fig. 20.8, tangents PT and PT′ are drawn from a point P at a distance of25 cm from the centre of the circle whose radius is 7 cm. Find the lengths of PT and PT′.

Solution : Here OP = 25 cm and OT = 7 cm

We also know that∠OTP = 90°∴ PT2 = OP2 – OT2

= 625 – 49 = 576 = (24)2

∴ PT = 24 cmWe also know that

PT = PT′∴ PT′ = 24 cm

Fig. 20.6

Fig. 20.8

Fig. 20.7

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Example 20.3 : In Fig. 20.9, A, B and C are three exterior points of the circle with centre O.The tangents AP, BQ and CR are of lengths 3 cm, 4 cm and 3.5 cm. Find the perimeter of∆ABC.

Solution : We know that the length of two tangents from an external point to a circle are equal

∴ AP = AR,

BP = BQ,

CQ = CR

∴ AP = AR = 3 cm,

BP = BQ = 4 cm

and CR = CQ = 3.5 cm

AB = AP + PB;

= (3 + 4) cm = 7 cm

BC = BQ + QC;

= (4 + 3.5) cm = 7.5 cm

CA = AR + CR

= (3 + 3.5) cm

∴ = 6.5 cm

∴ Perimeter of ∆ABC = (7 + 7.5 + 6.5) cm = 21 cm

Example 20.4 : In Fig. 20.10, ∠AOB = 50°.

Find ∠ABO and ∠OBT

Solution : We know that OA ⊥ XY

⇒ ∠OAB = 90°

∴ ∠ABO = 180° – (∠OAB + ∠AOB)

= 180° – (90° + 50°) = 40°

We know that ∠OBA = ∠OBT

⇒ ∠OBT = 40°

∴ ∠ABO = ∠OBT = 40°


1. Fill in the blanks with suitable words :

(i) A tangent is to the radius through the point of contact.

(ii) The lengths of tangents from an external point to a circle are .

(iii) A tangent is the limiting position of a secant when the two coincide.

Fig. 20.9

Fig. 20.10


(iv) From an external point tangents can be drawn to a circle.

(v) From a point in the interior of the circle, tangent(s) can be drawn to thecircle.

2. In Fig. 20.11, ∠POY = 40°, Find the ∠OYP and ∠OYT.

3. In Fig. 20.12, the incircle of ∆PQR is drawn. If PX = 2.5 cm, RZ = 3.5 cm and perimeterof ∆PQR = 18 cm , find the length of QY.

Fig. 20.11 Fig. 2012

4. Write an experiment to show that the lengths of tangents from an external point to a circleare equal.

20.8 INTERSECTING CHORDS INSIDE AND OUTSIDE A CIRCLE.

You have read various results about chords in the previous lesson. We will now verify someresults regarding chords intersecting inside a circle or outside a circle, when produced.

Let us perform the following activity :

Draw a circle with centre O and any radius. Draw twochords AB and CD intersecting at P inside the circle.

Measure the lengths of the line-segments PD, PC, PA andPB. Find the products PA × PB and PC × PD. You willfind that they are equal.

Repeat the above activity with another two circles afterdrawing chords intersecting inside. You will again findthat

PA × PB = PC × PDLet us now consider the case of chords intersecting outside the circle. Let us perform thefollowing activity :

Draw a circle of any radius and centre O. Draw two chords BA and DC intersecting each otheroutside the circle at P. Measure the lengths of line segments PA, PB, PC and PD. Find theproducts PA × PB and PC × PD.

Fig. 20.13

150 Mathematics

You will see that the product PA × PB is equal to theproduct PC × PD, i.e.,

PA × PB = PC × PD

Repeat this activity with two other circles with chordsintersecting outside the circle. You will again find that

PA × PB = PC × PD.Thus, we can say that

If two chords AB and CD of a circle intersect at a point P (inside or outside thecircle), then

PA × PB = PC × PD.

20.8 INTERSECTING SECANTS AND TANGENTS OF A CIRCLE

To see if there is some relation between the intersecting secantand tangent outside a circle, we conduct the following activity:

Draw a circle of any radius with centre O. From an external pointP, draw a secant PAB and a tangent PT to the circle.

Measure the length of the line-segments PA, PB and PT. Find theproducts PA × PB and PT × PT or PT2. What do you find ?

You will find that

PA × PB = PT2

Repeat the above activity with two other circles. You will again find the same result.

Thus, we can say

If PAB is a secant to a circle intersecting the circle at A and B, and PT is a tangentto the circle at T, then

PA × PB = PT2

Let us illustrate these with the help of examples :

Examples 20.5 : In Fig. 20.16, AB and CD are two chords ofa circle intersecting at a point P inside the circle. If PA = 3 cm,PB = 2 cm, PC = 1.5 cm, find the length of PD.

Solution : It is given that PA = 3 cm, PB = 2 cm and PC = 1.5 cm

Let PD = x

we know that PA × PB = PC × PD

⇒ 3 × 2 = (1.5) × x

Fig. 20.14

Fig. 20.15

Fig. 20.16


⇒ x =3 215×. =

3 2×1.5 = 4

∴ Length of the line-segment PD = 4 cm.Example 20.6 : In Fig. 20.17, PAB is a secant to the circle from a point P outside the circle.PAB passes through the centre of the circle and PT is a tangent. IfPT = 8 cm and OP = 10 cm, find the radius of the circle, using PA × PB = PT2.

Solution : Let x be the radius of the circle

It is given that OP = 10 cm∴ PA = PO – OA = (10 – x) cmand PB = OP + OB = (10 + x) cm

PT = 8 cmWe know that PA × PB = PT2

∴ (10 – x) (10 + x) = 82

or 100 – x2 = 64or x2 = 36 or x = 6i.e., radius of the circle is 6 cm.Example 20.7 : In Fig. 20.18, BA and DC are two chords of a circle intersecting each otherat a point P outside the circle. If PA = 4 cm, PB = 10 cm, CD = 3 cm, find PC.

Solution : We are given that PA = 4 cm, PB = 10 cm, CD = 3 cm,

Let PC = xWe know that PA × PB = PC × PDor, 4 × 10 = (x + 3) xor x2 + 3x – 40 = 0

(x + 8) (x – 5) = 0⇒ x = 5∴ PC = 5 cm


1. In Fig. 20.19, if PA = 3 cm, PB = 6 cm, PD = 4 cm, find the length of PC.

2. If in Fig. 20.19, PA = 4 cm, PB = x + 3, PD = 3 cm and PC = x + 5, find the valueof x.

Fig. 20.19 Fig. 20.20 Fig. 20.21

Fig. 20.17

Fig. 20.18

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3. In Fig. 20.20, if PA = 4 cm, PB = 10 cm, PC = 5 cm, find PD

4. In Fig. 20.20, if PC = 4 cm, PD = (x + 5) cm, PA = 5 cm and PB = (x + 2) cm, findx.

5. In Fig. 20.21, PT = 2 7 cm, OP = 8 cm, find the radius of the circle, if O is the centreof the circle.

20.9 ANGLES MADE BY A TANGENT AND A CHORD.

Let there be a circle with centre O and let XY be a tangent to the circle at point P. Draw achord PQ of the circle through the point P as shown in the Fig. 20.22. Mark a point R on themajor arc and let S be a point on the minor arc .

The segment formed by the major arc and chord PQ issaid to be the alternate segment of ∠QPY and the segmentformed by the minor and chord PQ is said to be thealternate segment to ∠QPX.

Let us see if there is some relationship between angles in thealternate segments and the angle between tangents and chord.

Join QR and PR.

Measure ∠PRQ and ∠QPY (See Fig. 20.22)

What do you find ? You will see that ∠PRQ = ∠QPY

Repeat this activity with another circle and same or different radius. You will again find that∠QPY = ∠PRQ.

Now measure ∠QPX and ∠QSP. You will again find that these angles are equal.

Thus, we can state that

The angles formed in the alternate segments by a chord through the point of contactof a tangent to a circle is equal to the angle between the chord and the tangent.

This result is more commonly called as “Angle in the Alternate Segment”.

Let us now check the converse of the above result.

Draw a circle, with centre O, and draw a chord PQ and letit form ∠PRQ in alternate segment as shown in Fig. 20.23.

At P, draw ∠QPY = ∠QRP. Extend the line segment PY toboth sides to form line XY. Join OP and measure ∠OPY.

What do you observe ? You will find that ∠OPY = 90° showing thereby that XY is a tangentto the circle.

Fig. 20.22

Fig. 20.23


Repeat this activity by taking different circles and you find the same result. Thus, we can statethat

If a line makes with a chord angles which are equal respectively to the anglesformed by the chord in alternate segments, then the line is a tangent to the circle.


Example 20.8 : In Fig. 20.24, XY is tangent to a circlewith centre O. If AOB is a diameter and ∠PAB = 40°,find ∠APX and ∠BPY.

Solution : By the Alternate-segment

theorem, we know that

∠BPY = ∠BAP

∴ ∠BPY = 40° [Q ∠BAP = 40° (Given)]

Again, ∠APB = 90° [Angle in a semi-circle]

And, ∠BPY + ∠APB + ∠APX = 180° (Angles on a line)

∴ ∠APX = 180° – (∠BPY + ∠APB)

= 180° – (40° + 90°) = 50°.

Example 20.9 : In Fig. 20.25, ABC is an isosceles trianglewith AB = AC and XY is a tangent to the circumcircle of∆ABC. Show that XY is parallel to base BC.

Solution : In ∆ABC, AB = AC∴ ∠1 = ∠2

Again XY is tangent to the circle at A.∴ ∠3 = ∠2 (Angle in the Alternate segment)∴ ∠1 = ∠3

But these are alternate angles∴ XY || BC


1. Explain with the help of a diagram, the angle formed by a chord in the alternate segmentof a circle.

2. In Fig. 20.26, XY is a tangent to the circle with centre O at a point P. If ∠OQP = 40°,find the value of a and b.

3. In Fig. 20.27, PT is a tangent to the circle from an external point P. Chord AB of thecircle, when produced meets TP in P. TA and TB are joined and TM is the angle bisectorof ∠ATB.

Fig. 20.24

Fig. 20.25

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Fig. 20.26 Fig. 20.27

If ∠PAB = 30° and ∠ATB = 60°, show that PM = PT.

LET US SUM UP

A line which intersects the circle in two points is called a secant of the circle.

A line which touches the circle at a point is called a tangent to the circle.

A tangent is the limiting position of a secant when the two points of intersection coincide.

A tangent to a circle is perpendicular to the radius through the point of contact.

From an external point, two tangents can be drawn to a circle, which are of equal length.

If two chords AB and CD of a circle intersect at a point P (inside or outside the circle),then

PA × PB = PC × PD

If PAB is a secant to a circle intersecting the circle at A and B, and PT is a tangent tothe circle at T, then

PA × PB = PT2

The angles formed in the alternate segments by a chord through the point of contact ofa tangent to a circle are equal to the angles between the chord and the tangent.

If a line makes with a chord angles which are respectively equal to the angles formedby the chord in alternate segments, then the line is a tangent to the circle.

TERMINAL EXERCISE

1. Differentiate between a secant and a tangent to a circle with the help of a figure.

2. Show that a tangent is a line perpendicular to the radius through the point of contact, withthe help of an activity.


3. In Fig. 20.28, if AC = BC and AB is a diameterof the circle, find ∠x, ∠y and ∠z.

4. In Fig. 20.29, OT = 7 cm and OP = 25 cm, findthe length of PT. If PT’ is another tangent to thecircle, find PT′ and ∠POT′.

5. In Fig. 20.30, the perimeter of ∆ABC equals27 cm. If PA = 4 cm, QB = 5 cm, find the lengthof QC.

6. In Fig. 20.30, if ∠BAC = 70°, find ∠BOC.

[Hint : ∠OBC + ∠OCB = 12 (∠ABC + ∠ACB)]

7. In Fig. 20.31, AB and CD are two chords of acircle intersecting at the interior point P of acircle. If PA = (x + 3) cm, PB = (x – 3) cm,

PD = 3 cm and PC = 5 13 cm, find x.

8. In Fig. 20.32, chords BA and DC of the circle,with centre O, intersect at a point P outside thecircle. If PA = 4 cm and PB = 9 cm, PC = x andPD = 4x, find the value of x.

Fig. 20.29

Fig. 20.30

Fig. 20.31

Fig. 20.28

Fig. 20.32

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9. In Fig. 20.33, PAB is a secant and PT is a tangentto the circle from an external point. If PT = x cm,PA = 4 cm and AB = 5 cm, find x.

10. In Fig. 20.34, O is the centre of the circle and∠PBQ = 40°. Find

(i) ∠QPY

(ii) ∠POQ

(iii) ∠OPQ Fig. 20.34

Fig. 20.33


ANSWERS


1. (i) Perpendicular (ii) equal (iii) points of intersection (iv) two (v) no

2. 50º, 50°

3. 3 cm


1. 4.5 cm 2. 3 cm

3. 8 cm 4. 10 cm

5. 6 cm


2. ∠a = ∠b = 50°

Terminal Exercise

3. ∠x = ∠y = ∠z = 45°

4. PT = 24 cm; PT′ = 24 cm, ∠POT′ = 60°

5. QC = 4.5 6. ∠BOC = 125°

7. x = 5 8. x = 3

9. x = 6

10. (i) 40° (ii) 80° (iii) 50°.

Constructions 157

21

Constructions

21.1 INTRODUCTION

One of the aims of the studying Geometry is to acquire the skill of drawing figures accurately.You have learnt how to construct geometrical figures namely triangles, squares and circles withthe help of ruler and compasses. You have constructed angles of 30°, 60°, 90°, 120° and 45°.You have also drawn perpendicular bisector of a line segment and bisector of an angle.

In this lesson we will extend our learning to construct some other important geometrical figures.

21.2 OBJECTIVES


divide a given line segment internally in a given ratio.

Construct a triangle from the given data

(i) SSS

(ii) SAS

(iii) ASA

(iv) RHS

(v) perimeter and base angles

(vi) base, sum/difference of the other two sides and one base angle.

(vii) two sides and a median corresponding to one of these sides.

Construct rectilinear figures such as parallelograms, rectangles, squares, rhombuses andtrapeziums.

Construct a quadrilateral from the given data

(i) four sides and a diagonal

(ii) three sides and both diagonals

(iii) two adjacent sides and three angles

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(iv) three sides and two included angles

(v) four sides and an angle.

Construct a triangle equal in area to a given quadrilateral.

Construct tangents to a circle from a point

(i) outside it

(ii) on it using the centre of the circle

Construct circumcircle of a triangle

Construct incircle of a triangle.


We assume that the learner already knows how to use a pair of compasses and ruler to construct

angles of 30°, 45°, 60°, 90°, 105°, 120°.

the right bisector of a line segment

bisector of a given angle.

parallelograms, rhombuses, rectangles, and squares

a circle

21.4 DIVISION OF A LINE SEGMENT IN THE GIVEN RATIO INTERNALLY

Construction 1 : To divide a line segment internally in a given ratio.

Given a line segment AB. You are required to divide it internally in the ratio2 : 3. We go through the following steps.

Step 1 : Draw a ray AC making an acute angle with AB.

Step 2 : Starting with A, mark off 5 points C1, C2, C3, C4 and C5 at equal distances from thepoint A.

Step 3 : Join C5 and B.

Step 4 : Through C2 (i.e. the second point), draw C2D parallel to C5B meeting AB in D.

Fig. 21.1

Then D is the required point which divides AB internally in the ratio 2 : 3 as shown inFig. 21.1

Constructions 159


1. Draw a line segment 7 cm long. Divide it internally in the ratio 3 : 4. Measure each part.Also write the steps of construction.

2. Draw a line segment PQ = 8 cm. Find the point R on it such that PR = 34 PQ .

[Hint : Divide the line segment PQ internally in the ratio 3 : 1].

21.5 CONSTRUCTION OF TRIANGLES

Construction 2 : To construct a triangle when three sides are given (SSS)

Suppose you are required to construct ∆ABC in which AB = 6 cm, AC = 4.8 cm andBC = 5 cm.

We go through the following steps :

Step 1 : Draw AB = 6 cm.

Step 2 : With A as centre and radius 4.8 cm, draw an arc.

Step 3 : With B as centre and radius 5 cm draw another arcintersecting the arc of Step 2 at C.

Step 4 : Join AC and BC.Then ∆ABC is the required triangle.[Note : You may take BC or AC as the base]Construction 3 : To construct a triangle, when two sides and the included angle is given (SAS)

Suppose you are required to construct a triangle PQR in which PQ = 5.6 cm,QR = 4.5 cm and ∠PQR = 60°

For constructing the triangle, we go through the following steps.

Step 1 : Draw PQ = 5.6 cm

Step 2 : At Q, construct an angle ∠PQX = 60°

Step 3 : With Q as centre and radius 4.5 cm draw an arccutting QX at R.

Step 4 : Join PR

Then ∆PQR is the required triangle.

[Note : You may take QR = 4.5 cm as the base instead of PQ]

Construction 4. To construct a triangle when two angles and the included side are given (ASA).

Let us construct a ∆ABC in which ∠B = 60°, ∠C = 45° and BC = 4.7 cm.

Fig. 21.2

Fig. 21.3

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To construct the triangle we go through the following steps :

Step 1 : Draw BC = 4.7 cm.

Step 2 : At B, construct ∠CBQ = 60°

Step 3 : At C, construct ∠BCR = 45° meeting BQ at A.

Then ∆ABC is the required triangle.

Note : To construct a triangle when two angles and any side(other than the included side) are given, we find thethird angle (using angle sum property of the triangle)and then use above method for constructing the triangle.

Construction 5 : To construct a right triangle, when its hypotenuse and a side are given.

Let us construct a right triangle ABC, right angled at B, sideBC = 3 cm and hypotenuse AC = 5 cm.

To construct the triangle, we go through the following steps.

Step 1 : Draw BC = 3 cm

Step 2 : At B, construct ∠CBP = 90°.

Step 3 : With C as centre and radius 5 cm draw an arc cuttingBP in A.

Step 4 : Join AC

∆ABC is the required triangle.

Construction 6 : To construct a triangle when its perimeter and two base angles are given.

Suppose we have to construct a triangle whose perimeter is 9.5 cm and base angle are 60°and 45°.

To construct triangle, we go through the following steps.

Step 1 : Draw XY= 9.5 cm

Fig. 21.6

Fig. 21.5

Fig. 21.4

Constructions 161

Step 2 : At X, construct ∠YXP = 30° [Which is 1/2 × 60°]

Step 3 : At Y, construct ∠XYQ = 22½° [Which is 1/2 × 45°]

Let XP and YQ intersect at A

Step 4 : Draw right bisector of XA intersecting XY at B.

Step 5 : Draw right bisector of YA intersecting XY at C.

Step 6 : Join AB and AC.


Construction 7 : To construct a triangle when sum of two sides, third side and one of theangles on the third side are given.

Suppose you are required to construct a triangle ABC.

When AB + AC = 8.2 cm, BC = 3.6 cm and ∠B = 45°.

To construct the triangle, we go through the following steps :

Step 1 : Draw BC = 3.6 cm.

Step 2 : At B, construct ∠CBK = 45°.

Fig. 21.7

Step 3 : From BK, cut off BP = 8.2 cm.

Step 4 : Join CP.

Step 5 : Draw right bisector of CP intersecting BP at A.

Step 6 : Join AC

∆ABC is required triangle.

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Construction 8 : To construct a triangle when difference of two sides, the third side and oneof the angles on the third side are given.

Suppose we have to construct a ∆ABC, in which BC = 4 cm ∠B = 60°, AB – AC = 1.2 cm.

To construct the triangle we go through the following steps :

Step 1 : Draw BC = 4 cm.

Step 2 : Construct ∠CBP = 60°.

Step 3 : From BP cut off BK = 1.2 cm.

Step 4 : Join CK

Step 5 : Draw right bisector of CK meeting BP producedat A.

Step 6 : Join AC


Construction 9 : To construct a triangle when its two sides and a median corresponding toone of these sides, are given.

Suppose you have to construct a ∆ABC in whichAB = 6 cm, BC = 4 cm and median CD = 3.5 cm.

We go through the following steps.

Step 1 : Draw AB= 6 cm.

Step 2 : Draw right bisector of AB meeting AB in D.

Step 3 : With D as centre and radius 3.5 cm draw an arc.

Step 4 : With B as centre and radius 4 cm draw anotherarc intersecting the arc of Step 3 in C.

Step 5 : Join AC and BC.

Then ∆ABC is the required triangle.


1. Construct a ∆DEF, given that DE = 5.1 cm, EF = 4 cm and DF = 5.6 cm. Write the stepsof construction as well.

Note : You are also required to write the steps of construction in each of the remainingproblems.

2. Construct a ∆PQR, given that PR = 6.5 cm, ∠P = 120° and PQ = 5.2 cm.

Fig. 21.8

Fig. 21.9

Constructions 163

3. Construct a ∆ABC given that BC = 5.5 cm, ∠B = 75° and ∠C = 45°.

4. Construct a right triangle in which one side is 3 cm and hypotenuse is 7.5 cm.

5. Construct a right angled isosceles triangle in which one of equal side is 4.8 cm.

6. Construct a ∆ABC given that AB + BC + AC = 10 cm, ∠B = 60°, ∠C = 30°.

7. Construct a ∆ABC in which AB = 5 cm, ∠A = 60°,

BC + AC = 9.8 cm.

8. Construct a ∆LMN, when ∠M = 30°, MN = 5 cm and LM – LN = 1.5 cm.

9. Construct a triangle PQR in which PQ = 5 cm, QR = 4.2 cm and median RS = 3.8 cm.

21.6 CONSTRUCTION OF RECTILINEAR FIGURES

You are advised to draw rough sketch for the given data in each of the following constructions.You will observe, that it helps you to visualise/understand the steps of construction.

Construction 10 : To construct a parallelogram when two adjacent sides and the included angleare given.

Suppose that you have to construct a parallelogram in which the adjacent sides are 4 cm and3 cm and included angle is 60°.

To construct the required parallelogram we go through the following steps :

Step 1 : Draw AB = 4 cm

Step 2 : At A, construct ∠BAK = 60°.

Step 3 : From AK cut off AD = 3 cm.

Step 4 : With B and D as centres and radii equal to 3 cm and4 cm respectively draw two arcs cutting each otherat C.

Step 5. Join CD and BC.

Then ABCD is the required parallelogram.

Construction 11 : To construct a rectangle when one of its diagonal and a side are given.

Suppose that you have to construct a rectangle ABCD in which AB = 4 cm and AC = 5.0 cm.

Recall that in a rectangle, each angle is 90° and opposite sides are equal.

To construct the rectangle we go through the following steps.


Fig. 21.10

164 Mathematics

Step 2 : At B, draw ∠ABK = 90°.

Step 3 : With A as centre and radius 5 cm, draw an arccutting BK at C.

Step 4 : With C as centre and radius 4 cm, draw an arc.

Step 5 : With A as centre and radius = BC, draw an arccutting the arc drawn in Step 4 at D.

Step 6 : Join DC and AD

ABCD is the required rectangle.

Construction 12 : To construct a square when its side is given.

Suppose you have to construct a square PQRS in which PQ = 4.4 cm.

We have to follow the following steps to construct the square :

Step 1 : Draw PQ = 4.4 cm.

Step 2 : Construct ∠PQT = 90° at Q.

Step 3 : From QT cut off QR = 4.4 cm.

Step 4 : From P and R, draw two arcs of radii 4.4 cm eachto cut each other at S.

Step 5 : Join PS and RS.

PQRS is the required square.

Construction 13 : To construct a parallelogram when two diagonals and the angle betweenthem is given.

Suppose that the lengths of two diagonals are 8 cm and 6 cm and the angle between them is60°.

Recall that diagonals of a parallelogram bisect each other.

To construct the parallelogram, we go through the following steps :

Step 1 : Draw AC = 8 cm.

Step 2 : Draw right bisector of AC meeting it at O.

Step 3 : Construct ∠COP = 60° and produce PO to Q.

Step 4 : Cut off OB = OD = 3 cm (1/2 × 6, length of seconddiagonal) from OP and OQ.

Step 5 : Join AB, BC, AD and CD.

ABCD is the required parallelogram.

Fig. 21.11

Fig. 21.12

Fig. 21.13

Constructions 165

Construction 14 : To construct a rhombus when one diagonal and side are given.

Suppose, you have to construct a rhombus, when one of its diagonal is 5.5 cm and the sideis 3.3 cm.

To construct the rhombus, we go through the following steps :

Step 1 : Draw AC = 5.5 cm.

Step 2 : With A as centre and radius 3.3 cm, drawtwo arcs one above AC and the other belowAC.

Step 3 : With C as centre and radius 3.3 cm draw twoarcs one above AC and the other below ACintersecting the arcs of Step 2 in B and Drespectively.

Step 4 : Join AB, BC, CD and AD.

ABCD is the required rhombus.

Construction 15 : To construct a trapezium in which one of parallel sides, two non-parallelsides and the distance between parallel sides are given.

Suppose you have to draw a trapezium in which one of parallel sides is 6 cm, two non-parallelsides are of length 4 cm and 5 cm and distance between parallel sides is 3 cm.

To construct the trapezium we go through the following steps :


Step 2 : At A, draw AP⊥ AB.

Step 3 : From AP cut off AK = 3 cm.

Step 4 : At K, draw KL⊥ AK.

Step 5 : With A and B as centres and radii 4 cm and5 cm respectively draw two arcs cutting KLat D and C respectively.

Step 6 : Join AD and BC.

Then ABCD is the required trapezium.


1. Construct a parallelogram if the lengths of its adjacent sides are 5.5 cm and 4 cm andthe included angle is 75°. Write steps of construction as well.

Note : You are also required to write the steps of construction in each of the followingproblems.

Fig. 21.13

Fig. 21.15

166 Mathematics

2. Construct a parallelogram if its sides are 4 cm and 6 cm and one of its diagonal is 7 cm.

3. Construct a parallelogram if its diagonals are 8 cm and 5 cm and the angle between themis 45°.

4. Construct a rectangle whose sides are 4.2 cm and 3.4 cm.

5. Construct a square whose side measures 5.1 cm.

6. Construct a rhombus whose side is 5 cm and one diagonal is 8 cm.

7. Construct a rhombus whose diagonals measure 5 cm and 4 cm.

8. Construct a trapezium one of whose parallel side is 7 cm, non-parallel sides are 4.2 cmand 5.3 cm and the distance between the parallel sides is 3.5 cm.

21.7 CONSTRUCTION OF QUADRILATERALS

You are advised to draw rough sketch from the given data in each of the following constructions.You will observe that it helps you to visualize/understand the steps of construction.

Construction 16 : To construct a quadrilateral when four sides and one diagonal are given.

Suppose you have to construct a quadrilateral ABCD in which AB = 3 cm, BC = 3.5 cm,CD = 4.1 cm, AD = 3.8 cm and diagonal BD = 5 cm.



Step 2 : With A as centre and radius 3.8 cm draw anarc.

Step 3 : With B as centre and radius 5 cm drawanother arc intersecting the arc of Step 2 inD.

Step 4 : Join AD and BD.

Step 5 : With B and D as centres and radii 3.5 cm and 4.1 cm respectively, draw two arcsintersecting each other at C.

Step 6 : Join BC and DC.

ABCD is the required quadrilateral.

Construction 17 : To construct a quadrilateral when three sides and both diagonals are given.

Suppose you are required to construct a quadrilateral ABCD when AB = 3.6 cm, BC = 2.7 cm,AD = 3cm, AC = 4.8 cm and BD = 5 cm.

To construct the required quadrilateral, we go through the following steps.

Step 1 : Draw AB = 3.6 cm.

Fig. 21.16

Constructions 167

Step 2 : With A and B as centres and radii 3 cm and5 cm respectively draw two arcs intersectingeach other at D.

Step 3 : Join AD.

Step 4 : With B and A as centres and radii 2.7 cm and4.8 cm draw two arcs intersecting each other atC.

Step 5. Join DC and BC.


Construction 18. To construct a quadrilateral when two adjacent sides and three angles aregiven.

Suppose you have to construct a quadrilateral ABCD in which

AB = 5 cm, BC = 4.2 cm, ∠B = 60°,

∠C = 90° and ∠A = 75°.

Step 1 : Draw AB = 5 cm

Step 2 : Construct ∠BAP = 75°.

Step 3 : Construct ∠ABQ = 60°

Step 4 : Cut off BC = 4.2 cm from BQ

Step 5 : At C, construct ∠BCR = 90° cutting AP atD.


Construction 19 : To construct a quadrilateral when three sides and two included angles aregiven.

Suppose you have to construct a quadrilateral PQRS in which PQ = 3 cm,

QR = 4 cm, RS = 6 cm, ∠Q = 120° and ∠R = 90°.

To construct the quadrilateral, we go through the followingsteps

Step 1 : Draw QR = 4 cm.

Step 2 : At Q and R, construct ∠RQK = 120° andconstruct ∠QRL = 90°

Step 3 : From QK, cut off QP = 3 cm

Fig. 21.18

Fig. 21.17

Fig. 21.19

168 Mathematics

Step 4 : From RL, cut off RS = 6 cm

Step 5 : Join PS.

Then PQRS is the required quadrilateral.

Construction 20 : To construct a quadrilateral when four sides and an angle are given.

Suppose that you have to construct a quadrilateral ABCD in which AB = 5.5 cm, BC = 3.5 cm,CD = 4 cm, AD = 5 cm, and ∠A = 45°.

To construct the quadrilateral ABCD, we go through the following steps.

Step 1 : Draw AB = 5.5 cm

Step 2 : At A, construct ∠BAK = 45°

Step 3 : Cut off AD = 5 cm from AK.

Step 4 : With B and D as centres and radii 3.5 cmand 4 cm respectively, draw two arcs cuttingeach other at C.

Step 5 : Join BC and DC



1. Construct a quadrilateral ABCD in which AB = 4 cm, BC = 5.2 cm, CD = 5.8 cm,DA = 6.5 cm and AC = 7.8 cm. Also write the steps of construction.


2. Construct a quadrilateral ABCD in which AB = BC = CD = 4 cm, AC = 6 cm andBD = 7 cm.

3. Construct a quadrilateral ABCD in which AB = 6 cm, BC = 6.5 cm, ∠A = 45°,∠B = 120° and ∠C = 90°.

4. Construct a quadrilateral PQRS in which PQ = QR = 5 cm, RS = 6 cm, ∠Q = 120° and∠R = 60°

5. Construct a quadrilateral ABCD in which AB = 4 cm, BC =3.6 cm, CD = 5 cm,AD = 5.2 cm and ∠B = 45°.

6. Construct a quadrilateral PQRS in which PQ = 5 cm, RS = 6 cm, PS = 6.2 cm,PR = 7 cm and the diagonal PR makes an angle of 30° with PQ.

21.8 CONSTRUCTION OF A TRIANGLE EQUAL IN AREA TO A GIVENQUADRILATERAL

Construction 21 : To construct a triangle equal in area to a given quadrilateral.

Suppose quad. ABCD is given

Fig. 21.20

Constructions 169

We have to construct a triangle equal in area to the quadrilateral ABCD.

For that we go through the following steps :

Step 1 : Join AC.

Step 2 : Through D, draw a line segment DE || AC intersecting BC produced at E.

Fig. 21.21

Step 3 : Join AE.

Then ∆ABE is the required triangle.

Construction 22 : To construct a quadrilateral and to construct a triangle equal in area to thisquadrilateral.

Suppose you have to construct a quadrilateral ABCD, in which AB = 3 cm, BC = 4.2 cm,CD = 3.6 cm, DA = 4.5 cm and ∠B = 135° and then to construct a triangle equal in area tothis quadrilateral.



Step 2 : Construct ∠ABK = 135° and cut offBC = 4.2 cm from BK.

Step 3 : With C and A as centres and radii 3.6 cmand 4.5 cm respectively draw two arcsintersecting each other at D.

Step 4 : Join AD and CD.

ABCD is the quadrilateral.

Step 5 : Join DB.

Step 6 : Through C draw CL || DB meeting AB produced in E.

Step 7 : Join DE

Then ∆DAE is the required triangle.

Fig. 21.22

170 Mathematics


1. Draw a quadrilateral PQRS and construct a triangle equal in area to this quadrilateral.Also write the steps of construction.

Note : You are required to write the steps of construction in each of the following problems.2. Construct a quadrilateral ABCD in which, AB = 4 cm, BC = 3.5 cm, CD = 4.2 cm,

DA = 3 cm and AC = 6 cm. Construct a triangle equal in area to this quadrilateral.

3. Construct a rectangle ABCD in which AB = 5 cm and BC = 3.5 cm. Construct a triangle,equal in area to the rectangle on AB as base.

21.9 CONSTRUCTION OF TANGENTS TO A CIRCLE

Construction 23 : To draw a tangent to a given circle at a given point on it using the centreof the circle.

Suppose C be the given circle with centre O and a point P on it. You have to draw a tangentto the circle. We go through the following steps.

Step 1 : Join OP.

Step 2 : At P, draw PT⊥ OP.

Step 3 : Produce TP to Q.

Then TPQ is the required tangent.

Construction 24 : To draw tangents to a circle from a given point outside it.

Suppose C be the given circle and a point A outside it. You have to draw tangents to the circlefrom the point A. For that, we go through the following steps :

Step 1 : Join OA.

Step 2 : Draw the right bisector of OA. Let R be midpoint of OA.

Step 3 : With R as centre and radius equal to RO, drawa circle intersecting the given circle at P and Q.

Step 4 : Join AP and AQ.

Then AP and AQ are the required tangents.


1. Draw a circle of 3 cm radius. Take a point A on the circle. At A, draw a tangent to thecircle by using the centre of the circle. Also write steps of construction.

2. Draw a circle of radius 2.5 cm. From a point P outside the circle, draw two tangents PQand PR to the circle. Verify that lengths of PQ and PR are equal. Also write steps ofconstruction.

Fig. 21.23

Fig. 21.24

Constructions 171

21.10. CONSTRUCTION OF CIRCUMCIRCLE AND INCIRCLE OF A TRIANGLE.

Construction 25 : To construct circumcircle of a triangle.

Suppose a ∆ABC is given. You have to draw a circumcircle of this triangle.

To construct it, we go through the following steps.

Step 1 : Draw the given ∆ABC.

Step 2 : Draw right bisectors of any two sides ray BCand AC which meet each other at O.

Step 3 : Join O with any one vertex say B.

Step 4 : With O as centre and radius OB, draw acircle.

This is the required circumcircle.

Construction 26 : Construct a triangle with sides 4 cm, 5 cm and 6 cm. Draw a circumcircleof this triangle.

We go through the following steps.

Step 1 : BC = 5 cm.

Step 2 : With B and C as centres and radii 4 and 6 cmrespectively draw two arcs intersecting each otherat A.

Step 3 : Join AB and AC. ABC is the required triangle.

Step 4 : Draw right bisectors of AB and BC which meet eachother at O.

Step 5 : Join OB.

Step 6 : With O as centre and radius OB, draw a circle.

This is the required circumcircle.

Construction 27 : To construct incircle of a triangle.

Suppose you have to construct a triangle ABC with AB = 4 cm,BC = 3.5 cm and ∠B = 60° and draw its incircle.

We go through the following steps

Step 1 : Draw BC = 3.5 cm

Step 2 : Draw ∠CBM = 60°

Step 3 : From BM, cut off BA = 4 cm

Fig. 21.25

Fig. 21.26

Fig. 21.27

172 Mathematics

Step 4 : Join AC

ABC is the required triangle.

Step 5 : Draw bisectors of ∠B and ∠C meeting each other at I.

Step 6 : From I, draw IK perpendicular to BC meeting BC in D.

Step 7 : With I as centre, and radius = ID, draw a circle.

This is the required incircle.


1. Construct a ∆ABC with AB = 5 cm, BC = 4.5 cm, ∠B = 75° and draw its circumcircle.Also write steps of construction.

Note : You are required to write the steps of construction in each of the following problems.

2. Construct an isosceles ∆ABC with base BC = 4 cm and one of the equal sidesAB = 3 cm and draw its circumcircle.

3. Construct a triangle with base 4 cm and base angles 60° and 75° and draw its incircle.

4. Construct an equilateral triangle of side 5 cm and draw its incircle.

TERMINAL EXERCISE

1. Draw a line segment PQ = 8 cm long. Divide it internally in the ratio 3 : 5. Also writethe steps of construction.


2. Draw a line segment AB = 6 cm. Find a point C on AB such that AC : CB = 3 : 2. MeasureAC and CB.

3. Construct a triangle with perimeter 14 cm and base angle 60° and 90°.

4. Construct a right angled triangle whose hypotenuse is 8 cm and one if its other two sidesis 5.5 cm.

5. Construct a ∆ABC in which BC = 3.5 cm, AB + AC = 8 cm and ∠B = 60°.

6. Construct a ∆ABC in which AB = 4 cm, ∠A = 45° and AC – BC = 1 cm.

7. Construct a parallelogram having its diagonals as 5 cm and 6 cm and angle between themis 45°.

8. Construct a rectangle with sides 8 cm and 6 cm. Measure the length of its two diagonals.

9. Construct a square with one side 4.2 cm. Measure its diagonals.

Constructions 173

10. Construct a square if its diagonal is 7 cm.

11. Construct a rhombus when the diagonals are 9 cm and 7 cm.

12. Construct a trapezium with one of parallel sides as 6 cm, two non-parallel sides as 4 cmand 4.5 cm and distance between parallel sides as 2.5 cm.

13. Construct a trapezium ABCD in which AB = 8 cm, BC = 4.5 cm, CD = 4 cm,∠B = 60° and AB || CD.

[Hint : ∠C = 180° – 60° = 120°]

14. Draw a quadrilateral ABCD in which AB = 4 cm, BC = 5 cm, CD = 4.5 cm,DA = 5 cm and AC = 7 cm. Construct a triangle equal in area to this quadrilateral.

15. Draw a circle of diameter 6 cm. From a point P outside the circle, draw two tangentsto the circle.

16. Construct triangle with sides 4 cm, 3 cm and 5.5 cm and draw it circumcircle.

17. Construct a right angled triangle with base 4 cm and hypotenuse 6 cm and draw its incircle.

174 Mathematics

22

Co-ordinate Geometry

22.1 INTRODUCTION

The problem of locating a village or a road on a large map can involve a good deal of searching.But the task can be made easier by dividing it into squares of managable size. Each squareis identified by a combination of a letter and a number, or of two numbers, one of which refersto a vertical division of the map into columns, and the other to a horizontal division into rows.

1 2 3 4 5 6(ii)

3

2

1

3

2

1

4

A B C D E F(i)

Fig. 22.1

Co-ordinate Geometry 175

In the above Fig. 22.1(i), we can identify the shaded square on the map by the coding, (B,2)or(4, 2) [See Fig. 22.1(ii))]. The pair of numbers used for coding is called ordered pair. If weknow the coding of a particular city, roughly we can indicate its location inside the shadedsquare on the map. But still we do not know its precise location. The method of finding theposition of a point in a plane very precisely was introduced by the French Mathematician andPhilosopher, Rene Descates (1596–1650).

In this, a point in the plane is represented by an ordered pair of numbers, called the Cartesianco-ordinates of a point.

In this lesson, we will learn more about cartesian co-ordinates of a point, distance betweentwo points in a plane, section formula and co-ordinates of the centroid of a triangle.

22.2 OBJECTIVES


fix the position of different points in a plane.

find the distance between two different points whose co-ordinates are given.

find the co-ordinates of a point, which divides the line segment joining two points, ina given ratio internally.

find the co-ordinates of the mid-point of the join of two points.

find the co-ordinates of the centroid of a triangle with given vertices.

solve problems based on the above concepts.


Idea of number line

Fundamental operations on numbers.

Properties of a right triangle.

22.4 CO-ORDINATE SYSTEM

Recall that you have learnt to draw the graph of a linear equation in two variable in Lesson 6.

The position of a point in a plane is fixed w.r.t to its distances from two axes of a reference,which are usually drawn by the two graduated number lines XOX′ and YOY′, at right anglesto each other at O (See Fig. 22.2).

The horizontal number line XOX′ is called x-axis and the vertical number line YOY′ is calledy-axis. The point O, where both axes intersect each other called the origin. The two axestogether are called rectangular coordinate system.

It may be noted that, the positive direction of x-axis is taken to the right of the origin O, OXand the negative direction is taken to the left of the origin O, i.e., the side OX′.

Similarly, the portion of y-axis above the origin O, i.e., the side OY is taken as positive andthe portion below the origin O, i.e., the side OY′ is taken as negative.

176 Mathematics

Fig. 22.2

22.5 CO-ORDINATES OF A POINT

The position of a point is given by two numbers, called co-ordinates which refer to the distancesof the point from these two axes. By convention the first number, the x-co-ordinate (or abscissa),always indicates the distance from the y-axis and the second number, the y-coordinate(or ordinate) indicates the distance from the x-axis.

Fig. 22.3

In the above Fig 22.3, the co-ordinates of the points A and B are (3, 2) and (–2, –4) respectively.

You can say that the distance of the point A(3, 2) from the y-axis is 3 units and from thex-axis is 2 two units. It is customary to write the co-ordinates of a point as an ordered pairi.e., (x co-ordinate, y co-ordinate).


It is clear from the point A(3, 2) that its x co-ordinate is 3 and the y-co-ordinate is 2. Similarlyx co-ordinate and y co-ordinate of the point B(–2, –4) are –2 and –4 respectively.

In general, co-ordinates of a point P(x, y) imply that distance of P from they-axis is x units and its distance from the x-axis is y units.

You may note that the co-ordinates of the origin O are (0, 0). The y co-ordinate of every pointon the x-axis is 0 and the x co-ordinate of every point on the y-axis is 0.

In general, co-ordinates of any point on the x-axis to the right of the origin is (a, 0) and thatto left of the origin is (–a, 0), where 'a' is a non-zero positive number.

Similarly, y co-ordinates of any point on the y-axis above and below the x-axis would be(0, b) and (0, –b) respectively where 'b' is a non-zero positive number.

You may also note that the position of points (x, y) and (y, x) in the rectangular co-ordinatesystem is not the same. For example position of points (3, 4) and (4, 3) are shown in Fig 22.4.

Fig. 22.4

Example 22.1 : Write down x and y co-ordinate for each of the following points :

(a) (1, 1) (b) (–3, 2) (c) (–7, –5) (d) (2, –6)

Solution : (a) x co-ordinate is 1 (b) x co-ordinate is –3

y co-ordinate is 1 y co-ordinate is 2.

(c) x co-ordinate is –7 (d) x co-ordinate is 2

y co-ordinate is –5. y co-ordinate is –6.

178 Mathematics

Example 22.2 : Write down distances from y and x axes respectively for each of the followingpoints :

(a) A(3, 4) (b) B(–5, 1) (c) C(–3, –3) (d) D(8, –9)

Solution : (a) The distance of the point A from the y-axis is 3 units and from the x-axis is4 units.

(b) The distance of the point B from the y-axis is 5 units to the left of the originand from the x-axis is 1 unit.

(c) The distance of the point C from the y-axis is 3 units to the left of the originand from the x-axis is also 3 units below the origin.

(d) The distance of the point D from the y-axis is 8 units to the right of the originand from the x-axis is 9 units below the origin.

22.6 QUADRANTS

The two axes XOX′ and YOY′ divide the plane into four parts called quadrants.

Fig. 22.5

The four quadrants (See Fig. 22.5) are named as follows :

XOY : I Quadrant ; YOX′ : II Quadrant

X'OY' : III Quadrant ; Y′OX : IV Quadrant

We have discussed in Section 22.4 that

(i) along x-axis, the positive direction is taken to the right of the origin and negative directionto its left.

(ii) along y-axis, portion above the x-axis is taken as positive and portion below the x-axisis taken as negative (See Fig. 22.6).


Fig. 22.6 Fig. 22.7

Therefore, co-ordinates of all points in the first quadrant are of the type (+, +) (See Fig. 22.7)

Any point in the II quadrant has x co-ordinate negative and y co-ordinate positive [(–, +)].Similarly, in III quadrant, a point has both x and y co-ordinates negative [(–, –)] and inIV quadrant, a point has x co-ordinate positive and y co-ordinate negative [(+, –)].

For example :

(a) P(5, 6) lies in the first quadrant as both x and y co-ordinates are positive.

(b) Q(–3, 4) lies in the second quadrant as its x co-ordiante is negative and y co-ordinateis positive.

(c) R (–2, –3) lies in the third quadrant as its both x and y co-ordinates are negative.

(d) S(4, –1) lies in the fourth quadrant as its x co-ordinate is positive and y coordinate isnegative.


1. Write down x and y co-ordinates for each of the following points :

(a) (3, 3) (b) (–6, 5) (c) (–1, –3) (d) (4, –2)

2. Write down distances of each of the following points from the y and x axis respectively.

(a) A(2, 4) (b) B(–2, 4) (c) C(–2, –4) (d) D(2, –4)

3. Group each of the following points quadrantwise :

A(–3, 2), B (2, 3), C(7, –6), D(1, 1), E(–9, –9),

F (–6, 1), G (–4, –5), H(11, –3), P(3, 12), Q(–13, 6),

180 Mathematics

22.7 PLOTTING OF A POINT WHOSE CO-ORDINATES ARE GIVEN

The point can be plotted by measuring its distancesfrom the axes. Thus, any point (h, k) can be plottedas follows :

(i) Measure OM equal to h along the x-axis(See Fig. 22.8).

(ii) Measure MP perpendicular to OM and equalto k.

Follow the rule of sign in both cases.

For example points (–3, 5) and (4, –6) would beplotted as shown in Fig 22.9.

Fig. 22.9

22.8 DISTANCE BETWEEN TWO POINTS

The distance between any two points P(x1, y1) andQ(x2, y2) in the plane is the length of the linesegment PQ.

From P, Q draw PL and QM perpendiculars on thex-axis and PR perpendicular on QM.

Then, OL = x1, OM = x2, PL = y1 and QM = y2

∴ PR = LM = OM – OL = x2 – x1

Fig. 22.8

Fig. 22.10


QR = QM – RM = QM – PL = y2 – y1

Since PQR is a right angled triangle

∴ PQ2 = PR2 + QR2

= (x2 – x1)2 + (y2 – y1)2 (By the Pythagoras Theorem)

∴ PQ = x x y y2 12

2 12− + −b g b g

Distance between two points = difference of abscissae difference of ordinates2 2b g b g+

Corollary : The distance of the point (x1, y1) from the origin (0, 0) is

x y12

120 0− + −b g b g = x y1

212+

Let us consider some examples to illustrate.

Example 22.3 : Find the distance between each of the following points :

(a) P(6, 8) and Q(–9, –12)

(b) A(–6, –1) and B(–6, 11)

Solution : (a) Here the points are P(6, 8) and Q(–9, –12)

By using distance formula, we have

PQ = − − + − −9 6 12 82 2b g b gm r= 15 202 2+ = 225 400+ = 625 = 25

Hence, PQ = 25 units.

(b) Here the points are A(–6, –1) and B(–6, 11)

By using distance formula, we have

AB = − − − + − −6 6 11 12 2b gm r b gm r

= 0 122 2+ = 12

Hence, AB = 12 units.

Example 22.4 : The distance between two points (0, 0) and (x, 3) is 5. Find x.

Solution : By using distance formula, we have the distance between (0, 0) and (x, 3) is

x − + −0 3 02 2b g b g

182 Mathematics

It is given that

x − + −0 3 02 2b g b g = 5

or x2 23+ = 5

squaring both sidesx2 + 9 = 25

or x2 = 16or x = ± 4Hence, x = + 4 or –4.

Example 22.5 : Show that the points (1, 1), (3, 0) and (–1, 2) are collinear.

Solution : Let P(1, 1), Q(3, 0) and R(–1, 2) be the given points

∴ PQ = 3 1 0 12 2− + −b g b g = 4 1 5+ = units

QR = − − + −1 3 2 02 2b g b g = 16 4 2 5+ = units

RP = − − + −1 1 2 12 2b g b g = 4 1 5+ = units

Now, PQ + RP = 5 5+d i units = 2 5 units = QR

∴ P, Q and R are collinear points.

Example 22.6 : Find the radius of the circle whose centre is at (0, 0) and which passes throughthe point (–6, 8)

Solution : Let A(0, 0) and B(–6, 8) be the given points.

Now, radius of the cirlce is same as the distanceof the line segment AB.

∴ AB = − − + −6 0 8 02 2b g b g= 36 64+ = 100= 10

Hence radius of the cirlce is 10 units.


1. Find the distance between each of the following pair of points :(a) (3, 2) and (11, 8) (b) (–1, 0) and (0, 3)

(c) (3, –4) and (–2, 5) (d) (2, –11) and (–9, –3)

Fig. 22.11


2. Find the radius of the cirlce whose centre is at (2, 0) and which passes through the point(7,–12)

3. Show that the points (–5, 6), (–1, 2) and (2, –1) are collinear.

22.9 SECTION FORMULA

To find the co-ordinates of a point, which divides the linesegment joining two points, in a given ratio internally.

Let A(x1, y1) and B(x2, y2) be the two given points andP(x, y) be a point on AB which divides it in the givenratio m : n. We have to find the co-ordinates of P.

Draw the perpendiculars AL, PM, BN on OX, and, AK,PT on PM and BN respectively. Then, from similartriangles APK and PBT, we have

APPB = AK

PTKPTB= ...(i)

Now, AK = LM = OM – OL = x – x1

PT = MN = ON – OM = x2 – x

KP = MP – MK = MP – LA = y – y1,

TB = NB – NT = NB – MP = y2 – y

∴ From (i), we have

mn =

x xx x

y yy y

−−

= −−

1

2

1

2

The first two relations give

mn =

x xx x−−

1

2

or mx2 – mx = nx – nx1

or x(m + n) = mx2 + nx1

or x =mx nx

m n2 1++

Similarly, from the relation APPB

KPTB

= , we get

mn =

y yy y−−

1

2 which gives on simplification,

Fig. 22.12

184 Mathematics

y =my ny

m n2 1++

∴ x =mx nx

m n2 1++ , and y =

my nym n2 1++ ...(i)

Hence co-ordinates of a point which divides the line segment joining the points(x1, y1) and (x2, y2) in the ratio m : n internally are :

mx nxm n , my ny

m n2 1 2 1++

++

FHG

IKJ

22.9.1 Mid-Point Formula

The co-ordinates of the mid-point of the line segment joining two points (x1, y1) and (x2, y2)can be obtained by taking m = n in the section formula above.

Putting m = n in (1) above, we have

x =nx nx

n n2 1++ =

x x2 12+

and y =ny ny

n n2 1++ =

y y2 12+

∴ The co-ordinates of the mid-point joining two points (x1, y1) and (x2, y2) are :

x x

2 , y y2

2 1 2 1+ +FHG

IKJ


Example 22.7 : Find the co-ordinates of a point which divides the line segment joining eachof the following points in the given ratio :

(a) (2, 3) and (7, 8) in the ratio 2 : 3 internally

(b) (–1, 4) and (0, –3) in the ratio 1 : 4 internally.

Solution : (a) Let A(2, 3) and B(7, 8) be the given points.

Let P(x, y) divide AB in the ratio 2 : 3 internally.

Using section formula, we have

x =2 7 3 2

2 3× + ×

+= 20

5 = 4


and y =2 8 3 3

2 3× + ×

+= 25

5 = 5

∴ P(4, 5) divides AB in the ratio 2 : 3 internally.

(b) Let A(–1, 4) and B(0, –3) be the given points.

Let P(x, y) divide AB in the ratio 1 : 4 internally

Using section formula, we have

x =1 0 4 1

1 445

× + × −+

= −b g

and y =1 3 4 4

1 4135

× − + ×+

=b g

∴ P −FHGIKJ

45

135

, divides AB in the ratio 1 : 4 internally.

Example 22.8 : Find the mid-point of the line-segment joining two points (3, 4) and (5, 12).

Solution : Let A(3, 4) and B(5, 12) be the given points.

Let C(x, y) be the mid-point of AB. Using mid-point formula, we have,

x =3 5

2+

= 4

and y =4 12

2+

= 8

∴ C(4, 8) are the co-ordinates of the mid-point of the line segment joining two points (3, 4)and (5, 12).

Example 22.9 : The co-ordinates of the mid-point of a line segment are (2, 3). If co-ordinatesof one of the end points of the line segment are (6, 5), find the co-ordiants of the other endpoint.

Solution : Let other the end point be A(x, y)It is given that C(2, 3) is the mid point

∴ We can write,

2 =x + 6

2 and 3 =y +5

2

or 4 = x + 6 or 6 = y + 5or x = –2 or y = 1

∴ A(–2, 1) be the co-ordinates of the other end point.

Fig. 22.13

186 Mathematics

22.10 CENTROID OF A TRIANGLE

To find the co-ordinates of the centroid of a triangle whose vertices are given.

Definition : The centroid of a triangle is the point of concurrency of its medians and divideseach median in the ratio of 2 : 1.

Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle ABC. Let AD be the medianbisecting its base. Then, using mid-point formula, we have

D =x x y y2 3 2 3

2 2+ +F

HGIKJ,

Fig. 22.14

Now, the point G on AD, which divides it internally in the ratio 2 : 1, is the centroid.

If (x, y) are the co-ordinates of G, then

x =2

21

2 1 3

2 31

1 2 3× + + ×

+= + +

x x x x x x

y =2

21

2 1 3

2 31

1 2 3× + + ×

+= + +

y y y y y y

Hence, the co-ordinates of the centroid are given by

x x x3 , y y y

31 2 3 1 2 3+ + + +FHG

IKJ

Example 22.10 : The co-ordinates of the vertices of a triangle are (3, –1), (10, 7) and (5, 3).Find the co-ordinates of its centroid.Solution : Let A(3, –1), B(10, 7) and C(5, 3) be the vertices of a triangle.Let G(x, y) be its centroid

then, x =3 10 5

3+ +

= 6


and y =− + +1 7 3

3 = 3

Hence G(6, 3) are the co-ordinates of the centroid.


1. Find the co-ordinates of the point which divides internally the line segment joining thepoints :

(a) (1, –2) and (4, 7) in the ratio 1 : 2

(b) (3, –2) and (–5, 4) in the ratio 1 : 1

2. Find the mid-point of the line joining :

(a) (0, 0) and (8, –5)

(b) (–7, 0) and (0, 10)

3. Find the centroid of the triangle whose vertices are (5, –1), (–3, –2) and (–1, 8).

LET US SUM UP

If (2, 3) are the co-ordinates of a point, then x co-ordinate (or abscissa) is 2 and they co-ordinate (or ordinate) is 3.

In any co-ordinates (x, y), ‘x’ indicates the distance from the y-axis and 'y' indicates thedistance from the x-axis.

The co-ordinates of the origin are (0, 0)

The y co-ordinate of every point on the x-axis is 0 and the x co-ordinate of every pointon the y-axis is 0.

The two axes XOX′ and YOY′ divides the plane into four parts called quadrants.

The distance of the line segment joining two points (x1, y1) and (x2, y2) is given by :

x x y y2 12

2 12− + −b g b g .

The distance of the point (x1, y1) from the origin (0, 0) is x y12

12+

The co-ordinates of a point, which divides the line segment joining two points (x1,y1)and (x2, y2) in a ratio m : n internally are given by :

mx nxm n

my nym n

2 1 2 1++

++

FHG

IKJ,

188 Mathematics

The co-ordinates of the mid-point of the line segment joining two points (x1, y1) and(x2, y2) are given by :

x x y y2 1 2 12 2+ +F

HGIKJ,

The co-ordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2) and(x3, y3) are given by :

x x x y y y1 2 3 1 2 33 3

+ + + +FHG

IKJ,

TERMINAL EXERCISE

1. In Fig 22.15, AB = AC. Find x.

Fig. 22.15

2. The length of the line segment joining two points (2, 3) and (4, x) is 13 units. Findx.

3. Find the lengths of the sides of the triangle whose vertices are A(3, 4), B(2, –1) andC(4, –6).

4. Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle.

5. Find the co-ordinates of a point which divides the join of (2, –1) and (–3, 4) in the ratioof 2 : 3 internally.

6. Find the centre of a circle, if the end points of a diameter are P(–5, 7) and Q(3, –11).

7. Find the centroid of the triangle whose vertices are P(–2, 4), Q(7, –3) and R (4, 5).


ANSWERS


1. (a) 3; 3 (b) –6; 5 (c) –1; –3 (d) 4; –2

2. (a) 2 units; 4 units

(b) 2 units to the left of the origin; 4 units above the x-axis

(c) 2 units to the left of the origin; 4 units below the origin

(d) 2 units; 4 units below the origin

3. Quadrant I : B(2, 3), D(1, 1) and P(3, 12)

Quadrant II : A(–3, 2), F(–6, 1) and Q(–13, 6)

Quadrant III : E(–9, –9) and G(–4, –5)

Quadrant IV : C(7, –6) and H (11, –3)


1. (a) 10 units (b) 10 units (c) 106 units (d) 185 units

2. 13 units.


1. (a) (2, 1) (b) (–1, 1) 2. (a) 4 52, −FH IK (b) −FH IK7

2 5,

3. 13

53,FH IK

Terminal Exercise

1. 3 units 2. 0 or 6

3. AB = 26 units, BC = 29 units and CA = 101 units

5. (0, 1) 6. (–1, –2) 7. (3, 2)

Area of Plane Figures 193

Module 4

Mensuration

With the advancement of civilization on this planet earth, the human beings were faced withthe following types of things in their immediate environment : part of the earth on which theylive (plane), fruit and vegetables trees, birds and animals, stone crests and other solids or planeobjects. The desire to know more about them, to have a measure of sizes and capacity andto acquire them was a natural phenomenon. With this came into being the science ofmeasurement of land and knowledge about rectilinear and solid figures, which was given thename of mensuration.

Old civilization like Greek, Egyptians, Arabian and especially Indian contributed a lot to thedevelopment of this science. In the old Indian scriptures, there are rules laid down to performspecial types of yajnas to fulfil specific desires in specially built altars (vedis) These vedis werein the shape of squares, rectangles, and circles made of cuboidal, cubical and spherical bricks.The whole lot of “Sulba Sutras” and Srauta Sutras contain the knowledge of methods ofconstruction of such vedis and the properties of the used shapes.

“Baudhayan Sulba” Apastamba Sulba and Katyayan Sulba which date back to 8th century B.C.are famous among all these sutras. After that even the Indian mathematician like Aryabhatt(476 AD), Brahmgupta (598 AD), Bhaskaracharya (1114 AD) and Varahmihir (1120 AD)devoted much of their works in providing rules for finding perimeters and areas of plane figuresand volumes of solids. One of the oldest text “Surya Sidhant” deals with the shapes (mostlyspherical) and orbits (mainly elliptical) of heavenly bodies.

In this text, we shall study about the perimeter and area of plane figures and surface area, totalsurface area and volumes of solids.

194 Mathematics

23

Area of Plane Figures

13.1 INTRODUCTION

In our day-to-day life, we have to estimate the amount of wood required for a rectangular tabletop or the area of cloth required for table cover, or to find the area of paths, parks, plots andfloor etc.

In this lesson, we shall study the methods of finding perimeters and areas of plane figures,especially square, rectangle, triangle, special quadrilaterals and circle. We will also studymethods to find areas of rectangular or circular paths.

23.2 OBJECTIVES

After studying this lesson the learner will be able to :

identify rectilinear and non-rectilinear plane figures.

explain the meaning of perimeter and area of closed figures.

find the perimeter and area of rectilinear figures, like a square, a rectangle, a triangle,a quadrilateral, a trapezium, a parallelogram and a rhombus.

find the area of a triangle using Hero’s formula.

find the area of rectilinear paths of different types in a rectangle (or a rectangularenclosure).

find the circumference and area of a circle.

find area and perimeter of a sector.

find the area of circular paths inside or outside a circle (or a circular enclosure).

Solve problems from day-to-day life situations based on the above concepts


Measurement of line-segments

Knowledge and conversion of units of measurements



Drawing plane geometric figures

Properties of special quadrilaterals

23.4 PERIMETER OF PLANE FIGURES

The distance covered to walk along the boundary of a plane figure is called its perimeter. Inconnection with the circle, this distance covered is called circumference of the circle. Theperimeter is measured in terms of linear units.

23.5 AREA OF PLANE FIGURES

The measure of the planar region enclosed by a plane figure is called its area.

This is measured in terms of square units.

23.5.1 A Unit Square

A square of side 1 cm is called a unit cm square and its area is taken to be 1 cm2 and its perimeteris taken to be 4 cm.

Fig. 23.1

Similarly a square of side 1 m is called a unit meter square and its area is taken to be 1 m2

and its perimeter as 4 m.

In general, area of a square of side ‘a’ unit is a2 sq. unit and its perimeter is 4aunit.

Example 23.1 : Find the area of a square whose perimeter is 80 meters.

Solution : Perimeter of square = 4a = 80 meters

∴ a = 804 or 20 metres

∴ Area of square = a2 sq. m

= 20 × 20 sq. m or 400 sq m.

196 Mathematics

Example 23.2. The area of a square park is 625 sq m. Find the length of the wire requiredto fence around the park.

Solution. Area of square park (a2) = 625 sq. m

∴ Side of square (a) = 625 m = 25 25× m or 25 m

Now, length of wire = Perimeter of square

= 4a = 4 × 25 m = 100 m

i.e. length of the wire required to fence the park is 100 m.

Example 23.3. Find the area of a square whose diagonal is 15 m.

Solution. Diagonal of the square ABCD = AC = AB BC2 2+

= a a2 2+ or 2 a m

∴ 2 a = 15 or a = 15

2 m

Area of the square ABCD = a2 sq m = 2252 or 112.5 sq m.

23.6. RECTILINEAR FIGURES

The figures whose boundaries are formed by line-segments are called rectilinear figures. Forexample square, rectangle, triangle, trapezium, quadrilateral, parallelogram and rhombus arerectilinear figures while a circle and a sector are non-rectilinear figures.

23.7 PERIMETER AND AREA OF A RECTANGLE

Take 4 centimetre squares and join them in a row as shown in Fig. 23.3(i). Add another rowof four centimetre square over the and above the first row to form the rectangle ABCD as shownin Fig 23.3(ii).

(i) (ii)

Fig. 23.3

Fig. 23.2


Rectangle ABCD is formed by 8 small unit squares. Therefore the planar region enclosed byit is 8 cm2.

You can see that 4 × 2 = 8.

If we take another rectangle PQRS with baseformed by 5 unit squares and has four such rows(See Fig. 23.4), we see that it contains 20 or(5 × 4) unit cm squares.

∴ Its area is 20 cm2

i.e. (5 × 4) cm2

or 20 cm2

From here we can generalise that the area of a rectangle of length l cm and breadth b cm canbe written as (l × b) cm2 or lb cm2.

In case of Fig 23.3(ii), the distance covered to go around the rectangle ABCD is(4 + 2 + 4 + 2) cm or 12 cm.

Similarly, the distance covered to go around the rectangle PQRS in Fig. 23.4 is(5 + 4 + 5 + 4) cm or 18 cm.

Do you observe that in each case the distance covered is obtained by adding 2 × length and2 × breadth?

i.e. Perimeter of a Rectangle = 2(l + b)

A square is a special rectangle in which length and breadth are equal.

Thus, the area of a

square = (side)2 square units

rectangle = (length × breadth) square units

And, the perimeter of a

square = 4 × side linear units

rectangle = 2 (length + breadth) linear units

Example 23.4 : The length and breadth of a rectangular field are 23.7 m and 14.5 m respecively.Find :

(i) the area of the field

(ii) the length of wire required to put a fence around the boundary of the field.

Solution : (i) Area of the field = Length × breadth

Fig. 23.4

198 Mathematics

= (23.7 × 14.5) sq m

= 343.65 sq m

(ii) Length of wire = Perimeter of the field

= 2(length + breadth)

= 2(23.7 + 14.5) m

= 2(38.2) or 76.4 m

Example 23.5 : A rectangular plot measures 400 m × 121 m. Find the side of a square plothaving area equal to the rectangular plot.

Solution : Area of rectangular plot = (400 × 121) sq m = (20)2 × 112 sq m

∴ Area of square plot = (20)2 × 112 sq m

∴ Side of square plot = 20 112 2× m

= 220 m.

23.8. AREA OF A PARALLELOGRAM

We know from our study of geometry thatparallelograms on the same base and between thesame parallels are equal in area.

∴ Area (||gm ABCD) = Area Rect. PBCQ

= (BC × PB) sq units.

∴ Area of a Parallelogram = Base × Altitude square units

Example 23.6 : Find the area of a parallelogram whose base and altitude are 12 cm and8 cm respectively.Solution : We know that Area of ||gm ABCD = Base × altitude

= (12 × 8) cm2

= 96 cm2.

23.9. AREA OF A TRIANGLEWe know that a diagonal of a parallelogram divides it into two triangles of equal area

∴ Area of ∆ DBC = 12 Area of ||gm ABCD

= 12 Base × Altitude

∴ Area of a triangle = 12 Base × Altitude square units

Fig. 23.5

Fig. 23.6


Example 23.7 : The base of a triangular field is three times its altitude. If the cost of cultivatingthe field at the rate of Rs 15 per square metre is Rs 20250, find the base and the altitude ofthe field.

Solution : Area of the field = 2025015 sq m

= 1350 sq mLet the altitude of the triangular field be x m

∴ Base = 3x m

∴ Area of the field = 12 .3x.x = 3

22x sq m

As given in the question,

32

2x = 1350

or x2 = 1350 23× = 900

or x = 30Thus, the base of the field is 90 m and its altitude is 30 m.

23.10 AREA OF A TRAPEZIUM

In Fig. 23.7, ABCD is a trapezium in which AB || DC.

∴ Area of trapezium ABCD

= Area of ∆ ABD + Area ∆ BCD

= 12 .a.h + 1

2 .b.h

= 12 (a + b).h

Thus, the area of a trapezium

= 12 (sum of the parallel sides) × Distance between them square units

Example 23.8 : Find the area of a trapezium lengths of whose parallel sides are 20 cm and12 cm and the distance between them is 5 cm.

Solution : We know that

Area of Trapezium = 12 (Sum of the Parallel sides) × Distance between them

Fig. 23.7

200 Mathematics

= 12 [20 + 12] × 5 cm2

= 12 × 32 × 5 cm2

= 80 cm2.

23.11 AREA OF A QUADRILATERAL

Area of quadrilateral ABCD

= Area ∆ ABC + Area ∆ ACD

(See Fig. 23.8)

= 12

121 2AC h AC h. .+

= 12 1 2AC h h+b g ,

where h1 and h2 are lengths of perpendicularsfrom vertices B and D on diagonal AC

Example 23.9 : Find the area of a field in the shape of a quadrilateral, one of whose diagonalsis 40 m long and the perpendiculars from the other two vertices on this diagonal are 32 mand 18 m.

Solution : Area of quadrilateral field ABCD

= Area of ∆ ABC + Area of ∆ ACD

= 12 40 32 1

2 40 18× × + × ×LNM OQP sq m

= (640 + 360) sq m = 1000 sq m.

23.12 AREA OF RHOMBUS

In Fig. 23.10, ABCD is a rhombus whose diagonal AC and BD bisect each other at right angles.Thus : Area of rhombus ABCD = Area ∆ ABD + Area ∆ BCD

= 12

12BD AP BD PC. .+

= 12 BD (AP + PC) = 1

2 (BD) (AC)

Thus, the area of rhombus

= 12 × product of its diagonals square units.

Fig. 23.8

Fig. 23.9

Fig. 23.10


Example 23.10. The diagonals of a rhombus are of length 12 cm and 8 cm. Find the area ofthe rhombus.

Solution. We know that

Area of a Rhombus = 12 × Product of Diagonals

= 12 (12 × 8) cm2

= 48 cm2

Example 23.11. Find the area of a rhombus, one of whose diagonals is 12 cm and the sideis 10 cm.

Solution : In Fig.23.11, let ABCD be the given rhombus with AB = BC = CD = AD = 10 cmand the diagonal BD = 12 cm.

AC and BD bisect (at right angles) at O (See Fig. 23.11)

∴ OB = OD =6 cm

∴ OA = 10 62 2− cm = 8 cm

∴ Diagonal AC = 16 cm

Area of rhombus = 12 (product of diagonals)

= 12 × 12 × 16 cm2

= 96 cm2

23.13 AREA OF A TRIANGLE USING HERO’S FORMULA

The area of triangle ABC whose sides are given by a, b and c is given by

∆ = s s a s b s c− − −b gb gb g

where s = a b c+ +2

This formula is known as Hero’s formula after the name of Greek Mathematician Heron ofAlexendria. The formula was also obtained by the Indian Mathematicians Brahmagupta andAryabhata.

Example 23.12 : The sides of a triangular field are 165 m, 143 m and 154 m. Find the areaof the field.

Solution : Here we use the Hero’s formula for finding the area of the field

Fig. 23.11

202 Mathematics

∆ = s s a s b s c− − −b gb gb g

where a, b, c are sides and s = a b c+ +2

∴ s = 165 143 1542

4622

+ + = or 231 m

∴ ∆ = 231 231 165 231 143 231 154− − −b gb gb g sq m

= 231 66 88 77b gb gb g sq m

= 11 7 3 3 2 11 2 2 2 11 7 11× × × × × × × × × × × sq m

= 11 × 11 × 7 × 3 × 2 × 2 = 10164 sq m.Thus area of the triangular field = 10164 sq m

Example 23.13 : Find the area of a field in the shape of a trapezium whose parallel sides areof length 11 m and 25 m and the non-parallel sides are of length 15 m and 13 m.

Solution : Let ABCD be a field in the shape of a trapezium with AB || CD in whichAB = 25 m, CD = 11 m, AD = 13 m and BC = 15 m [Fig. 23.12]

Draw CE || DA and let CF be perpendicular to AB

Now,

Area of ∆ BCE = 21 21 14 21 15 21 13− − −b gb gb g

because s = 13 14 152

+ + or 21 m

= 21 7 6 8× × × or 84 sq m

Also, area of ∆ BCE = 12 14b g.CF = 7 CF

∴ 7 CF = 84 or CF = 12 m

∴ Area of trapezoidal field ABCD

= 12 (sum of parallel sides) × distance between them

= 12 (11 + 25) × 12 sq m

= 216 sq m

23.14 AREA OF RECTANGULAR PATHS

In Fig. 23.13, ABCD is a rectangular park of dimensions a × b and let there be a path of uniformwidth c all around the park as shown

Fig. 23.12


∴ Area of path = Area of Rectangle PQRS –Area of rectangle ABCD

= (PQ × QR) – (AB × BC)

= [(a + 2c). (b + 2c)] – (a × b)

= ab + 2bc + 2ac + 4c2 – ab

= 4c2 + 2c(a + b)

In Fig. 23.14, ABCD is a rectangular park of dimensions a × b, in which PQRS and EFGHare two perpendicular paths of width c, parallel to the sides of the park.

Fig. 23.14

Area of paths = Area of path EFGH + Area of PQRS – Area of path QLMN

= a.c + b.c – c.c

= (a + b – c)c


Example 23.14 : A rectangular hall 15 m long and12 m broad is surrounded by a verandah 2 m wide.Find the area of verandah.

Solution : Dimension of rectangular enclosure PQRSare (15 + 2 × 2) m and (12 + 2 × 2)m or 19 m and16 m

From Fig. 23.15, we see that

Area of Verandah = Area of rectangle PQRS – Area of rectangle ABCD

= (19 × 16 – 15 × 12) sq m

= (304 – 180) sq m or 124 sq m

Fig. 23.13

Fig. 23.15

204 Mathematics

Example 23.15 : A rectangular piece of landmeasures 100 m by 80 m. From the centre ofeach side a path 5 m wide runs across up to theopposite side. Find the area of the paths.

Solution : Area of paths = Area of rectangleABCD + Area ofrectangle PQRS –Area of squareQLMN

= [(100 × 5) +(80 × 5) – (5 × 5)] sq m

= [500 + 400 – 25] sq m

= 875 sq m


1. The area of a square field is 225 square metres. Find its perimeter.

2. What will be the diagonal of a square whose perimeter is 60 m?

3. The length and breadth of a rectangular field are 22.5 m and 12.5 m. Find

(i) the area of the field, (ii) the length of the wire required to put a fence along theboundary of the field.

4. The sides of a rectangular field of area 726 sq metres are in the ratio3 : 2. Find the length of its diagonal correct up to one decimal place.

5. Find the area of a triangular field whose sides are 50 m, 78 m and 112 m. Also, find thelength of the perpendicular from the opposite vertex to the side measuring 112 m.

6. Find the area of a field in the form of a quadrilateral ABCD in whichAB = 165 m, BC = 143 m, CD = 85 m, AD = 85 m and the diagonal AC = 154 m.

7. Find the area of a parallelogram with base and altitude of length 20 cm and 12 cmrespectively.

8. The parallel sides of a trapezium are 20 metres and 16 metres long respectively and thedistance between them is 12 m. Find its area.

9. The perimeter of a rhombus is 146 cm and one of its diagonals is 48 cm, find the otherdiagonal and the area of the rhombus.

10. Find the area of a quadrilateral one of whose diagonals is 30 metres long and the lengthsof perpendiculars from the other two vertices to this diagonal are 10 m and 14 mrespectively.

Fig. 23.16


11. A rectangular courtyard 120 m long and 90 m broad has a path of uniform width 5 mon the inside running round it. Find the area of the path.

12. A rectangular lawn 80 metres by 60 metres has two roads, each 2 metres wide runningin the middle of it, one parallel to the length and the other parallel to the breadth. Findthe area of the roads.

23.15 CIRCUMFERENCE AND AREA OF A CIRCLE

Recall that a circle is the locus of a point which moves in a plane in such a way that its distancefrom a fixed point (in the same plane) always remains constant. The fixed point (O) is calledthe centre of the circle and the constant distance is called the radius of the circle. Any straightline drawn through the centre of circle, whose end point lie on the circle is called a diameter,whose length is equal to twice the radius of the circle.

We state without any logical proof that the ratio of the circumference of a circle to its diameteris always constant which is also equal to the ratio of the area enclosed by a circle to the squareof its radius (r).

i.e. Cr2 = A

r2 = constant.

The value of this constant is nearly 3 17 and is usually

denoted by the Greek letter π (pie), which is an irrationalnumber but its value, correct to four places of decimalsis 3.1416, to avoid lengthy calculations the value of π

is often taken to be 227 .

∴ Cr2 = π or C = 2πr

andAr2 = π or A = π r2 .

Thus, circumference of a circle = 2πππππr linear units

and, the area of the circle = πππππr2 square units.

Example 23.16 : Find the circumference and area of a circle of radius 3.5 cm[Use π = 22/7]

Solution : Circumference of the circle = 2πr

= 2 227 35× × . cm

= 22 cmAlso, area of the circle = π r2

=227× ×3.5 3.5 cm2

= 38.5 cm2

Fig. 23.17

206 Mathematics

Example 23.17 : The radius of a wheel is 42 cm. How many revolutions will it make in going26.4 km ?

Solution : Distance travelled in one revolution = Circumference of the wheel

= 2πr = 2 227 42× × cm

= 264 cm

∴ No. of revolutions required to travel 26.4 km

= 26 4 1000 100264

. × ×

= 10000

Example 23.18 : The difference between the circumference and the radius of a circle is74 metres. Find the diameter of the circle.

Solution : We have

2π r – r = 74

or r (2π – 1) = 74

or r 447 1−FH IK = 74

or r 377FH IK = 74

or r = 74 737× or 14 m

∴ The diameter of the circle = 2(14) m = 28 m.

23.16 PERIMETER AND AREA OF SECTOR OF A CIRCLE

Two radii OA and OB enclose a portion of the circular region making central angle θ . Theregion is called a sector of the circle.

In Fig. 23.18, sector AOB is the sector with central angleθ. Let ‘l’ be the length of arc AB. The,

ABcircumference = θ

360°[Corresponding arcs subtend proportional central

angles]

or l2πr = θ

360° or l = 2 360π θr. ° Fig. 23.18


∴ Perimeter of the sector = OA + OB + AB

= 2r + 2 360π θr

Also area of sector is given by

Area of sector AOBπr2 =

θ360°

or Area of sector AOB = π θr2360° .

23.17. AREA OF CIRCULAR PATHS

If we have a circular field of radius ‘r’, surrounded by a path ofuniform width (d) so that r + d = R (say)

then, the area of the circular path

= Area of the outer circle – Area of the inner circle

= (π R2 – π r2) sq unit

= π (R2 – r2) sq unit

Let us take some examples based on the above formula :

Example 23.19 : A circular road runs round a circular garden. If the difference between thecircumference of outer circle and inner circle is 66 meters, find the width of the road.

Solution : Let the radius of inner circle be rand that of outer circle be R

∴ Width of the road = (R – r) = d (say)Now, we have 2 π R – 2 π r = 66

or 2π (R – r) = 66

∴ d = (R – r) = 66 722

12× × m = 10.5 m.

Thus width of the road is 10.5 mExample 23.20 : A path of width 2 meters runs around a circular plot whose circumference

is 75 37 metres. Find

(i) the area of path.(ii) the cost of gravelling the path at Rs 7 per square meter.

Solution : Here 2π r = 75 37

i.e. 2 227× × r = 528

7

Fig. 23.20

Fig. 23.21

Fig. 23.19

208 Mathematics

or r = 5287

12

722× × = 12

i.e., the radius of the plot is 12 m.

(i) Area of path = π (R2 – r2) = 227 14 122 2−d i

= 227 52× m2 = 1144

7 sq m

(ii) Cost = Rs 1144

77×F

HGIKJ

= Rs 1144.

Example 23.21. Find the area of sector of a circle whose radiusis 6 cm when

(a) the angle at the centre is 35°(b) when the length of arc is 22 cm

Solution. (a) Area of sector = π θr2360. °

= 227 6 6 35

360× × × cm2

= 11 sq cm(b) Here length of arc l = 22 cm

∴ 2 360π θr ° = 22 cm

Area of sector = π θr2360. °

= 12 2 360r r. π θ

°

= 12 r.l = 1

2 6 22× × sq cm = 66 sq cm


1. Find the areas and circumference of the circles with radius (i) 1.4 m (b) 4.9 m.

2. The difference between the circumference and diameter of a circle is 210 m. Find the radiusof the circle.


3. The difference between the area of a circle and the square of its radius is 105 sq m. Findthe circumference of circle.

4. The driving wheel of a locomotive engine, 1.4 m in radius, makes 70 revolutions in oneminute. Find the speed of the train in km/hr.

5. A circular park of radius 21 m has a road 7 m wide on its outside all around. Find thearea of the road.

6. Find the areas of sectors of a circle with radius 3.5 m having central angle

(i) 60° (ii) 90°.

7. The length of the minute hand of a clock is 7 cm. Find the area covered by the minute

hand in 6 minutes. Use π =LNM OQP227 .

LET US SUM UP

Area of a square of side a = a2 sq units and perimeter = 4a units

Perimeter of rectangle = 2(length + breadth) units

Area of rectangle = (length × breadth) sq units

Area of a triangle = 12 × base × height sq units

= s s a s b s c− − −b gb gb g sq units

where a, b anc c are sides of the triangle and

s = a b c+ +2 units

Area of a parallelogram = Base × Altitude sq units

Area of a trapezium = 12 (sum of parallel sides) × Distance between them sq units

Area of a rhombus = 12 (Product of the diagonals) sq units

Area of rectangular path = Area of outer rectangular enclosure – Area of inner rectangularenclosure

Area of perpendicular paths in the middle of a rectangular field

= Area of path parallel to length + Area of path parallel tobreadth – Area of common square.

210 Mathematics

Circumference of a circle = 2π r units

Area of a circle = π r2 sq units

Area of circular path = π (R2 – r2), where R > r

Area of a sector of a circle = π θr2360

.° sq units

Perimeter of a sector = 2 2 360r r+ °LNM OQPπ θ units

TERMINAL EXERCISE

1. The side of a square park is 37.5 m. Find its area.

2. The perimeter of a square is 480 m. What is its area ?

3. How long will a man take to walk round the boundary of a square field of area40000 sq meters at the rate of 4 km an hour ?

4. The length of a room is three times its breadth. If its breadth is 4.5 m, find the area ofits floor.

5. The perimeter of a rectangle is 980 meters; and the length is to breadth is 5 : 2. Findits area.

6. Find the area of each of the following parallelograms :

(i) side 20 m and the corresponding altitude 12 m.

(ii) one side is 13m, second side is 14 m and the diagonal is 15 m.

7. The area of a rectangular field is 27000 sq meters and the ratio between its length andbreadth is 6 : 5. Find the cost of wire required to go four times round the field at therate of Rs 7 per 10 metres of length of wire.

8. Find the area of the following trapezia :

Lengths of parallel sides Perpendicular distance between them

(i) 30 m and 20 m 15 m

(ii) 15.5 m and 10.5 m 7.5 m

(iii) 17 m and 40 m 14.6 m

(iv) 40 m and 22 m 12 m

9. Find the area of a piece of land in the shape of a quadrilateral one of whose diagonalsis 20 m and lengths of perpendiculars from the other two vertices on the diagonal areof length 12 m and 18 m respectively.


10. Find the area of a field in the form of a trapezium whose parallel sides measure 48 mand 160 m and the non-parallel sides measure 50 m and 78 m respectively.

11. Find the area of a quadrilateral ABCD in which AB = 85 meters, BC = 143 metres,CD = 165 metres, DA = 85 metres and DB = 154 metres.

12. Find the areas of the following triangles whose sides are :

(i) 25 m, 60 m, 65 m

(ii) 60 m, 111 m, 153 m

13. The sides of a triangle are 51 m, 52 m, and 53 m. Find the perpendicular from the oppositeside on the side of length 52 m. Also find the areas of the two triangles into which theoriginal triangle is divided.

14. Find the area of a rhombus, one of whose diagonals measures 8 m and the side is 5 m.

15. The difference between the two parallel sides of a trapezium is 8 metres, the perpendicularbetween them is 24 metres and the area of the trapezium is 312 sq metres. Find the lengthsof the two parallel sides.

16. A rectangular piece of land measures 200 m by 150 m. From the centre of each side apath of 10 m wide runs across up to the opposite side. Find the area of paths.

17. A rectangular plot of grass measures 65 m by 40 m. It has a path of uniform width8 m all around inside it. Find the cost of spreading red sand stone on the path at the rateof Rs 5.25 per sq m.

18. A rectangular plot of land measuring 30 m by 20 m has two paths each 2 m wide on boththe sides (inside and outside) of the boundary. Find the total area of the two paths.

19. A path 3 metres wide runs around a circular park whose radius is 9 meters. Find the areaof the path.

20. The difference between the circumference and diameter of a circle is 30 m. Find the radiusof the circle.

21. From the circular piece of cardboard with radius 1.47 m, a sector with central angle 60°has been removed. Find the area of the portion removed.

22. A circular plot with a radius of 15 m has a road 2 m wide running all around inside it.Find the area of the road.

212 Mathematics

ANSWERS


1. 60 m 2. 15 2 m 3. 281.25 sq m, 70 m

4. 39.7 m 5. 1680 sq m, 30 m 6. 12936 sq m

7. 240 sq m 8. 216 sq m 9. 55 m, 1320 sq m

10. 360 sq m 11. 2000 sq m 12. 276 sq m


1. (i) 6.16 sq m, 8.8 m (ii) 75.46 sq m, 30.8 m

2. 49 m 3. 44 m 4. 36.96 km/hr

5. 1078 sq m 6. (i) 6.41 sq m (ii) 9.6 sq m

7. 15.4 cm.

Terminal Exercise

1. 1406.25 sq m 2. 14400 sq m 3. 12 min

4. 60.75 sq m 5. 49000 sq m 6. (i) 240 sq m (ii) 168 sq m

7. Rs 1848

8. (i) 375 sq m (ii) 97.5 sq m (iii) 416.1 sq m (iv) 372 sq m

9. 300 sq m 10. 3120 sq m 11. 12936 sq m

12. (i) 750 sq m (ii) 2754 sq m 13. 45 m, 540 sq m, 630 sq m

14. 24 sq m 15. 17 m, 9 m 16. 3400 sq m

17. Rs 7476 18. 400 sq m 19. 198 sq m

20. 7 m 21. 1.1319 sq m 22. 176 sq m

Surface Area and Volume of Solids 213

24

Surface Area and Volume of Solids

24.1 INTRODUCTION

In the previous lesson you have studied about plane figures i.e., figures which completely liein a plane like squares, rectangles, triangles, circles etc.

But the objects like a brick, a glass tumbler, a box, a football, an ice cream cone, etc. are notplane figures. They are called Solids.

In this lesson we will study about these types of solids.

24.2 OBJECTIVES


identify different solids

explain the meaning of surface area of a solid

find the surface area of a cube, cuboid, cylinder, cone, sphere and a hemisphere usingthe respective formulae.

find the area of four walls of a room

explain the meaning of volume of a solid

find the volume of a cube, cuboid, cylinder, cone, sphere and a hemisphere using respectiveformulae

solve problems from day to day life situations based on the above concepts.


Area of plane and rectilinear figures

Circumference and area of a circle


24.4. SOLID FIGURES

In Fig 24(i), we have a paper cut in the form as shown. It is a plane figure.

214 Mathematics

But when we fold the paper along the dotted lines, we can make a box (like a chalk box) asshown in Fig 24.1(ii).

(i) (ii)

Fig. 24.1

As the box occupies some part of the space, it has more than two dimensions. Such objectswhich occupy space (i.e., they have three dimensions) are called solids.

The sum of the areas of the plane figures making up the boundary of a solid figure is calledits surface area, [For example, the area of paper in Fig 24.1(i) is the surface area of box] andthe measure of part of space occupied by a solid is called its Volume.

Now, we shall take some solids like a cuboid, a cylinder, a cone and a sphere and learn tofind their surface areas and volumes.

24.5 CUBOID

A brick, chalk box, geometrical box, match box and a book areall examples of cuboids. Fig 24.2 represents a cuboid. It can beeasily seen from the figure that a cuboid has six rectangular planesurfaces called faces. [ABCD, ABEF, BCHF, EFHG, ADGE andDCHG], and the opposite faces [like ABCD and EFHG] arecongurent.

Two adjacent faces meet along a line segment called an edge.For example, faces ABCD and BCHF meet along the edge BC.A cuboid has 12 edges in all.

A cuboid has 8 corners called the vertices. A, B, C, D, E, F, Gand H are the vertices of the cuboid represented by Fig 24.2.

It can be seen that at every vertex, there are three edges meeting [called coterminous edges].One of these edges is taken as length, the other the breadth and the third as height and aredenoted by ‘l’, ‘b’ and ‘h’ respectively. The line-segment joining the vertex A to the vertexH is called a main diagonal of the cuboid.

Fig. 24.2


Let AB= l, AE = b and AD = h,

then the total surface area of cuboid = sum of the areas of six faces ABCD, EFGH, ADGE,BCHF, ABFE and DCHG

= (lh + lh + bh + bh + lb + lb)

= 2(lb + bh + hl)

∴ Total surface area of a cuboid = 2(lb + bh + hl)

Main diagonal of a cuboid= l b h2 2 2+ +

Cube : A cube is a special case of a cuboid when all the edges are equal i.e. l = b = h = a(say)

∴ The surface area of a cube = 6a2

Main Diagonal of a cube = 3 a

To find the volume of a cuboid, we first define the unit of measurement of volume, whichis unit cube.

A unit cube is the volume of a cube of side 1 unit. So, if the side of a cube is 1 cm, itsvolume is 1 cubic centimeter (or 1 cu. cm) and if the side is 1 m, the volume is 1 cu m.

To find the volume of a cuboid, we are to find the number of unit cubes contained in it.

Let us have a cuboid with sides 5 cm, 4 cm and 3 cm. In Fig. 24.3, you can easily find thatthe number of unit cubes in the cuboid are 5 × 4 × 3 = 60. So, the volume of the cuboid =60 cu. cm.

Fig. 24.3

You can see that

Volume = 60 cu. cm. = 5 cm × 4 cm × 3 cm

= length × breadth × height

216 Mathematics

In general, we can deduce that

Volume of a cuboid = length × breadth × height

and Volume of cube = (edge)3

Let us now take some examples to illustrate the above formulae :

Example 24.1 : Find the surface area and volume of a slab of stone measuring 3 m in length,2 m in breadth and 25 cm in thickness.

Solution : Here l = 3 m, b = 2 m and h = 25 cm = 25100 m = 1

4 m

Surface area = 2(lb + bh + hl)

= 2 3 2 2 14 3 1

4× + × + ×FH IK sq m

= 2 6 12

34+ +FH IK sq m = 14.5 sq m.

Volume = l.b.h = 3 × 2 × 14 cu m = 1.5 cu m.

Example 24.2 : If the surface area of a cube is 96 sq cm, find its volume.

Solution : The surface area of cube = 6 a2, where a is the side of cube

∴ 96 = 6 a2

⇒ a2 = 966 = 16

∴ a = 4 cm

∴ Volume of the cube = a3 = (4)3 cu cm = 64 cu cm.

Example 24.3 : A tank contains 60,000 cu. m of water. If the length and breadth are 50 mand 40 m respectively, find its depth.

Solution : Volume of water in tank = 60,000 cu m

Length of tank = 50 m

Breadth of tank = 40 m

Let depth of the tank be x m

∴ 50 × 40 × x = 60,000

⇒ x = 60 00050 40

,× = 30 m

Hence depth of the tank = 30 m.


Example 24.4 : If the volume of cube is 2197 cu cm, find the surface area and the length ofthe main diagonal of the cube.

Solution : Volume of cube = (side)3 = 2197 cu cm = (13)3 cu cm

∴ Side of cube = 13 cm

Surface area of cube = 6 (side)2 sq units

= 6(13)2 sq cm

= 1014 sq cm

In Fig 24.4, DF is diagonal of a cube

∴ Length of the diagonal of a cube = 3 (side)

= 3 .13 cm = 13 3 cm or 22.516 cm.

Example 24.5 : Five cubes each of edge 16 cm, are joined end to end. Find the surface areaof the resulting cuboid.

Solution : Let the cuboid obtained by joining five cubes be as shown in Fig 24.5

Fig. 24.5

∴ l = 16 × 5 = 80 cm

b = 16 cm

h = 16 cm

∴ Surface area of the resulting cuboid = 2(lb + bh + hl)

= 2(80 × 16 + 16 × 16 + 16 × 80) sq cm

= 2(1280 + 256 + 1280) sq cm

= 5632 sq cm.

Example 24.6 : A wooden box 1.5 m long, 1.25 m wide and 65 cm deep and open at the topis to be made. Determine the cost of wood required for it, if one sq m of wood costs Rs 10.

Solution : Surface area of wood required

= lb + 2bh + 2lh [Q The box has five faces]

Fig. 24.4

218 Mathematics

= 1.5 × 1.25 + 2(1.25 × .65 + 1.5 × .65) sq cm

= 1.875 + 2 × 0.65 × 2.75 sq cm

= 1.875 + 3.575 = 5.450 sq cm.

∴ Cost of wood required for the box = Rs (5.45 × 10)

= Rs 54.50.

Example 24.7 : A closed wooden box measures externally as 42 cm by 32 cm by 27 cm. Thewood used is 1 cm thick. Find the internal capacity (volume) of the box.

Solution : Here external dimensions are 42 cm, 32 cm, 27 cm

Since the wood is 1 cm thick, so the internal dimension will be (42 – 2) cm, (32 – 2) cm,(27 – 2) cm

∴ Volume = (40 × 30 × 25) cu cm = 30000 cu cm.

Example 24.8 : A river 10 meters deep and 100 meters wide is flowing at the rate of 4.5 kman hour. Find how many cubic meter of water runs into the sea per second.

Solution : Rate of flow = 4.5 km/hr

= 45003600 m/sec = 5

4 m/s

∴ Length of river bed per second = 54 m/s

Breadth = 100 m and depth = 10 m

∴ Volume of water flowing/sec = 54 100 10× × cu m

= 1250 cu m.

Example 24.9 : A field is 600 m long and 50 m broad. A tank 30 m long, 20 m broad and12 m deep is dug in the field. The earth taken out of it is spread evenly over the field. Findthe height of the field raised by it.

Solution : Area of the field = (600 × 50) sq m = 30000 sq mArea of the tank = (30 × 20) sq m = 600 sq m

Volume of earth taken out of tank= (30 × 20 × 12) cu m= 7200 cu m

Area of the field, where the earth is to be spread= (30,000 – 600) sq m = 29400 sq m

∴ Height of the field raised

= 720029400 m =

1249 m.


Example 24.10 : A cuboidal beam is 8 meters long, 50 cm broad and 20 cm thick. What isits cost at Rs 7000 per cubic meter ?

Solution : Volume of beam 8 50100

20100× ×FH IK cu m

= 45 cu m

Cost of the beam = Rs 45 7000×FH IK = Rs 5600


1. Fill in the blanks to make each of the following statements true :

(i) Total surface area of a cuboid = ...............

(ii) Surface area of a cuboid open at the top = ...............

(iii) Total surface area of a cube = ...............

(iv) Volume of a cuboid = ...............

(v) Volume of a cube = ...............

(vi) Length of diagonal of a cuboid = ...............

2. Find the volume and surface area of each of the following cuboids :

Length Breadth Height

(i) 6 m 3 m 2.5 m

(ii) 15 cm 10 cm 5 cm

(iii) 10 m 4 m 75 cm

(iv) 3 m 2.5 m 1.5 m

3. Find the surface areas and the volumes of the cubes with edges

(i) 5 cm (ii) 3.6 cm (iii) 1.6 m

4. Find the edges of a cube whose volume is

(i) 3375 cu cm (ii) 2.197 cu m (iii) 15.625 cu cm.

5. A closed wooden box measures externally as 50 cm by 40 cm by 30 cm. The wood usedin 2 cm thick. Find the capacity of the box.

6. The length of a room is 12 meters, width 8 meters and height 6 meters. How many boxesit can hold if each box occupies a space of 1.5 cu meters ?

220 Mathematics

7. A cuboidal box whose external dimensions including the lid are 32 dm, 27 dm, 12 dmis made of wood 1 dm thick. What is the capacity of the box and what is the volumeof wood used in it ?

8. Find the total surface area of a wooden plank of width 3 m, thickness 0.75 m and of volume33.75 cubic meters.

9. Three cubes each of side 8 cm are joined end to end. Find the surface area of the resultingcuboid.

10. The areas of three adjacent faces of a cuboid are a, b and c. It its volume is V, provethat V2 = abc.

24.6 RIGHT CIRCULAR CYLINDER

A right circular cylinder is a solid generated by the revolution of a rectangle about one of itssides which remains fixed.

Thus if the rectangle ABCD revolves about the side AB, itdescribes the cylinder shown in Fig 24.6.

In our daily life, we come across many solids like water pipes,thin cans, beakers in the laboratory, powder box, which are rightcircular cylinders.

We note that the ends (or bases) of a right circular cylinder arecongurent circles and the line joining the centres A and B of thesecircles is perpendicular to the two ends.

24.6.1 Volume of a Cylinder

= (area of the base) × height= π r2h cubic units

where, r is the radius of the base and h is the height of the cylinder.

24.6.2 Curved Surface of a Cylinder

Area of the rectangle obtained by cutting a hollow cylinder along any line on its surface parallelto the axis is called its curved surface area.

Fig. 24.7

Fig. 24.6


∴ Curved surface area of a cylinder = 2π rh

If the cylinder is closed at both ends, then the total surface area = curved surface area + areaof two ends

= 2π rh + 2π r2 sq units

= 2π r (r + h) sq units

And if the cylinder is closed from one end and open form the other, then its curved surfacearea = 2π rh + π r2 sq units.

Let us now take some examples to illustrate the above formulae. In all proboems use π = 227

unless stated otherwise.

Example 24.11 : Find the volume and the total surface area of a closed right cylinder whoseradius is 7 m and height is 10 m.

Solution : Here r = 7 m and h = 10 m

∴ Volume = π r2h

= 227 × 7 × 7 × 10 cu m

= 1540 cu m

Total surface area = 2 π r (r + h)

= 2 × 227 × 7 × (7 + 10) sq m

= 22 × 34 = 748 sq m.

Example 24.12 : A hollow cylindrical tube, open at both ends is made of iron 1 cm thick.If the external diameter is 12 cm and the length of tube is 70 cm, find the volume of iron usedin making the tube.

Solution : Here external radius (R) = 6 cm

and internal radius (r) = 6 – 1 = 5 cm

and height (h) = 70 cm

∴ Volume of iron = External volume – Internal volume

= π R2h – π r2h

= π h(R2 – r2)

=227 × 70 × (36 – 25) cu cm

= 220 × 11 cu cm = 2420 cu cm.

222 Mathematics

Example 24.13 : The diameter of a roller 1 m long is 70 cm. If it takes 200 revolutions tolevel a playground, find the cost of levelling at the rate of 75 paise per sq m.

Solution : Here r = 35 cm = 0.35 m and h = 1 m

∴ Curved surface area = 2π rh

= 2 227

35100 1× × × sq m

= 2.2 sq m

∴ Area swept in 200 revolution = 2.2 × 200 = 440 sq m

∴ Cost of levelling = 440 × 0.75 = Rs 330.

Example 24.14 : A field is 150 m long and 70 m broad. A circular tank of radius 5.6 m anddepth 20 cm is dug in the field and the earth taken out of it is spread evenly over the field.Find the height of the field raised by it.

Solution : Volume of earth dug out = π r2h

= 227

5610

5610 20× × ×FH IK cu m

= 1971.2 cu m.

Area of field = (150 × 70) sq m = 10500 sq m

Area of base of tank =227

5610

2× FH IK

LNM

OQP sq m

= 98.56 sq m

∴ Area of the field where the earth is to be spread

= 10500 – 98.56 sq m

= 10401.44 sq m

∴ Height of the field raised

= Volumearea = 1971.2

10401.44 = 0.1895 m

= 18.95 cm.

Example 24.15 : A cubic meter of iron is drawn into a wire of diameter 3.5 mm. Find the

length of the wire. = Use π =LNM OQP227 .

Solution : Volume of iron melted = 1 cu m

= 100 × 100 × 100 cu cm


Let the length of the wire be x cm

∴ Volume of wire = π 740

2FH IK × x cu cm

= 227

740

740× × × x cu cm

By the problem,

227

740

740× × × x = 100 × 100 × 100

x = 100 × 100 × 100 × 722

407

407× × cm

= 10 122 40 40

7× × × km

= 103896 m (approx).

Example 24.16 : A cylindrical bucket of diameter 28 cm and height 12 cm, is full of water.The water is emptied into a rectangular tub of length 66 cm and breadth 28 cm. Find the heightto which water rises in the tub.

Solution : Volume of water in the bucket

= π r2h = 227 × 14 × 14 × 12 cu cm

= 7392 cu cm.

Let h be the height to which water rises in the tub.

∴ Volume of water in the tub = 66 ×28 × h cu cm

∴ By the problem,

66 × 28 × h = 7392

or h = 739266 28× = 4 cm

i.e. water rises to a height of 4 cm in the tub.



(i) Volume of a right circular cylinder = ...............

(ii) Surface area of a closed cylinder = ...............

224 Mathematics

(iii) Surface area of a cylinder open at the top = ...............

(iv) Surface area of a cylinder open at both ends = ...............

2. Find the volumes and total surface area of the following cylinders :

Radius Height

(i) 7 cm 12 cm

(ii) 10 cm 3.5 m

(iii) 5 m 1.4 m

3. A water storage tank has a cylindrical shape. If it is 2.1 m high and has a diameter 1 m,find its volume.

4. A rectangular piece of paper 33 cm long and 16 cm wide is rolled along its breadth toget a cylinder of height 16 cm. Find the volume of the cylinder.

5. The volume of right circular cylinder is 3080 cu cm and the radius of base is 7 cm. Findthe curved surface area of the cylinder.

6. A hollow cylindrical tube, open at both ends, is made of iron 2 cm thick. If the externaldiameter be 50 cm and the length of tube be 140 cm, find volume of iron used in makingthe tube.

7. A well, with 10 meters inside diameter, is dug 14 meters deep. Earth taken out of it hasbeen spread all-around it to a width of 5 meters to form an embankment. Find the heightof the embankment.

8. The radii of two cylinders are in the ratio 3 : 2 and their heights are in the ratio 7 : 4.Calculate the ratios of their volumes and of the cured surface areas.

9. The diameter of a garden roller is 2.8 m and it is 1.5 m long. How much area will it coverin 100 revolutions ?

10. Find the whole surface area of a hollow cylinder open at the ends, if its length is 8 cm,the external diameter is 10 cm and the thickness is 2 cm (π = 3.1416).

24.7 SURFACE AREA AND VOLUME OF RIGHT CIRCULAR CONE

A right circular cone is a solid generated by the revolution ofa right angled triangle about one of the its sides containing theright angle as axis.

In Fig 24.8 right triangle AOB revolves along AO to generatethe cone of height ‘h’ and radius ‘r’ and slant height l = AC.

24.7.1 Surface Area of a Right circular cone

To find the total surface of a cone, let us cut it along its slantheight, and spread it into a sector ABC as shown in Fig 24.9.

Fig. 24.8


Fig. 24.9

Thus curved surface area = 12 . radius AC . arc BC.

= 12 2. .l πr = π rl sq. units

Since area of the base is πr2, so total surface area

= πrl + πr2 = πr(l + r)

24.7.2 Volume of a Right circular cone

Volume of a cone =13 × area of base × height

=13

2πr h cu units.

24.8 SURFACE AREA AND VOLUME OF A SPHERE

A sphere is a solid generated by the revolution of a semicircle about its diameter. It can alsobe defined as under :

A sphere is the locus of a point which moves in space such thatits distance from a fixed point in space remains constant. Thefixed point is called the centre of sphere and the constant distanceis called the radius of sphere. In Fig. 24.10, O is the centre andOP = r is the radius.

To find the surface area and the volume of a sphere, we use thefollowing formulae.

Surface area of sphere = 4π r2

Volume of sphere = 43

3πr

Fig 24.10

226 Mathematics

24.8.1 Hemisphere

A hemisphere is obtained by cutting a sphere into two equal halves by a plane passing throughits centre [Fig. 24.11]

Thus, curved surface area of a hemisphere

= 2π r2

Volume of a hemisphere = 23

3πr

Total surface area of a solid hemisphere

= 2π r2 + π r2

= 3π r2

Let us now take some examples :

Example 24.17 : Find the volume, curved surface area and the total surface area of a rightcircular cone, the radius of whose base is 14 m and the height is 9 m.

Solution : Here r = 1.4 m and h = 9 m

∴ Volume of cone = 13

2πr h

= 13

227

1410

1410 9× × × × cu m

= 18.48 cu m

l = r h2 2+ = 1.4b g2 29+

= 196 81 82 96. .+ = = 9.1 m

Curved surface area = π rl

= 227

1410 91× × . sq m

= 40.04 sq m

Total surface area = π rl + π r2

= 40 04 227

1410

1410. + × × sq m

= 40.04 + 6.16 sq m

= 46.20 sq m.

Fig 24.11

Fig 24.12


Example 24.18 : Find the surface area and the volume of a sphere of radius 10.5 cm.

Solution : Here radius = 10.5 cm

∴ Volume = 43

3πr = 43

227

212

212

212× × × × cu cm

= 4851 cu cm

Surface area = 4π r2

= 4 227

212

212× × × sq cm

= 1386 sq cm.

Example 24.19 : If the radius of a sphere is tripled, what is the ratio of the volume of originalsphere to that of the second ?

Solution : Let the radius of sphere = ‘r’ units

∴ Volume of sphere V1 = 43

3πr cu units

When the radius is tripled, new radius = 3 r units

∴ Volume of sphere V2 = 43 3 3π rb g = 4

3 33 3πr b g cu units

∴VV

1

2=

43

43 27

127

3

3

π

π

r

r .=

∴ Ratio is 1 : 27.

Example 24.20 : Find the diameter of a sphere whose volume is 38808 cu cm.

Solution : Let r be the radius of the sphere

Then its volume = 43

3πr

By question,43

3πr = 38808

∴ r3 = 38808 3 74 22

× ××

= 441 × 21= (21)3

∴ r = 21

∴ Radius of the sphere = 21 cm.Thus diameter of the sphere is (21 × 2) or 42 cm.

228 Mathematics

Example 24.21 : Find the radius of the base of right circular cone of height 10.5 cm and volume176 cu cm.

Solution : Let r be the radius of the base of the cone. Its height = 10.5 cm

Then its volume = 13

2πr h

= 13

227

2× × ×r 10.5 cu cm

∴ 13

227

2× × ×r 10.5 = 176

∴ r2 = 176 3 722 10.5

× ×× = 16

∴ r = 4

Hence radius of the base = 4 cm.

Example 24.22 : Rain water which fills a tub of dimensions 6 m × 4 m × 2.75 cm is transferredto a cylindrical vessel of radius 20 cm. Find the height of water in the cylindrical vessel.

Solution : Volume of rain water collected in the tub

= 6 4 27510000× × cu m = 0.66 cu m

Let the height of the water raised in the cylindrical vessel be h m

∴ Volume of water in cylindrical vessel = π r2h

= 227

20100

20100× × × h cu m

= 22175

h cu m

∴ 22175 h = 0.66

⇒ h = 66100

17522× = 5.25 m.



(i) Curved surface area of a cone = ...............

(ii) Total surface area of a cone = ...............

(iii) Volume of a cone = ...............


(iv) Surface area of a sphere = ...............

(v) Volume of a sphere = ...............

(vi) Volume of a hemisphere = ...............

(vii) Total surface area of a solid hemisphere = ...............

2. Find the volume, the curved surface area and the total surface area of right circular conewith the following dimensions :

Radius (r) Height (h)

(i) 21 cm 28 cm

(ii) 14 cm 12 cm

(iii) 3.5 m 12 m

3. How many meters of cloth 3 m wide will be required to make a conical tent, the radiusof whose base is 5 m and whose height is 12 m ?

4. The area of the base of a right circular cone is 616 sq cm and its height is 9 cm. Findits volume.

5. A conical cup of height 15 cm has a base radius of 12.6 cm. The cup is full of water.The water is poured into a cylinder of base radius 7 cm. Find the height to which waterwill rise in the cylinder.

6. A conical tent is 6 m high with radius of base as 8 m

(a) Find the cost of cloth required to make the tent, if one square meter of cloth costsRs 30.

(b) How many persons can sit in the tent if each person requires 4 sq m of space on theground and 20 cu meter of air to breath in. [Use π = 3.14]

7. Find the slant height and curved surface area of cone whose volume is 12936 cu cm andthe diameter of whose base is 42 cm.

8. Find the volume and the surface area of a sphere of radius 2.1 cm.

9. The radius of an iron sphere is 3.5 cm. It is melted to form smaller spheres of diameter1.75 cm. Find the number of smaller spheres formed.

10. A cone, a hemisphere and a cylinder stand on equal bases and have same height. Showthat their volumes are inthe ratio 1 : 2 : 3.

11. Find the radius of the base of a right circular cone of height 21 cm and volume 550 cucm.

12. Find the diameter of a sphere of volume 4851 cu cm.

230 Mathematics

24.8 AREA OF FOUR WALLS OF A ROOM

Let Fig 24.15(i) represents the four walls of a room of length (l), breadth (b) and height (h).

Suppose a box open on both sides, is to be made of thin cardboard and cut along its heightAB. We can spread out into one continuous rectangle as shown in Fig 24.15(ii).

(i) (ii)

Fig. 24.15

Length of this rectangle = Perimeter of room

= 2(length + breadth) = 2(l + b)

Breadth of this rectangle = height of room = h

∴ Area of four walls of room = Perimeter × Height

= 2(l + b) × h.Let us take some examples and illustrate :

Example 24.23 : A room, 7 meters long, 4 meters broad and 3 meters high, has two windows

1 12 m × 1 m and two doors 2 m × 1 1

2 m. Find the cost of papering the walls with paper 50

cm wide at Rs 15 per meter.

Solution : Area of four walls 2(l + b) h= 2(7 + 4) 3 = 66 sq m

Area of two windows = 2 × 1 12 × 1 = 3 sq m

Area of two doors = 2 × 2 × 1 12 = 6 sq m

∴ Area of walls to be papered= 66 – (3 + 6) = 57 sq m

∴ Length of the paper =575. = 114 m

∴ Cost = Rs 114 × 15 = Rs 1710.


Example 24.24 : The length of a room is 6.5 m. The cost of painting the walls at Rs 28 persquare meter is Rs 2464, and the cost of carpeting the room at Rs 112 per square meter isRs 3276. Find the height and width of the room.

Solution : Length of the room = 6.5 m

Let breadth of the room be b m

and height of the room = h m

∴ Area of four walls = 2(6.5 + b).h sq m

Cost of painting the walls = 2(6.5 + b).h × 28

∴ 2(6.5 + b) × h × 28 = 2464

or (6.5 + b) × h = 44 ...(i)

Area of floor = l × b = 6.5 × b sq m

Cost of carpeting = Rs 6.5 × b × 12

∴ 6.5 × b × 112 = 3276

or, b = 4.5 ...(ii)

From (i), we have (6.5 + 4.5) h = 44

or h = 4 m

Hence, the height of room = 4 m and its width = 4.5 m.

Example 24.25 : The area of the floor of a room is 77 square meters. The area of the twolarger walls together is 88 square meters; and the two shorter walls together is 56 square meters.Find the dimensions of the room.

Solution : Let length of room be ‘l’, breadth ‘b’ and height ‘h’

Then l × b = 77 ...(i)

2(l × h) = 88

or l ×h = 44 ...(ii)

and 2 (b × h) = 56

or b × h = 28 ...(iii)

∴ lb = 44

28117=

or l = 117 b

Putting in (i) we get

117 b2 = 77

232 Mathematics

b2 = 77 711×

or b = 7 m

∴ l = 117 7× = 11 m and h = 44

11 = 4 m

∴ The length of the room is 11m, its breadth is 7 m and its height is 4 m.


1. Find the area of four walls of the room in each of the following cases :

(i) l = 8 m, b = 6 m, h = 3m.

(ii) l = 20 m, b = 12 m, h = 8 m

2. A room is 6 m long, 5 m wide and 4 m high. The doors and windows in the room occupya space of 5 square meters. Find the cost of papering the remaining portion of the wallswith paper 75 cm wide, at the rate of Rs 1.20 per meter.

3. Find the cost of painting the walls and the ceiling of a room measuring 10 m × 6m ×3 m atthe rate of Rs 1.50 per square meter.

4. A room measures 9 m × 7m × 3m. It has three doors, each having an area of 3 squaremeters and 4 windows each measuring 1.25 × 1 m. Find the cost of papering the remainingportion of the walls at the rate of Rs 1.50 per square meter.

5. The area of two side walls of a room is 5250 sq dm and the area of the two end walls4550 sq dm, and the area of the floor is 4875 sq dm. Find the dimensions of the room.

LET US SUM UP

The figures, which occupy space and have more than two dimensions, are called Solidfigures.

The sum of the areas of the plane figures making up the boundary of a solid object iscalled its surface area.

The amount of space occupied by the solid object is called its Volume.

Total surface area of a cuboid = 2(lb + bh + hl)

Total surface area of a cube = 6a2, where a is its side.

Volume of a cuboid = l × b × h.

Volume of a cube = a3.

A right circular cylinder is a solid generated by revolution of a rectangle about one ofits sides, which remains fixed.


Volume of a cylinder = π r2h.

Total surface area of an cylinder open at one end = 2π rh + π r2

Total surface area of a cylinder closed at both ends = 2π rh + 2 π r2 = 2π r (h + r)

A right circular cone is a solid generated by the revolution of a right triangle about oneof its sides containing the right angles as axis.

Curved surface area of a cone = π rl, where l2 = r2 + h2

Total surface area of a solid cone = π rl + π r2

Volume of a cone = 13

2πr h

A sphere is a solid generated by the revolution of a semicircle about its diameter.

Surface area of a sphere = 4π r2

Volume of a sphere = 43

3πr

Curved surface area of a hemisphere = 2π r2

Total surface area of a solid hemisphere = 3π r2

Volume of a hemi-sphere = 23

3πr

Area of four walls of a room = 2(l + b) × h.

TERMINAL EXERCISE

1. Find the edge of a cube of volume equal to the volume of a cuboid of dimensions63 cm × 56 cm × 21 cm.

2. Find the number of 5 cm cubes that can be cut out of a 15 cm cube.

3. Three cubes of metals whose edges are 3, 4 and 5 cm respectively are melted and formedinto a single cube. If there is no waste in the process, find the edge of the new cube soformed.

4. A school room is to be built to accommodate 70 children, so as to allow 2.2 sq m offloor area and 11 cu m of space for each child. If the room is to be 14 m long, what mustbe its breadth and height ?

5. How many bricks 20 cm × 10 cm × 7.5 cm be carried by a truck whose capacity tocarry load is 6 metric tons ? One cubic meter of bricks weighs 2000 kg. [1 metric ton= 1000 kg]

234 Mathematics

6. A field is 200 m long and 75 m broad; and a tank 40 meter long, 20 meter broad and10 meter deep is dug in the field, and the earth taken out of it is spread evenly over thefield. How much is the level of field raised ?

7. Four cubes each of sides 5 cm are joined end to end. Find the surface area of the resultingcuboid.

8. The sides of an open box are 0.5 cm thick and bottom is 1 cm thick. If the internal length,breadth and depth are respectively 14 cm, 10 cm ad 8 cm, find the quantity of materialused in the construction of the box.

9. Find the whole surface area of a hollow cylinder open at the ends, if its length is 10 cm,the internal diameter is 8 cm and the thickness is 1 cm [Use π = 3.14]

10. A cubic cm of gold is drawn into a wire 1/5 mm in diameter; find the length of wire.(π = 3.14)

11. A well with 8.4 meter inside diameter is dug 10 meter deep. Earth taken out of it hasbeen spread all around it to a width of 4 meters to form an embankment. Find the heightof the embankment.

12. Find the slant height of a cone whose volume is equal to 12936 cubic meters and thediameter of whose base is 42 meters.

13. The volume of a cone is 616 cubic meters. If the height of cone is 27 meters, find theradius of its base.

14. A conical vessel of internal radius 14 cm and height 36 cm is full of water. If this wateris poured into a cylinder with internal radius 21 cm, find the height to which the waterrises in the cylinder.

15. Find the diameter of a sphere whose volume is 606.375 cubic meter.

16. A room 12 meters long, 4 meters broad and 3 meters high has two windows 2 m × 1m and a door 2.5 m × 2 m. Find the cost of papering the walls with paper 50 cm wideat Rs 20 per meter.

17. A hall, whose length is 15 m and breadth is twice its height, takes 250 meters of paper2 meters wide for its four walls. Find the area of the floor.

18. The length of a room is 1.5 times its breadth. The cost of carpeting it at Rs 150 per squaremeter is Rs 14400 and the cost of white washing the four walls at Rs 5 per square meteris Rs 625. Find the dimensions of the room.


ANSWERS


1. (i) 2(lb + bh + hl) (ii) 2(bh + hl) + lb (iii) 6x2

(iv) lbh (v) x3 (vi) l b h2 2 2+ +2. (i) 45 cu cm, 81 sq cm (ii) 750 cu cm, 550 sq cm

(iii) 30 cu m, 101 sq m (iv) 11.25 cu m, 31.5 sq m3. (i) 150 sq cm, 125 cu cm (ii) 77.76 sq cm, 46.65 cu cm

(iii) 15.36 sq m, 40.96 cu m4. (i) 15 cm (ii) 1.3 cm (iii) 2.5 cm5. 43056 cm3 6. 3847. 7500 cu m, 2868 cu m 8. 117 sq m 9. 896 sq cm.


1. (i) π r2h (ii) 2π r(r + h) (iii) 2π rh + π r2

(iv) 2π rh2. (i) 1848 c cm, 836 sq cm

(ii) 1100 cu cm, 848.57 sq cm(iii) 110 cu cm, 201.14 sq cm

3. 1.65 cu m 4. 1386 cu cm 5. 880 sq m6. 48000 cu cm 7. 3.68 m 8. 63 : 16, 21 : 89. 1320 sq m 10. 433.54 sq cn,


1. (i) π rl (ii) π rl + π r2 (iii)13

2πr h

(iv) 4π r2 (v)43

3πr (v)23

3πr

(v) 3π r2

2. (i) 12936 cm3, 2310 cm2, 3696 cm2

(ii) 2464 cm3, 811.3 cm2, 1427.3 cm2,(iii) 154 m3, 134.5 m2, 176 m2

3. 68 m (approx) 4. 1848 cm3 5. 16.2 cm6. 50 7. 35 cm, 2310 cm2 8. 38.8 cm3, 55.44 cm2

9. 64 10. 12 cm.

236 Mathematics


1. 84 m2, 512 m2 2. Rs 132.80 3. Rs 234

4. Rs 123 5. 75 dm, 65 dm, 35 dm.

Terminal Exercise

1. 42 cm 2. 27 3. 6 cm4. 11 m, 5 m 5. 2000 6. 1.9 m7. 450 cm2 8. 365 cm3 9. 2574.8 cm2

10. 31.84 m 11. 3.55 m 12. 35 m

13.143 m 14.

163 m 15. 42 cm

16. Rs 3480 17. 300 sq m 18. 12m, 8 m, 3.125 m.

240 Mathematics

Module 5Trigonometry

Imagine a man standing near the base of a hill, looking at the temple on the top of the hill.Before venturing to start climbing the hill, he wants to have an approximation of the distancebetween him and the temple. We know that problems of this and related problems can be solvedonly with the help of a science called trigonometry.

The first introduction to this topic was done by Hipparcus in 140 B.C., when he hinted atthe possibility of finding distances and heights of inaccessible objects. In 150 A.D. Tolemyagain raised the same possibility and suggested the use of a right triangle for the same. Butit was Aryabhatta (476 A.D.) whose introduction to the name “Jaya” lead to the name “sine”of an acute angle of a right triangle. The subject was completed by Bhaskaracharya(1114 A.D.) while writing his work on Goladhayay. In that, he used the words Jaya, Kotijyaand “sparshjya” which are presently used for sine, cosine and tangent (of an angle). But it goesto the credit of Neelkanth Somstuvan (1500 A.D.) who developed the science and used termslike elevation, depression and gave examples of some problems on heights and distance.

In this chapter, we shall define an angle-positive or negative, in terms of rotation of a ray fromits initial position to its final position, define trigonometric ratios of an acute angle of a righttriangle, in terms of its sides develop some trigonometric identities, trigonometric ratios ofcomplementary angles and solve simple problems on height and distances, using at the mosttwo right triangles, using angles of 30°, 45° and 60°.

Introduction to Trigonometry 241

25

Introduction to Trigonometry

25.1 INTRODUCTION

The word ‘Trigonometry’ is derived from the word “Trigon” meaning “a triangle” and “metron”meaning “measurement”. Thus, trigonometry is that branch of Mathemetics which deals withthe measurement of sides and angles of a triangle. This branch of Mathematics has beeninstrumental in the development of architecture, surveying, navigation, astronomy etc. The greatastronomer Hipparchus is said to have developed this branch of Mathematics. The Indiantraditional texts like Surya Sidhant have developed and used the knowledge of trigonometryextensively.

25.2 OBJECTIVES

After studying this lesson, the learner will be able to

write the trigonometric ratios of an acute angle of a right triangle.

find the sides and angle of a right triangle when some of its sides and trigonometric ratiosare known.

write the relationships among trigonometic ratios.

establish trigonometric identities.

solve problems based on trigonometric ratios and identities.


Concept of an angle

Construction of right triangles

Drawing parallel and perpendicular lines

Types of angles–acute, obtuse and right

Types of triangles–acute, obtuse and right

Types of triangles–isosceles, equilateral.

242 Mathematics

25.4 TRIGONOMETRIC RATIOS OF AN ACUTE ANGLE

Let XOX' and YOY' be rectangular axes of co-ordinates.Let A be a point on OX. Let the ray OA start rotating inthe plane in an anti-clockwise direction from the initialposition OA about the point O till it reaches its finalposition OP after some interval of time. (See Fig. 25.1).Thus, an angle POB is formed with x-axis. Let∠POB = θ. (θ is a Greek letter, and we read it as “theta”).Draw PB ⊥ OX. Now clearly ∆PBO is right angledtriangled. Let PB = p, OB = b and OP = h. we define thefollowing trigonometric ratios for angle θ.

sine of θ = Side opposite to angle θHypotenuse

PBOP

ph= =

cosine of θ = Adjacent side to angle θHypotenuse

OBOP

bh= =

tangent of θ =Side opposite to angle Adjacent side to angle

θθ = =PB

OBpb

cotangent θ =Adjacent side to angle Side opposite to angle

θθ = =OB

PBbp

secant θ =Hypotenuse OP

OBhbAdjacent side to angle θ = =

cosecant θ = Hypotenuse OPPB

hpSide opposite to angle θ = =

The above trigonometric ratios are written below in an abbreviated form :

sine of θ is abbreviated as sin θ

cosine of θ is abbreviated as cos θ

tangent of θ is abbreviated as tan θ

cotangent of θ is abbreviated as cot θ

secant of θ is abbreviated as sec θ

and cosecant of θ is abbreviated as cosec θ

Notes 1.Throughout the study of trigonometry we shall be using only abbreviated form of thesetrigonometric ratios.

2. sin θ is an abbreviation for “sine of angle θ”and not the product of sin and θ.

Fig. 25.1


So far you have defined six trigonometic ratios. You must have observed that thesetrigonometric ratios are the ratios of the sides of a right angled triangle. Thus, these trigonometicratios depend only on the magnitude of angle θ and not on the size of the right triangle. Letus verify this by taking right triangles of different sizes as shown in Fig. 25.2.

In Fig. 25.2, let ∠POM = θ

∆PMO ~ ∆P'M'O ...(AAA)

∴PM

P M' ' =MOM O

POP O

k (say)' '

= =

Thus, PM = k P'M'

OP = k P'O

and MO = k M'O

Now in ∆P'M'O,

sin θ =P MOP

kP MkOP

PMOP

' ''

' ''

= =

i.e., the value of sin θ is the same for both the triangles PMO and P'M'O.

Thus the trigonometric ratios are independent of the length of the sides of the right triangle.

Example 25.1 : In Fig. 25.3, ∆ABC has a right angle at B. If AB = BC = 1 cm and

AC = 2 cm, find sin C, cos C and tan C.

Solution : In right angled ∆ABC,

sin C =opposite sidehypotenuse

= =ABAC

12

cos C =adjacent sidehypotenuse

= =BCAC

12

and tan C =opposite sideadjacent side

= = =ABBC

11

1

Example 25.2 : In Fig. 25.4 ∆ABC is a right angled triangle, right angled at B. IfAB = 5 cm, AC = 13 cm and BC = 12 cm, find cosec C, cot C and sec C.

Solution : ∆ABC is a right angled triangle

∴ cosec C =hypotenuse

opposite side = ACAB

=135

Fig. 25.2

Fig. 25.3

2 cm

Fig. 25.4

244 Mathematics

cot C =adjacent sideopposite side = =BC

AB125

and sec C =hypotenuse

adjacent side = =ACBC

1312

Example 25.3 : In Fig. 25.5, ABC is a right angled triangle. If AB = 21 cm,BC = 20 cm and CA = 29 cm and ∠A = θ, find sin θ, cot θ and sec θ.


sin θ =opposite sidehypotenuse

= BCAC = 20

29

cot θ =adjacent sideopposite side

= ABBC = 21

20

and sec θ =hypotenuse

adjacent side

= ACAB

= 2921

Example 25.4 : In right angled triangle ABC in Fig. 25.6, AB = 9 cm, BC = 40 cm andCA = 41 cm. Find the values of cos C, tan A, cot C and cosec A.


cos C =adjacent sidehypotenuse = 40

41

tan A =opposite sideadjacent side = 40

9

cot C =adjacent sideopposite side = 40

9

and cosec A =hypotenuse

opposite side = 4140

Fig. 25.6

Fig. 25.5



In Fig. 25.7, write all the six trigonometric ratios for the angle C :

(i) (ii)

(iii) (iv)

Fig. 25.7

25.5 GIVEN TWO SIDES OF A RIGHT TRIANGLE, TO FIND TRIGONOMETRICRATIOS

When any two sides of a right triangle are given, its third side can be found out by using thePythagoras theorem. We can find trigonometric ratios of the given angle as learnt in previoussection. Let us take some examples to illustrate :

Example 25.5 : In Fig. 25.8, ∆PQR is right angled at Q. If PQ = 5 cm and QR = 12 cm findthe value of sin R, cos R and tan R.

Solution : ∆PQR is a right angled triangle.∴ PR2 = PQ2 + QR2 ...[By Pythagoras theorem]

= (5)2 + (12)2

= 25 + 144= 169 = (13)2

∴ PR = 13 cmNow, for ∠R, opposite side is PQ = 5 cm, adjacent sideQR = 12 cm and the hypotenuse PR = 13 cm

∴ sin R = PQPR = 5

13

cos R = QRPR = 12

13

and tan R =PQQR = 5

12

Fig. 25.8

246 Mathematics

Example 25.6 : In Fig. 25.9, ∆ABC is a right triangle right angled at C. If BC = 3 cm,AC = 4 cm, find the values of cot A, sec A and cosec A.

Solution : ∆ABC is a right angled triangle

∴ AB2 = BC2 + AC2 ...[By Pythagoras theorem]

= (3)2 + (4)2

= 9 + 16

= 25 = (5)2

⇒ AB = 5 cm

∴ cot A = ACBC = 4

3

sec A =ABAC

= 54

and cosec A =ABBC

= 53

Example 25.7 : In triangle PQR, in Fig. 25.10, right angled at Q, PR = 25 cm, QR = 7 cmand ∠RPQ = θ, find tan θ, cosec θ and sec θ.

Solution : ∆PQR is a right angled triangle

∴ (PR)2 = (RQ)2 + (PQ)2

or (25)2 = (7)2 + (PQ)2

or 625 = 49 + (PQ)2

or (PQ)2 = 625 – 49 = 576

= (24)2

∴ PQ = 24 cm

∴ tan θ =RQPQ = 7

24

cosec θ = PRRQ = 25

7

and sec θ = PRPQ = 25

24


1. For ∆ABC right angled at B, if AC = 10 cm, BC = 8 cm and AB = 6 cm, find sin C,cos C and tan C.

Fig. 25.9

Fig. 25.10


2. ∆ABC is a right angled triangle, right angled at C. If BC = 24 cm and AC = 7 cm, findsin A, cosec A and cot A.

3. In ∆PQR, right angled at Q, PR = 10 2 cm and QR = 10 cm. Find sec P, cot P andcosec P.

4. In right angled ∆PQR, PQ = 3 cm, QR = 1 cm and PR = 2 cm. Findtan R, cosec R, sin P and sec P.

5. In right angled triangle ABC, AC = 25 cm, AB = 7 cm and ∠ACB = θ, findcot θ, sin θ, sec θ and tan θ.

25.6 GIVEN ONE TRIGONOMETRIC RATIO, TO FIND THE OTHERS

If one trigonometric ratio is given, the values of the other trigonometric ratios can be easilyfound out. Let us take some example to illustrate this :

Example 25.8 : If sin θ = 725 , find the value of cos θ and tan θ.

Solution : Draw a right angled triangle ABC in which ∠B = 90° and ∠C = θ, as shown inFig. 25.11.

We know that

sin θ =opposite sidehypotenuse = AB

AC

= 725

Let AB = 7 and AC = 25

By the Pythagoras theorem, we have

AC2 = AB2 + BC2

or (25)2 = (7)2 + BC2

or BC2 = 625 – 49

= 576 = (24)2

∴ BC = 24

Now in ∆ABC,

cos θ = BCAC = 24

25

and tan θ = ABBC = 7

24

Fig. 25.11

248 Mathematics

Example 25.9 : If cot θ = 125 , find the value of cot .sin

tan2 2θ θ

θ

Solution : In Fig. 25.12, ABC is a right angled triangle and ∠C = θ

We have

cot θ =adjacent sideopposite side = =BC

AB125

Let BC = 12 and AB = 5

In right angled triangle ABC,

AC2 = AB2 + BC2

= (5)2 + (12)2

= 25 + 144

= 169 = (13)2

∴ AC = 13

Now sin θ = ABAC = 5

13

cot θ = BCAB = 12

5

and tan θ = ABBC = 5

12

Hence, cot .sintan

2 2θ θθ =

14425

25169

512

.

= 144169

125

1728845× =

Example 25.10 : In ∆ABC, right angled at B, if tan C = 13 , fiind the values of

sin A cos C + cos A sin C.

Solution : In Fig. 25.13, ∆ABC is a right angled triangle and ∠B = 90°

We know that,

tan C =13

= ABBC

Fig. 25.12


∴ AB = 1 and BC = 3

Also, AC2 = AB2 + BC2 ...(By Pythagoras theorem)

= (1)2 + ( 3 )2

= 1 + 3 = 4 = (2)2

∴ AC = 2

∴ sin C = ABAC = 1

2

cos C = BCAB = 3

2

cos A = ABAC = 1

2

sin A = BCAC = 3

2

Hence, sin A cos C + cos A sin C = 32

32

12

12

. .+

= 34

14+

= 44 1=


1. If sin θ = 2029 , find the values of cos θ and tan θ.

2. If tan θ = 247 , find the values of sin θ and cos θ.

3. If cos A = 725 , determine the values of sin A and tan A.

4. If cos θ = mn , determine the values of cot θ and cosec θ.

5. If cos θ = 45 , evaluate the following expression :

cos .cotsecθ θ

θ1 2−

Fig. 25.13

3

250 Mathematics

6. If cosec θ = 23 , find the value of

sin2θ cos θ + tan2θ

7. If cot B = 54 , show that cosec2B = 1 + cot2B.

8. Triangle ABC is a right triangle with ∠C = 90°. If tan A = 32 find sin B and tan B.

9. If tan A = 13 and tan B = 3 , show that cos A cos B – sin A sin B = 0

10. If cot A = 125 , show that tan2A – sin2A = sin4A sec2A

25.7 RELATIONSHIP BETWEEN TRIGONOMETRIC RATIOS

We know that in a right triangle ABC, right angled at B, we have

sin θ = ABAC

cos θ = BCAC

and tan θ = ABBC

= ABAC

ACBC

.

= ABAC

BCAC÷

= sin cosθ θ÷

= sincos

θθ

Also, 1sinθ = 1

ABAC

ACAB ec= = cos θ

1cosθ = 1

BCAC

ACBC= = secθ

Fig. 25.14


and 1tanθ = 1

ABBC

BCAB= = cot θ

Thus, we have

tan θ = sincos

θθ

cosec θ = 1sinθ

sec θ = 1cosθ

and cot θ = 1tan

cossinθ

θθ=


Example 25.11 : If cos θ = 12 and sin θ = 3

2 , find the values of cosec θ, sec θ and tan θ.


cosec θ = 1 13

2

23sinθ = =

sec θ = 1 112

2cosθ = =

and tan θ =sincos

θθ =

3212

= 32

21 3× =

Example 25.12 : For a ∆ABC, right angled at C, if tan A = 1, find the value of cos B.

Solution : In Fig. 25.15, ∆ABC is a right triangle, right angled at C

We have

tan A = 1 = BCAC

Let AC = 1 = BC

∴ AB = 2

Now cos B =BCAB = 1

2

Hence cos B =12 Fig. 25.15

252 Mathematics


1. If sin θ = 12 and cos θ = 3

2 , calculate the value of cot θ and sec θ.

2. If sin θ = 32 and tan θ = 3 , find the value of cos2θ + sin θ cot θ.

3. In a right angled triangle ABC, right-angled at C. If cos A = 32 , then find the value

of sin A sin B + cos A cos B.

4. If cosec A = 2, find the value of sin A and tan A.

5. In a right angled triangle ABC right angled at B, tan A = 3 , find the value oftan2B sec2A – (tan2A + cot2B)

25.8 TRIGONOMETRIC IDENTITIES

In section 25.1, you have learnt to define an angle with the help of the rotation of a ray frominitial to the final position. You have also learnt to define all the trigonometric ratioscorresponding to that angle. Let us recall them here also.

Let XOX' and YOY' be the rectangular axes. Let A be any point on OX. Let the ray OA startrotating in the plane in an anti-clockwise direction about the point O till it reaches the finalposition OA' after some interval of time. Let ∠A'OA = θ. Take any point P on a ray OA'. DrawPM ⊥ OX.

In right angled ∆PMO,

sin θ =PMOP ,

cos θ =OMOP

Squaring and adding, we get

sin2θ + cos2θ =PMOP

OMOP

2

2

2

2+

=PM OM

OP

2 2

2+

=OPOP

2

2 ... QPM OM OP2 2 2+ =

= 1

Fig. 25.16


Hence sin2θθθθθ + cos2 θθθθθ = 1 ...(1)

Also, we know that

sec θ =OPOM and tan θ =

PMOM

Squaring and subtracting these equations, we get

sec2θ – tan2θ =OPOM

PMOM

2

2

2

2−

=OP PM

OM

2 2

2−

=OMOM

2

2 ... QOP PM OM2 2 2− =

= 1

Hence, sec2θθθθθ – tan2θθθθθ = 1 ...(2)

Again, we know that

cosec θ =OPPM and cot θ =

OMPM

Squaring and subtracting, we get

cosec2θ – cot2θ =OPPM

OMPM

2

2

2

2−

=OP OM

PM

2 2

2−

=PMPM

2

2 ... QOP PM OM2 2 2= +

= 1

Hence cosec2θθθθθ – cot2θθθθθ = 1 ...(3)

Note : You may also write identities 1, 2 and 3 in the following forms by making use ofelementary algebraic operations.

sin2θ + cos2θ = 1 can be written as

sin2θ = 1 – cos2θ

254 Mathematics

or cos2θ = 1 – sin2θ

and sec2θ – tan2θ = 1 can be written as

sec2θ = 1 2+ tan θ

or tan2θ = sec2θ – 1

Also, cosec2θ – cot2θ = 1 can be written as

cosec2θ = 1 + cot2θ

or cot2θ = cosec2θ –1

Let us make use of these identities to solve some examples :

Example 25.13 : Prove that

tan θ + cot θ =1

sin cosθ θ

Solution : L.H.S. = tan cotθ θ+

=sincos

cossin

θθ

θθ

+

=sin cos

sin .cos

2 2θ θθ θ+

=1

sin .cosθ θ ... Qsin cos2 2 1θ θ+ =

= R.H.S.


sincos

cossin

AA

AA1

1+

+ += 2 cosec A

Solution : L.H.S =sin

coscos

sinA

AA

A11

++ +

=sin cos

sin cos

2 211

A AA A+ +

+b gb g

=sin cos cos

sin cos

2 21 21

A A AA A

+ + ++b g

=sin cos cos

sin cos

2 2 1 21

A A AA A

+ + ++b g


=2 2

12 1

1+

+=

++

cossin cos

cossin cos

AA A

AA Ab gb gb g

= 2cosec A = R.H.S.Example 25.15 : Prove that

11

−+

sinsin

AA = (sec A – tan A)2

Solution : L.H.S. =11

−+

sinsin

AA

=11

11

−+

× −−

sinsin

sinsin

AA

AA

=11

2

2−−

sinsin

AA

b g

=1 2

2− sincos

AA

b g ... Q1 2 2− =sin cosA A

=1 2−FHG

IKJ

sincos

AA

=1 2

cossincosA

AA

−FHG

IKJ

= (sec A – tan A)2

= R.H.S.Example 25.16 : Prove that

11

−+

sinsin

θθ

=cos

sinθ

θ1+

Solution : L.H.S. =11

−+

sinsin

θθ

=11

11

−+

× ++

sinsin

sinsin

θθ

θθ

= 11

2−+

sinsin

θθ

256 Mathematics

= cossin

2

1θθ+

... Q1 2 2− =sin cosθ θ

=cos

sinθ

θ1+

= R.H.S.


cos4A – sin4A = cos2A – sin2A

= 1 – 2sin2A

Solution : L.H.S. = cos4A – sin4A

= (cos2A)2 – (sin2A)2

= (cos2A – sin2A) (cos2A + sin2A)

= cos2A – sin2A ... Qsin cos2 2 1A A+ =

= R.H.S.

Also, cos2A – sin2A = (1 – sin2A) – sin2A ... Qcos sin2 21A A= −

= 1 2 2− sin A= R.H.S.


tan sectan sec

θ θθ θ

+ −− +

11 =

1+ sincos

θθ

Solution : L.H.S. =tan sectan sec

θ θθ θ

+ −− +

11

=tan sec sec tan

tan secθ θ θ θ

θ θ+ − −

− +

2 2

1d i

... Q 1 2 2= −sec tanθ θd i

=tan sec sec tan sec tan

tan secθ θ θ θ θ θ

θ θ+ − + −

− +b g b g b g

1

=tan sec sec tan

tan secθ θ θ θ

θ θ+ − − −

− +b g b g1

1

=tan sec sec tan

sec tanθ θ θ θ

θ θ+ − − +

− +b g b g1

1


= tan θ + sec θ

= sincos cos

θθ θ+ 1

= 1+ sincos

θθ

= R.H.S.


Prove each of the following identities :

1. (cosec2θ – 1) sin2θ = cos2θ

2. sin4A + sin2A cos2A = sin2A

3. cos2θ (1 + tan2 θ) =1

4. (1 + tan2θ) sin2θ = tan2θ

5. sincos

sincos cosA

AA

A ecA1 1 2+ + − =

6. 11

1+− = +cos

coscos

sinAA

AA

7. sec tansec tan

cossin

A AA A

AA

−+ = +1

8. (sin A – cos A)2 + 2sinA cosA = 1

9. cos4θ + sin4θ – 2sin2θ cos2θ = (2cos2θ – 1)2

10. sin sincos cos

cos cossin sin

A BA B

A BA B

−+ + −

+ = 0

11. (cosec θ – sin θ) (sec θ – cos θ) (tan θ + cos θ) = 1

12. sin A(1 + tan A) + cosA(1 + cot A) = sec A + cosec A

13. 11

2−+ = −cos

cos cos cotAA ecA Ab g

14.tan

cotcot

tansec .cosA

AA

AA ecA

1 11

−+

−= +

15.cot coscot cos

cossin

A ec AA ec A

AA

+ −− + = +1

11

258 Mathematics

LET US SUM UP

In a right angled triangle, we define trigonometric ratios as under :

sin θ =Side opposite to angle

Hypotenuseθ = AB

AC

cos θ =Adjacent side to angle

Hypotenuseθ = BC

AC

tan θ =Side opposite to angle Adjacent side to angle

θθ = AB

BC

cot θ =Adjacent side to angle Side opposite to angle

θθ = BC

AB

sec θ =Hypotenuse AC

BCAdjacent side to angle θ =

cosec θ =Hypotenuse AC

ABSide opposite to angle θ =

The following relationships exist between the trigonometic ratios

(i) tan θ = sincos

θθ (ii) cot θ =

cossin

θθ

(iii) cot θ = 1

tanθ (iv) cosec θ = 1

sinθ

(v) sec θ = 1

cosθ

The trigonometric identities are

(i) sin2θ + cos2θ =1 (ii) sec2θ – tan2θ = 1

(iii) cosec2θ – cot2θ = 1

TERMINAL EXERCISE

1. If sin A = 45 , detemine the value of cos A and tan A.

2. If tan A = 2021 , determine the value of cosec A and sec A.

Fig. 25.17


3. If tan θ = 43 , find the value of sin θ + cos θ.

4. If sec θ = mn , find the value of sin θ and tan θ.

5. If cos = 35 , find the value of sin tan

tanθ θ

θ−1

2 2

6. If sec θ = 54 , find the value of tan

tanθ

θ1+

7. If tan A = 1 and tan B = 3 , find the value of cos A cos B – sinA sin B

8. Prove that

(sec θ + tan θ) (1 – sin θ) = cos θ

9. Prove that

cottan

cossec

θθ

θθ

−− =1

1ec

10. Prove that

11

1 2−+ = −FH IKsin

sinsin

cosθθ

θθ

11. Prove that

tan sintan sin

secsec

θ θθ θ

θθ

+− = +

−11

12. Prove that

tan cotcot tan tan cotA B

A B A B++ =

13.coscos

cossin

ec xec x

xx

+− = −

11 1

14. Prove that

sin3A – cos3A = (sin A– cos A) (1 + sin A cos A)

15. costan

sincot cos sinA

AA

A A A1 1− + − = +

260 Mathematics

ANSWERS


(i) sin C = 513 ; cos C = 12

13 ; tan C = 512

cosec C = 135 ; sec C = 13

12 and cot C = 125

(ii) sin C = 35 ; cos C = 4

5 ; tan C = 34

cosec C = 53 ; sec C = 5

4 and cot C = 43

(iii) sin C = 2425 ; cos C = 7

25 ; tan C = 247

cosec C = 2524 ; sec C = 25

7 and tan C = 724

(iv) sin C = 35 ; cos C = 4

5 ; tan C = 34

cosec C = 53 ; sec C = 5

4 and cot C = 43


1. sin C = 35 ; cos C = 4

5 and tan C = 34

2. sin A = 2425 ; cosec A = 25

24 and cot A = 724

3. sec P = 2 ; cot P = 1 and cosec P = 2

4. tan R = 3 ; cosec R = 23 ; sin P =

12 and sec P =

23

5. cot θ = 247 ; sin θ = 7

25 ; sec θ = 2524 and tan θ = 7

24


1. cos θ = 2129 and tan θ = 20

21 2. sin θ = 2425 and cos θ = 7

25


3. sin A = 2425 and tan A = 24

7 4. cot θ = m

n m2 2− and cosec θ = n

n m2 2−

5. −256135 6. 27

8

7. sin B = 213 , tan B = 2

3


1. cot θ = 3 and sec θ = 23 2. 3

4

3. 32 4. sin A =

12 and tan A =

13

5. − 143

Terminal Exercise

1. cos A = 35 , tan A = 4

3 2. cosec A = 2920 , sec A = 29

21

3.75 4. sin θ = m n

m2 2− and tan θ = m n

n2 2−

5. 3160 6. 3

7

7. 1 32 2−

262 Mathematics

26

Trigonometric Ratios of Some Special Angles

26.1 INTRODUCTION

In the previous lesson, we have defined trigonometric ratios for acute angles and developedsome relationship between them. In this lesson, we shall find the values of trigonometric ratiosof angels of 30°, 45° and 60° by using our knowledge of geometry. We will also use theknowledge of trigonometry to solve simple problems based on heights and distances taken fromday to day life.

26.2 OBJECTIVES


find geometrically the trigonometric ratios for the angles of 30°, 45° and 60°.

find trigonometric ratios of complementary angles.

solve daily life problems of heights and distances, using trigonometric ratios.


The student must know before starting the lesson that :

when a ray rotates in an anti-clockwise direction about the origin, the angle formed withx-axis is positive.

if the sum of two angles is 90°, then the angles are said to be complementary.

a triangle, with one angle of 90°, is called a right-angled triangle.

in a right-angled triangle ABC right-angled at B, AC2 = AB2 + BC2.

a triangle with three sides equal, is said to be an equilateral triangle.

recalls the conditions of congruency of two triangles.

in a right triangle ABC, right-angled at B

sin C =opposite sidehypotenuse cosec C =

hypotenuseopposite side

Trigonometric Ratios of some Special Angles 263

cos C =adjacent sidehypotenuse sec C =

hypotenuseadjacent side

tan C =opposite sideadjacent side cot C =

adjacent sideopposite side

26.4 TRIGONOMETRIC RATIOS FOR AN ANGLE OF 45°

Let a ray OA rotate in the anti-clockwise direction and make an angle of 45° with x-axis (theinitial position of the rotating line) at any interval of time as shown in Fig. 26.1.

Take any point P on OA. Draw PM ⊥ OX.

Now ∆PMO is a right-angled triangle

We know that

∠POM + ∠OPM + ∠PMO = 180°

45° + ∠OPM + 90° = 180°

⇒ ∠OPM = 180° – 135°

= 45°

In ∆PMO, ∠OPM = ∠POM = 45°

⇒ OM = PM

Let OM = a

Now in the right-angled triangle PMO

OP2 = OM2 + PM2 (By Pythagorus theorem)

= a2 + a2

⇒ OP2 = 2a2

⇒ OP = 2a

Now, sin 45° =PMOP 2

= =aa

12 ⇒ cosec 45°= 2

cos 45° =OMOP 2

= =aa

12 ⇒ sec 45°= 2

and tan 45° = PMOM = =a

a 1 ⇒ cot 45°= 1

Fig. 26.1

264 Mathematics

Table for Trigonometric ratios for an angle of 45°

sin 45° =12 ; cosec 45° = 2

cos 45° =12 ; sec 45° = 2

tan 45° = 1; cot 45° = 1

26.5 TRIGONOMETRIC RATIOS FOR AN ANGLE OF 30°

Let a ray OA rotate in the anticlockwise direction and make an angle of 30° with its initialposition OX.

Take any point P on OA

Draw PM ⊥ OX.

Produce PM to P′ such that PM = P′M. Join OP′.

Now, in ∆PMO and ∆P′MO

OM = OM (Common)

∠PMO = ∠P′MO (Each is equal to 90°)

PM = P′M (By construction)

∴ ∆PMO ≅ ∆P′MO (SAS congruence)

∴ ∠OPM = ∠OP′M = 60° (c.p.c.t.)

⇒ ∆OPP′ is an equilateral triangle.

∴ OP = OP′

Let PM = a

Also PP′ = PM + MP′

= a + a = 2a

⇒ OP = OP′ = PP′ = 2a (sides of an equilateral triangle)

Now, in right-angled ∆PMO

OP2 = PM2 + OM2 (Pythagoras theorem)

Fig. 26.2


or (2a)2 = (a)2 + OM2

⇒ OM2 = 4a2 – a2

= 3a2

⇒ OM = 3a

Hence, sin 30° =PMOP 2

= =aa

12 ⇒ cosec 30° = 2

cos 30° =OMOP 2

= =3 32

aa

⇒ sec 30° =23

and tan 30° =PMOM

= =aa3

13 ⇒ cot 30° = 3

Table for Trigonometric ratios for an angle of 30°

sin 30° = 12 ; cosec 30° = 2

cos 30° = 32 ; sec 30° = 2

3

tan 30° = 13

; cot 30° = 3

26.6 TRIGONOMETRIC RATIO OF AN ANGLE OF 60°

Let a ray OA rotate in an anti-clockwise direction andmake an angle of 60° with its initial position OX at anyinterval of time. Take any point P on it. Draw PM ⊥ OX.Now ∆PMO is a right-angled triangle.

Now produce OM to M′ such that

OM = MM′.

Join PM′, Let OM = a

Now, in ∆PMO and ∆PMM′

PM = PM (Common)

∠PMO = ∠PMM′ (Each is equal to 90°)

OM = MM′ (Construction)

∴ ∆PMO ≅ ∆PMM′ (SAS congruence)

Fig. 26.3

266 Mathematics

∴ ∠POM = ∠PM′M = 60°

⇒ ∆POM′ is an equilateral triangle

∴ OP = PM′ = OM′ = 2a

In right-angled ∆PMO

OP2 = PM2 + OM2 (By Pythagoras theorem)

(2a)2 = (PM)2 + (a)2

⇒ PM2 = 4a2 – a2

= 3a2

⇒ PM = 3aHence,

sin 60° = PMOP = =3

23

2a

a ⇒ cosec 60° =23

cos 60° = OMOP = =a

a212 ⇒ sec 60° = 2

and tan 60° = PMOM = 3 ⇒ cot 60° =

13

Table for Trigonometric Ratios for an angle of 60°

sin 60° = 32 ; cosec 60° = 2

3

cos 60° = 12 ; sec 60° = 2

tan 60° = 3 ; cot 60° = 13

26.7 TRIGONOMETRIC RATIOS FOR ANGLES OF 0° AND 90°

We have defined trigonometric ratios for acute angles 30°, 45° and 60°. For angles of 0° and90° we shall use the following results, as axioms and will not give any logical proof for them

Table for Trigonometric Ratios for Angles 0° and 90°

sin 0° = 0; cosec 0° = Not defined

cos 0° = 1; sec 0° = 1

tan 0° = 0; cot 0° = Not defined

sin 90° = 1; cosec 90° = 1

cos 90° = 0; sec 90° = Not defined

tan 90° = Not defined; cot 90° = 0


26.8 AN IMPORTANT TABLE : AN AID TO MEMORY

The values of trigonometric ratios for angles 0°, 30°, 45°, 60° and 90° are quite often usedin solving problems in our day-to-day life. Thus, the following table will enable us to rememberthe values of sin θ and cos θ more easily and we will be able to write the values of othertrigonometric ratios by using the existing relation between them.

θ 0° 30° 45° 60° 90°

Trig. Ratio

sin θ 04

14

24

34

44

cos θ 44

34

24

14

04

tan θ 04 0−

14 1−

24 2−

34 3− Not defined

cot θ Not defined 34 3−

24 2−

14 1−

04 0−

cosec θ Not defined 41

42

43

44

sec θ 44

43

42

41 Not defined

Let us now take some examples to illustrate the use of these trigonometric ratios.

Example 26.1: Find the value of

tan260° – sin230°


tan 60° = 3

and sin 30° =12

∴ tan260° – sin230° = 3 12

2 2d i − FH IK

= 3 14− = 11

4

268 Mathematics

Hence, tan260° – sin230° = 114 .

Example 26.2 : Find the value of

tan260° cosec245° + sec245° sin30°


tan 60° = 3 ; cosec 45° = 2

sec 45° = 2 ; sin 30° = 12

∴ tan260° cosec245° + sec245° sin 30°

= 3 2 2 12

2 2 2d i d i d i+

= 3 2 2 12

× + ×

= 6 + 1 = 7

Hence, tan260° cosec245° + sec245° sin 30° = 7

Example 26.3 : Verify that

tancos

seccot

sincos

4530

6045

2 900

°° +

°° −

°°ec =

12


tan 45° = 1, cosec 30° =2, sec 60° = 2, cot 45° = 1

sin 90° = 1 and cos 0° = 1

∴ L.H.S. = tancos

seccot

sincos

4530

6045

2 900

°° +

°° −

°°ec

= 12

21

2 11+ − ×

= 12 2 2+ − = 1

2 = R.H.S.


43 30 3 60 2 60 3

4 302 2 2 2cot sin cos tan°+ °− °− °ec = 103

Solution : We know that,

cot 30° = 3 , sin 60° = 32 ,


cosec 60° = 23

and tan 30° = 13

L.H.S. = 43 30 3 60 2 60 3

4 302 2 2 2cot sin cos tan°+ °− °− °ec

= 43 3 3 3

2 2 23

34

13

2 2 2 2× + FHG

IKJ − FHG

IKJ − FHG

IKJd i

= 43 3 3 3

4 2 43

34

13× + × − × − ×

= 4 94

83

14+ − − = 10

3 = R.H.S.


4 60 30 2 4560 45

2 2 2

2 2cot sec sin

sin cos°+ °− °

°+ ° =

43


cot 60° =13 , sec 30° =

23 , sin 45° =

12

sin 60° =3

2 and cos 45° =

12

L.H.S. =4 60 30 2 45

60 45

2 2 2

2 2cot sec sin

sin cos°+ °− °

°+ °

=4 1

323

2 12

32

12

2 2 2

2 2

× FHGIKJ + FHG

IKJ − × FHG

IKJ

FHGIKJ + FHG

IKJ

=4 1

343 2 1

234

12

× + − ×

+ = 4

3 = R.H.S.

Hence, L.H.S. = R.H.S.Example 26.6 : If θ = 30°, verify that

tan 2θ = 21 2

tantan

θθ−

Solution : We have θ = 30°L.H.S = tan 2θ

= tan 60°

= 3

270 Mathematics

R.H.S. = 21 2

tantan

θθ− =

2 301 302

tantan

°− °

=2 1

3

1 13

2

×

− FHGIKJ

=

23

1 13

−

=2 33 2×× =

33

= 3Hence, L.H.S. = R.H.S.

Example 26.7 : Taking A = 30°, verify that

sin 3A = 3 sinA – 4 sin3A

Solution : We know that A = 30°

L.H.S. = sin 3A

=sin 90°

=1

R.H.S. = 3 sin A – 4 sin3A

= 3 sin 30° – 4 sin3 30°

= 3 12 4 1

23

× − × FH IK

= 32

12− = 1

Hence, L.H.S. = R.H.S.



(i) sin260° + cos245°

(ii) 2 sin230° – 2 cos245° + tan260°

(iii) 4 sin260° + 3 tan230° – 8 sin 45°cos45°

(iv) 4 (sin430° + cos460°) – 3 (cos245° – 2sin245°)

(v)tan sec

cotsincos

4530

6045

5 902 0

°° +

°° −

°°cosec


2. Verify each of the following :

(i) cosec330° × cos 60° × tan345° × sin290° + sec245° × cot30° = 8 3

(ii) tan sin cos cot2 2 2 230 12 45 1

3 30 60°+ °+ °+ ° = 76

(iii) 5 30 45 4 302 30 30 45

2 2 2sin cos tansin cos tan

°+ °− °° °+ ° = 5

6 2 3−d i

3. If ∠A = 30°, verify that

(i) tan 2A = 21 2

tantan

AA−

(ii) cos 2A = cos2A – sin2A

(iii) cos 3A = 4 cos3A – 3 cos A

4. Taking 2A = 60°, find sin 30° and cos 30° using cos 2A = 2 cos2A – 1.

26.9 TRIGONOMETRIC RATIOS FOR COMPLEMENTARY ANGLES

Let XOX' and YOY' be a rectangular system of coordinates. Let A be any point on OX. Leta ray OA rotate in an anti-clockwise direction and trace an angle θ from its initial position(x-axis) at any interval of time. Let ∠POM = θ.

Draw PM ⊥ OX.

∆PMO is a right triangle

Also, ∠POM + ∠OPM + ∠PMO = 180°

∠POM + ∠OPM + 90° = 180°

∴ ∠POM + ∠OPM = 90°

⇒ ∠OPM = 90° – ∠θ

i.e. ∠OPM and ∠POM are complementary angles.

In right-angled ∆PMO, we know that

sin θ = PMOP , cos θ = OM

OP and tan θ = PMOM

cosec θ = OPPM , sec θ = OP

OM and cot θ = OMPM

For reference angle (90° – θ), we have in right-angled ∆OPM

sin (90° – θ) = OMOP = cos θ

Fig. 26.4

272 Mathematics

cos (90° – θ) = PMOP = sin θ

tan (90° – θ) = OMPM = cot θ

cot (90° – θ) = PMOM = tan θ

cosec (90° – θ) = OPOM = sec θ

and sec (90° – θ) = OPPM = cosec θ

Let us take some examples to illustrate the above

Example 26.8 : Prove that tan 11° = cot 79°.


tan (90° – θ) = cot θ

∴ R.H.S. = cot 79° = cot (90° – 11°)

= tan 11°

= L.H.S.

Example 26.9 : Evaluate sin240° – cos250°.

Solution. We know that,

cos (90° – θ) = sin θ

cos 50° = cos (90 – 40)°

= sin 40°

Hence, sin240° – cos250° = sin240° – sin240° = 0

Example 26.10 : Evaluate cossin

seccos

4347

3258

°° +

°°ec


cos (90° – θ) = sin θ

sin 47° = sin (90° – 43°) = cos 43°

Also cosec 58° = cosec (90° – 32°) = sec 32°

∴ cossin

seccos

4347

3258

°° +

°°ec = cos

cos4343

3232

°° +

°°

ssecec

= 1 + 1 = 2


Example 26.11 : Show that tan 7°.tan23°.tan60°.tan67°.tan83° = 3


tan (90° – θ) = cot θ

and tan 60° = 3

∴ tan 83° = tan (90° – 7°) = cot 7°

Also, tan 67° = tan (90° – 23°) = cot 23°

L.H.S. = tan 7°.tan23°.tan60°.tan67°.tan83°

= tan 7°.tan 23°.tan 60°.cot 23°.cot 7°

= 3 7 7 23 23tan .cot tan cot° ° ° °b gb g= 3 1 1× ×

= 3

= R.H.S.

Example 26.12 : Prove that cossin

sincos

θθ

θθ90 90°−

+°−b g b g = 2


sin (90° – θ) = cos θ

and cos (90° – θ) = sin θ

L.H.S. = cossin

sincos

θθ

θθ90 90°−

+°−b g b g

= coscos

sinsin

θθ

θθ+ = 1 + 1 = 2 = R.H.S.

Example 26.13 : Prove that sin

coscossec

9090

9090

°−°−

+°−°−

θθ

θθ

b gb g

b gb gec = 1


sin (90° – θ) = cos θ

cosec (90° – θ) = sec θ

cos (90° – θ) = sin θ

274 Mathematics

sec (90° – θ) = cosec θ

L.H.S. =sin

coscossec

9090

9090

°−°−

+°−°−

θθ

θθ

b gb g

b gb gec

=cossec

sincos

θθ

θθ

+ec

= cos2θ + sin2θ

= 1

= R.H.S.

Example 26.14 : Express tan 68° + sec 68° in terms of angles between 0° and 45°.


tan (90° – θ) = cotθ

Also, sec (90° – θ) = cosec θ

∴ tan 68° = tan (90° – 22°) = cot 22°

and sec 68° = sec (90° – 22°) = cosec 22°

Hence, tan 68° + sec 68° = cot 22° + cosec 22°


1. Show that

(i) cos 35° = sin 55°

(ii) sin211° – cos279° = 0

(iii) cos251° – sin239° = 0


(i) cossin

sincos

cossin

7515

1278

1872

°° +

°° −

°°

(ii) sincos

cossin cos47

434347 4 45

2 22°

°FH IK + °

°FH IK − °

(iii) cos cossin sin

cos2 22 2

220 7059 31

0°+ °°+ °

− °


3. Prove that

(i) sin θ cos(90° – θ) + cos θ sin (90° – θ) = 1

(ii) cos θ cos (90° – θ) – sin θ sin (90° – θ) = 0

(iii)cos

sinsin

cos90

1 901 90

90°−

+ °−+

+ °−°−

θθ

θθ

b gb g

b gb g = 2 cosec θ

(iv) sin (90° – θ) cos (90° – θ) = tantan

901 902

°−+ °−

θθ

b gb g

4. Express each of the following in terms of angles between 0° and 45°.

(i) cos 55° + sin 68°

(ii) cot 75° + cosec 75°

(iii) sec262° + sec269°

26.10 APPLICATIONS OF TRIGONOMETRY

We have so far learnt to define trigonometric ratios of an angle. Also, we have learnt todetermine the values of trigonometric ratios of the angles of 30°, 45° and 60°. In this lesson,we will learn how trigonometry can be used to determine the distance between the objects(particularly inaccessible ones) or the heights of the objects by taking some examples fromday-to-day life. We shall first define some terms which will be needed in the study of heightsand distances.

26.10.1 Angle of Elevation

Whenever an observer is looking at an object which is at a greaterheight than the observer, he has to lift his eyes to see the object;and an angle of elevation is formed between the line of sightjoining the observers eye to the object. and the horizontal line.

In Fig. 26.5 , ∠θ is the angle of elevation.

26.10.2 Angle of Depression

On the contrary, if the observer, at a height, is looking at an objectat a lesser height, the angle formed between the line of sight andthe line joining eye of the observer to the object is called an angleof depression. In Fig. 26.6, α is the angle of depression.

Example 26.15 : A ladder leaning against a window of a house makes an angle of 60° withthe ground. The length of the ladder is 8 m. Find the distance of the foot of the ladder fromthe well.

Solution : Let AC be a ladder leaning against the wall AB making an angle of 60° with thelevel ground BC.

Fig. 26.5

Fig. 26.6

276 Mathematics

Let BC = x m

Now, in right-angled triangle ABC

cos 60° =BCAC

= x8

⇒12 =

x8

⇒ x = 4

Hence, the foot of the ladder is 4 m away from the wall.

Example 26.16 : A balloon is connected to a meteorological ground station by a cable of length100 m inclined at 60° to the horizontal. Determine the height of the balloon from the ground.Assume that there is no slack in the cable.

Solution : Let A be the position of the balloon, attached to a string AC of length 100 m whichmakes an angle of 60° with the level ground BC.

Let AB = x mNow, in right-angled ∆ABC

sin 60° = ABAC = x

100

32 = x

100

⇒ x = 50 3× m= 86.6 m

Hence, the balloon is at a height of 86.6 m.

Example 26.17 : The upper part of a tree is broken by the action of wind. The top of the treemakes an angle of 30° with the horizontal ground. The distance between the base of a treeand the point where it touches the ground is 10 m. Find the height of the tree.

Solution : Let AB represent a tree. Let C be a point from where the tree wasbroken by the action of the wind in two parts upper part makes an angle of 30°at D with level ground such that BD = 10 m

Let BC = x m

Now, in right-angled ∆CBD

tan 30° = BCBD = x

10

⇒13 = x

10

Fig. 26.7

Fig. 26.8

Fig. 26.9


x = 103 3× ...(i)

We know that

The height of the tree = BC + CD

Now we shall find CD

In right-angled ∆CBD

sin 30° = BCDC DC= x

⇒ 12 = x

DC⇒ DC = 2x

= 2 103 3× [From (i)]

= 203 3

∴ Height of the tree = 103 3 m + 20

3 3 m

= 10 3 = 17.32 m.

Example 26.18 : A vertical flagstaff AB stands on a horizontal plane. At a point P, which is200 m away from its foot, the top of the flagstaff makes an angle APB = 30°, where A is thetop of the flagstaff. Find the length AB of the flagstaff.

Solution : In right-angled ∆ABP

tan 30° = ABPB = x

200

⇒13 = x

200

⇒ x =200

3

= 2003 3× = 200 3

32

3= ×173.2

= 346.43 = 115.5 m (app.)

Hence, the length AB of the flagstaff is 115.5 m. (app.)

Fig. 26.10

278 Mathematics

Example 26.19 : At a point on level ground, the angle θ made by the top of the tower withit is found to be such that tanθ = 5 12 . On walking 192 m towards the foot of the tower, thetangent of the angle becomes 3/4. Find the height of the tower.

Solution : Let AB be a tower and C, D be the two positions of the observer such thatCD = 192 m. Let AB = h m and BC = x m.

We are given that tan θ = 512 and tan C = 3

4 .

Now, in right-angled ∆ABC

tan C = hx

34 = h

x

⇒ x = 43 h ...(i)

Again, in right-angled ∆ABD

tan θ = hx192 +

512 = h

h192 43+

=3

576 4h

h+

⇒ 36 h = 2880 + 20 h

⇒ 16 h = 2880

⇒ h = 180

Hence, the height of the tower is 180 m.

Example 26.20 : Standing on the top of a tower 100 high, Swati observes two cars on theopposite side of the tower. If their angles of depression are 45° and 60°, find the distancebetween the two cars.

Solution : Let PQ be a tower which is 100 m high. Let A and B be the position of the twocars. Let the angle of depression of the car at A be 60° and for the car at B be 45° as shownin Fig. 26.12

Now, ∠RPA = ∠PAB = 60°

and ∠SPB = ∠PBQ = 45°

Fig. 26.11


In right-angled triangle PQB,

tan 45° =PQQB QB= 100

⇒ 1 = 100QB ⇒ QB = 100 m

Also, in right-angled ∆PQA

tan 60° =PQQA QA= 100

3 = 100QA

⇒ QA =100

3 = 1003 3 = 100

3 ×1.732 = 57.74

Hence, the distance between the two cars

= AQ + QB = (100 + 57.74) m = 157.74 m


1. A ladder leaning against a vertical wall makes an angle of 60° with the ground. The footof the ladder is at a distance of 3 m from the wall. Find the length of the ladder.

2. At a point 50 m away from the base of a tower, an observer measures the angle of elevationof the top of the tower to be 60°. Find the height of the tower.

3. The angle of elevation of the top of the tower is 30° from a point 150 m away from itsbase. Find the height of the tower.

4. The string of a kite is 100 m long. It makes an angle of 60° with the horizontal ground.Find the height of the kite, assuming that there is no slack in the string.

5. A kite is flying at a height of 100 m from the level ground. If the string of kite makesan angle of 60° with a point on the ground, find the length of the string, assuming thatthere is no slack in the string.

6. The angle of elevation of tower at a point is 45°. After going 40 m towards the foot ofthe tower, the angle of elevation becomes 60°. Find the height of the tower.

7. Two men are on either side of a cliff which is 80 m high. They observe the angles ofelevation of the top of the cliff to be 30° and 60° respectively. Find the distance betweenthe two men.

8. One of the equal sides of an isosceles triangle is 18 2 m. If the angle of its vertex is90°, find the length of the base.

Fig. 26.12

280 Mathematics

9. From the top of a building 60 m high, the angles of depression of the top and bottomof a tower are observed to be 45° and 60° respectively. Find the height of the tower andits distance from the building.

10. A ladder of length 4 m makes an angle of 30° with the level ground while leaning againsta window of a room. The foot of the ladder is kept fixed on the same point of the levelground. It is made to lean against a window of another room on its opposite side, makingan angle of 60° with the level ground. Find the distance between these two rooms.

LET US SUM UP

1. The following are the relation between trigonometrical ratios for the complementaryangles :

(i) sin (90° – θ) = cos θ

(ii) cos (90° – θ) = sin θ

(iii) tan (90° – θ) = cot θ

(iv) cosec (90° – θ) = sec θ

(v) sec (90° – θ) = cosec θ

(vi) cot (90° – θ) = tan θ

2. The following table illustrates the values of trigonometric ratios for the angle θ such that0 90°≤ ≤ °θ :

Table of Values of Trigonometric Ratios

θ 0° 30° 45° 60° 90°Trig.Ratio

sin θ 0 12

12

32 1

cos θ 1 32

12

12 0

tan θ 013 1 3 not defined

cot θ not defined 3 113 0

cosec θ not defined 2 223 1

sec θ 123 2 2 not defined


TERMINAL EXERCISE

1. Find the value of each of the following :

(i) 4cos260° + 4sin245° – sin230°

(ii) sin245° – tan245° + 3(sin290° + tan230°)

(iii)5 30 45 4 30

2 30 30 45

2 2 2

2 2sin cos tan

sin cos tan°+ °− °° °+ °

2. Prove that

(i) 2cot230° – 2cos260° – 34 sin245° – 4sec230° = –

524

(ii) 2 sin230° + 2 tan260° – 5 cos245° = 4

(iii) cos 60° cos 45° + sin 60° sin 45° = sin 45° cos 30° + cos 45° sin 30°

3. It θ = 30°, verify that

(i) tan 2θ = 2

1 2tantan

θθ−

(ii) sin 2θ = 2

1 2tantan

θθ+

(iii) cos 2θ = 2 cos2θ – 1

4. If ∠A = 60° and ∠B = 30°, verify that

(i) sin (A – B) ≠ sin A – sin B

(ii) sin (A + B) = sin A cos B + cos A sin B

(iii) cos (A + B) = cos A cos B – sin A sin B

5. Prove that

(i) sin 20° sin 70° – cos 20° cos 70° = 0

(ii) sin A sin (90° – A) – cos A cos (90° – A) = 0

(iii)sin cos

tan90 90°− °−A A

Ab g b g = sin2(90° – A)

(iv) tan 20°. tan 35°. tan45°. tan55°.tan70° = 1

(v)cos

sin90

1 90°−

+ °−A

Ab gb g +

1 9090

+ °−°−

sincos

AA

b gb g = 2 sec (90° – A)

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6. The length of a string without slack, between a kite and a point on the ground is 150m. If the string makes an angle θ with the horizontal plane such at sinθ = 4 15 , determinethe height of the kite.

7. A ladder leaning against a vertical wall makes an angle θ with the ground such thattanθ = 4 3. The foot of the ladder is 3 m away from the wall. Determine the length ofthe ladder.

8. Two pillars of equal height stand on either side of a roadway which is 150 m wide. Ata point on the roadway between the pillars, the elevation of the top of the pillars are 60°and 30°. Find the height of the pillars and the position of the point.

9. An observer standing 40 m from a building notices that the angles of elevation of thetop and bottom of a flagstaff, which is surmounted on the building, are 60° and 45°respectively. Find the height of the tower and the flag staff.

10. From the top of a hill, the angle of depression of two consecutive kilometer stones dueeast are found to be 30° and 60°. Find the height of the hill.


ANSWERS


1. (i) 54 (ii) 5

2 (iii) 0 (iv) 2 (v) 0

4. sin 30° = 12 ; cos 30° = 3

2


2. (i) 1 (ii) 0 (iii) 0

4. (i) sin 35° + cos 22°

(ii) tan 15° + sec 15°

(iii) cosec228° + cosec221°


1. 6 m 2. 86.6 m 3. 86.6 m 4. 86.6 m 5. 115.46 m

6. 94.64 m 7. 184.75 m 8. 36 m 9. 25.35 m 10. 5.46 m.

Terminal Exercise

1. (i) 114 (ii) 7

2 (iii) 40121

6. 40 m 7. 5 m 8. 64.95 m, 37.5 m

9. 40 m, 29.28 m

10. 433 m

Data and Their Representation 287

Module 6Statistics

Since ancient times, it has been the practice by the house holders, shopkeepers, individuals,etc. to keep records of their income, expenditure and other resources. In fact, this record keepingwas started by ancient kings to keep account of their warriors, armoury and other fightingmaterials, etc. To make readers acquainted with the methods of recording, condensing andtaking out relevant information from given records (or data), they would be exposed to thelesson on “Data and their Representation”.

Everyday through communication media like radio, television, newspapers, periodicals,magazines, etc., we come across data in the form of tables, charts, graphs, etc. on differentaspects of a number of variables. The information presented is eye-catching and important. Inorder to read and interpret these correctly, the learners would be introduced to lesson on“Graphical Representation of Data”.

Sometimes, we are required to describe the data arithmetically, like describing mean age ofa class, mean height of a group, median score of a class or modal shoe size of a group. Thelearner will be introduced to these in the lesson on “Measures of Central Tendency”. Theywould also learn about the characteristics and limitations of these measures.

In the last lesson of this module, the learner will get acquainted to the concept of probabilityas measure of uncertainty, through games of chance like tossing a coin, throwing a die, etc.

288 Mathematics

27

Data and Their Representation

27.1 INTRODUCTION

In this lesson, we shall first learn about statistics as science dealing with collection, recordingand condensing data to take out relevant information from them. We shall learn about differenttypes of data-primary and secondary, raw/ungrouped and grouped. We shall also learn aboutsome important concepts like classes, frequency, class mark, class interveral, frequency table,cumulative frequency table, etc.

27.2 OBJECTIVES


define statistics – both in singular and plural form

differentiate between primary and secondary data

differentiate between raw/ungrouped and grouped data and cite examples

define frequency and cumulative frequency of a class

condense raw data into frequency table

form cumulative frequency tables

define class-mark, class-interval, class limits and true class limits in case of grouped data.


knowledge of writing numbers in increasing/decreasing order

finding averages of given numbers

classification of numbers into different classes

making/using tools to collect data.

27.4 STATISTICS AND STATISTICAL DATA

Statistics is the science which started with collection of data about different aspects. It maybe data regarding number of students presented in a school, classwise literacy rate in different


states of the country, unemployment data, income-tax rates etc. With the passage of time besidescollection of data, their tabulations, interpretation and drawing of inferences came under thepurview of statistics.

The word statistics is used both in plural and singular sense. In its singular form, it refers tothe science of statistics, while in plural sense it means the numerical facts or data/observationscollected with some specific purpose in view.

27.5 PRIMARY AND SECONDARY DATA

Data are of two types – primary and secondary. In any study, if the investigator himself isresponsible for the collection of data according to the desired plan and objective, they are calledprimary data.

Sometimes, it is not possible for the investigator to collect data himself due to paucity of time,finances and other reasons. In that case, the investigator uses the data collected by otherinvestigators in some other context or published data from government records, journals, booksetc. Since the data collected from other sources, may have been collected with an objectivedifferent from what it is being used, they are called secondary data.


1. Fill in blanks with suitable word(s) so that the following sentences give the proper meaning:

(a) Statistics in plural sense refers to ... collected for some definite purpose.

(b) Statistics in singular sense refers to the ... .

(c) The data which are collected by the investigator himself are called ... data.

(d) The data taken from other sources and not collected by the investigator himself arecalled ... data.

27.6 RAW/UNGROUPED AND GROUPED DATA

Consider the marks obtained by 20 students of a class in a class test (out of 25):

15, 6, 8, 20, 14, 12, 12, 25, 20, 22,

10, 9, 25, 25, 14, 18, 19, 17, 16, 25 ...(i)

The data in this form are called raw/ungrouped data.

The data given in (i), do not give much information about the standard achievement of students.Let us try to arrange the data in ascending order. We will have

6, 8, 9, 10, 12, 12, 14, 14, 15, 16,

17, 18, 19, 20, 20, 22, 25, 25, 25, 25 ...(ii)

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Data arranged in the above form (ii) are called arrayed data. The arrayed data gives somewhatbetter perspective of the situation. They present the minimum and maximum of the data at aglance.

The difference between the maximum and minium observations of the data iscalled the range.

A point may be made about arrayed data. When the number of observations is very large, itis very difficult and time consuming to arrange data in arrayed form.

To further simplify and overcome this difficulty, we can arrange the data in tabular form asfollows :

Table–1

Marks Number of Students

6 18 19 110 112 214 215 116 117 118 119 120 222 125 4

Total 20

Table–1 shows the number of students getting a particular number of marks. We can say thatfour students have secured 25 marks each, two have secured 20 marks each, and so on. Wecan also say that seven students have got 20 or more marks. Similarly, eleven students havegot 16 or more marks, and so on. The above data is called a frequency distribution tablefor ungrouped data.

The quantity about which the observations (or data) are collected is called variable(or variate) and the number of times an observation is repeated is called thefrequency of that observation.


To bring out some more salient features of data, we further simplify the presentation of datain condensed form into classes (or groups). Let us form the groups as follows :

Table – 2

Classes 5–7 8–10 11–13 14–16 17–19 20–22 23–25

Frequencies 1 3 2 4 3 3 4

Thus, for the class 20–22, 20 is called the Lower Class Limit and 22 is called the Upper ClassLimit. Table 2 is called a frequency table for Grouped Data.

The data presented in classes (or groups) is called Grouped Data. From this type ofpresentation, we can draw better conclusions about the data than before. Some of theseconclusions are :

(i) The number of students getting marks between 23 and 25 is 4.

(ii) No student has got marks less than 5

(iii) Four students have got marks between 14 and 16 and so on.

27.7 NEED FOR CONTINUOUS CLASSES

From Table 2, we see that the classes are non-overlapping because there are no fractional markshere. This may happen with marks, but there may be variables like weight, height, etc. whichadmit fractional measurements like 2.5 kg, 1.65 m, etc. In such cases, we need to havecontinuous classes so that all observations can be entered. This can be done as follows for datain Table 2.

Classes : 4.5–7.5, 7.5–10.5, 10.5–13.5, 13.5–16.5, 16.5–19.5, 19.5–22.5, 22.5–25.5

i.e., the lower limit of each class is decreased by 0.5 and upper limit increased by 0.5, to havesame class-interval (i.e., length of the class). The changed limits are called True class Limits.Thus for the class 7.5-10.5,7.5 is the True Lower class limit and 10.5 is the True upper classlimit. This changes the frequency distribution table as follows :

Table – 3

Classes Frequency

4.5 – 7.5 1

7.5 – 10.5 3

10.5 – 13.5 2

13.5 – 16.5 4

16.5 – 19.5 3

19.5 – 22.5 3

22.5 – 25.5 4

TOTAL 20

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For making Table 3, the following two assumptions have been made:

(i) The frequency in a class is centred at its mid-point, called class-mark of that class

∴ Class mark of a class = Upper Class Limit + Lower Class Limit

2

orTrue Upper Class Limit + True Lower Class Limit

2

∴ The class marks of distribution in Table 3 are

6, 9, 12, ..., 24

(ii) In each class, the upper class boundary is not included. Thus, in the class 7.5–10.5, 10.5is not included and in the class 22.5–25.5, 25.5 is not included. The observations 10.5and 25.5 are taken in respective next higher classes.


1. Differentiate between raw data and arrayed data

2. Find the range of the following data :

6, 8, 9, 12, 25, 3, 14, 18, 22, 27

3. Give one example each for frequency distribution table for ungrouped data and groupeddata.

4. Define :

(a) class limits

(b) true class limits

(c) class interval

(d) class-mark

5. What assumptions do you make while making a frequency table for grouped data ?

6. Write the lower and upper class limits for each of the following classes :

(a) 11–15, 16–20, 41–45

(b) 10.5–15.5, 15.5–20.5, 40.5–45.5

7. Explain one of the reasons for condensing raw data to grouped data.

27.8 CONDENSING RAW DATA INTO FREQUENCY TABLE

For condensing raw data into grouped data we follow the following steps :

Step 1 : Arrange the data in ascending order and find the range of raw data.


Step 2 : We decide upon number of classes to be formed. For that we remember thefollowing :

(a) There should be classes to accommodate the minimum and maximum of data

(b) The classes should not be open ended

(c) There is no definite rule for number of classes. The golden rule mostly used is

“Not fewer than 5 and not more than 10 classes”

For our example, we took 7 classes, the class interval being 3. This is obtainedby dividing the range by class-interval and increase that to next integral value.

Step 3 : Take each observation from raw data, one at a time and put a tally mark (|) againstthe class to which the observation belongs, till all observations are over. (The tallymarks are taken in bunches of five, the fifth one crossing the other four diagonally.For example, 6 is denoted by tally marks as |||| |)

Step 4 : Count the tally marks in each class to get the frequency of that class.

Step 5 : The resulting table is called Frequency Table.

Let us take an example to illustrate.

Example 27.1 : The marks obtained by 20 students in a class test are given below:

10, 15, 9, 16, 25, 30, 14, 8, 29, 15, 4, 7, 12, 13, 14, 18, 19, 16, 29, 7

Construct a frequency distribution tables with a class size of 4.

Solution : The data in ascending order is :

4, 7, 7, 8, 9, 10, 12, 13, 14, 14, 15, 15, 16, 16, 18, 19, 25, 29, 29, 30

Here the range of data is 26. Taking the class size as 4, we get the number classes as 7. (Next

integral value of 264 )

Let the classes be

4–8, 8–12, 12–16, 16–20, 20–24, 24–28, 28–32

Frequency Table for Marks obtained by Students

Table 4

Classes Tally Marks Frequency

4 – 8 ||| 38 – 12 ||| 3

12 – 16 |||| | 616 – 20 |||| 420 – 24 — —24 – 28 | 128 – 32 ||| 3

Total 20

294 Mathematics

Example 27.2 : The weights of apples (in grams) picked up at random from a basket are givenbelow :

110, 100, 105, 110, 190, 200, 210, 218, 250, 170

75, 80, 95, 100, 160, 215, 225, 270, 210, 270

Construct a frequency table from the above data, with one of the classes as 75–100 (100 notincluded).

Solution : The range of data s 200 grams. Taking one of the classes as 75–100, the classesare

75–100, 100–125, 125–150, 150–175, 175–200, 200–225, 225–250, 250–275

The frequency table is given below :

Table – 5

Frequency table for weights of Apples

Class Tally Marks Frequency

75–100 ||| 3100–125 |||| 5125–150 — —150–175 || 2175–200 | 1200–225 |||| 5225–250 | 1250–275 ||| 3

Total 20

27.9 CUMULATIVE FREQUENCY OF A CLASS

The total of frequencies of a particular class and of all classes prior to that class is called thecumulative frequency of that class. A table which shows cumulative frequencies of differentclasses is called a Cumulative Frequency Table. Thus for Table 4, the corresponding CumulativeFrequency Table is presented in Table 6

Table – 6

Class Frequency Cumulative Frequency

4–8 3 38–12 3 612–16 6 1216–20 4 1620–24 0 1624–28 1 1728–32 3 20Total 20


Similarly, for Table 5, the Cumulative Frequency Table is given in Table 7.

Table – 7

Class Frequency Cummulative Frequency

75–100 3 3100–125 5 8125–150 0 8150–175 2 10175–200 1 11200–225 5 16225–250 1 17250–275 3 20

Total 20

Let us take another example :

Example 27.3 : The class marks of a distribution are 34, 38, 42, 46, 50, 54 and 58, determinethe class size, classes and construct frequency and cummulative frequency table. If thefrequencies from first class onward are 2, 5, 7, 9, 8, 5, 4.

Solution : The class size is the difference between the class marks of two adjacent classes.Therefore,

Class size = 38–34 = 4

∴ The classes are

[(34 – 2) – (34 + 2)], [(38 – 2) – (38 + 2)], ... ... [(58 – 2) – (58 + 2)] i.e., 32–36,36–40, 40–44, 44–48, 48–52, 52–56, 56–60

∴ The frequency table and cummulative frequency table is given below :

Table–8

Classes Frequency Cummulative Frequency

32–36 2 236–40 5 740–44 7 1444–48 9 2348–52 8 3152–56 5 3656–60 4 40

Total 40

296 Mathematics


1. Enumerate the steps needed to condense raw data to grouped data.

2. Form a frequency table and cumulative frequency table for the data on runs scored bya batsman in 20 innings, one class being 0–20

12, 2, 15, 25, 75, 35, 45, 16, 28, 70

102, 79, 52, 71, 15, 5, 69, 72, 15, 9

3. The class marks and their corresponding frequencies are given below :

Class mark : 23 28 33 38 43 48 53 58

Frequency : 1 2 5 8 14 6 3 1

Form a cumulative frequency table from the above data.

27.11 LET US SUM UP

Statistics is that branch of Mathematics which deals with collection, classification andinterpretation of data. Statistics has meaning both in singular and plural sense.

The data collected from the respondents “as it is”, is called raw data.

The data collected by experimenter himself through his own designed tools is calledprimary data.

The data taken from other sources and not collected by the experimenter is calledsecondary data.

The data arranged in ascending/descending order is called “arrayed data”

When raw data are arranged with frequencies, they are said to form a frequency table forungrouped data.

When the data are divided into groups/ classes, they are called grouped data.

The classes have to be decided according to the range of data and size of class.

In a class 10–15, 10 is called the lower limit and 15 is called the upper limit of that class.

The number of observations lying in a particular class is called its frequency and the tableshowing classes with frequencies is called a frequency table.

The mid-points of classes are called class-marks.

Sometimes the classes have to be changed to make them continuous. In that case, the classlimits are called true class limits.

The total of frequencies of a particular class and of all classes prior to that is called thecumulative frequency of that class and the table showing cumulative frequencies is calleda cumulative frequency table.


TERMINAL EXERCISES

1. Define statistics as a term used in singular and a term used in plural sense. Give examplesto clarify the difference.

2. What are primary data ? Why are they more reliable than secondary data ?

3. Fill in the blanks by appropriate words/phrases to make each of the following statementstrue :

(i) The data which are taken from government records, magazines, etc. is called ... data.

(ii) The data, which are not collected by investigatory himself should be used ... .

(iii) When the raw data are arranged in ascending/descending order, they are called ... data.

(iv) When the data are condensed in classes of equal size with frequencies, they are called... data and the table is called ... table.

(v) The mid-points of a class are called ... .

(vi) When the class limits are adjusted to make them continuous, the class limits arerenamed as ... .

(vii) The number of observations falling in a particular class is called its ... .

(viii) The difference between the upper limit and lower limit is called the ... .

(ix) The sum of frequencies of a class and all classes prior to that class is called ...frequency of that class.

(x) Class size = Difference between successive .................. .

4. Enumerate different steps needed to convert a given raw data to grouped data and to forma frequency table.

5. Find the range of the following raw data and put it as arrayed data :

7, 13, 5, 3, 4, 12, 13, 4, 3, 4, 18, 19, 12, 4, 13, 8, 4, 9, 8, 24

Also change the above data into a frequency distribution of class size 3, one of the classesbeing 6–9 (9 not included)

6. The marks obtained by 30 students in a class test are given below :

10, 18, 42, 21, 17, 19, 28, 43, 13, 17

25, 23, 47, 18, 17, 21, 19, 25, 27, 19

15, 16, 48, 46, 31, 34, 18, 19, 12, 4

(i) Form a frequency table, with one of the classes being 14–21 (21 not included)

(ii) Form a cummulative frequency table for table formed in (i)

(iii) Can you interpret the data and give some salient observations on it ?

298 Mathematics

7. The heights of 20 students of a class (in cm) are given below :

140, 137, 146, 152, 160, 157, 153, 140, 142, 145

150, 152, 153, 155, 160, 160, 162, 145, 148, 150

(i) From a frequency distribution table for the above data using equal classes, one of thembeing 136–140 (140 not included)

(ii) Form a cumulative frequency table for the above data.

8. For the following frequency table, answer the following questions :

Classes Frequency

15–20 220–25 325–30 530–35 735–40 440–45 345–50 1Total 25

(i) Write the lower limit of the first class.

(ii) Write the class limits of the third class.

(iii) Find the class mark of the fourth class.

(iv) Determine the class size.

(v) Form a cumulative frequency table.

9. The class-marks, in order, of a distribution and the corresponding frequencies are givenbelow :

Class marks : 5 15 25 35 45 55 65 75

Frequencies : 2 6 10 15 12 8 5 2

(i) find the frequency table

(ii) find the cummulative frequency table.


ANSWERS


1. (a) numerical facts or observations (b) science of statistics

(c) primary (d) secondary


2. 24 6. (a) Lower class limits 11, 16, 41; (b) lower limit 10.5, 15.5, 40.5

Upper class limits : 15, 20, 45 Upper limt 15.5, 20.5, 45.5


2. Classes Frequency Cumulative Frequency

0–20 8 820–40 3 1140–60 2 1360–80 6 1980–100 0 19100–120 1 20

Total 20

Terminal Exercise

3. (i) secondary (ii) carefully (iii) arrayed(iv) grouped; grouped frequency table (v) class marks (vi) true class limits

(vii) frequency (viii) class limit interval or class size(ix) cummulative frequency (x) class marks.

5. (i) xi : 3 4 5 7 8 9 12 13 18 19 24 Total

fi : 2 5 1 1 2 1 2 3 1 1 1 20

(ii) Classes : 3–6 6–9 9–12 12–15 15–18 18–21 21–24 24–27 Total

Frequency : 8 3 1 5 0 2 0 1 20

6. (i) Classes Frequency Cummulative frequency0–7 1 17–14 3 414–21 12 1621–28 6 2228–35 3 2535–42 0 2542–49 5 30Total 30

300 Mathematics

7. Classes Frequency CummulativeFrequency

135–140 1 1

140–145 3 4

145–150 4 8

150–155 6 14

155–160 2 16

160–165 4 20

Total 20

8. (i) 15 (ii) 25, 30 (iii) 32.5 (iv) 5

(v) Classes Frequency Cumulative Frequency

15–20 2 2

20–25 3 5

25–30 5 10

30–35 7 17

35–40 4 21

40–45 3 24

45–50 1 25

Total 20

9. Classes Frequency Cumulative Frequency

0–10 2 210–20 6 820–30 10 1830–40 15 3340–50 12 4550–60 8 5360–70 5 5870–80 2 60

Total 60

Graphical Representation of Data 301

28

Graphical Representation of Data

28.1 INTRODUCTION

Whenever verbal problems involving a certain situation is presented visually before thelearners, it makes easier for the learner to understand the problem and attempt its solution.Similarly, when the data are presented pictorially (or graphically) before the learners, it makesthe presentation eye-catching and more intelligible. The learners can easily see the salientfeatures of the data and interpret them.

There are many forms of representing data graphically. They are

(i) Bar graphs

(ii) Histograms

(iii) Frequency polygons

(iv) Ogive

(v) Pictographs

(vi) Pie charts

In this lesson, we shall learn to read and draw Bar graphs, Histograms and Frequency polygons,and graphs related to day-to-day use, like temperature-time graph, velocity-time graph,pressure-volume graph, etc. Other form of graphs are beyond the scope of the present lesson.

28.2 OBJECTIVES


draw bar charts for given data

draw a histogram and frequency polygon for given data

read and interpret given bar charts and histograms

read the relevant information from graphs relating to day-to-day activities, like

(i) temperature-time graph

302 Mathematics

(ii) velocity-time graph

(iii) pressure volume graph, etc.

draw graph relating to day-to-day activities, like the ones above.


Knowledge of drawing and marking axes

Knowledge of drawing rectangles and plotting points

Practice of reading graphs.

28.4 BAR GRAPHS

A bar graph is a graphical representation of frequency distributions of ungrouped data. It isa pictorial representation of the numerical data by a number of bars (rectangles) of uniformwidth erected vertically (or horizontally) with equal spacing between them.

28.4.1 Construction of Bar Graphs

For the construction of bar graphs, we go through the following steps :

Step 1 : We take a graph paper and draw two lines perpendicular to each other and call themhorizontal and vertical axes.

Step 2 : Along the horizontal axis, we take the values of the variables and along the verticalaxis, we take the frequencies.

Step 3 : Along the horizontal axis, we choose the uniform (equal) width of bars and theuniform gap between the bars, according to the space available.

Step 4 : Choose a suitable scale to determine the heights of the bars. The scale is chosenaccording to the space available.

Step 5 : Calculate the heights of the bars, according to the scale chosen and draw the bars.

Step 6 : Mark the axes with proper labelling


Example 28.1 : The number of trees planted by an agency in different years is givenbelow :

Years 1997 1998 1999 2000 2001 2002 Total

Number of 400 450 700 750 900 1500 4700trees planted

Solution : The bar graph is given below in Fig. 28.1 :


Fig. 28.1

Step 1 : We draw two perpendicular lines OX and OY.

Step 2 : On OX, we represent years, from 1997–2002 and on OY we represent the numberof trees planted.

Step 3 : On OY, we start with 400 and marks points at equal intervals of 200.

Step 4 : The height of the bars are calculated according to the number of trees.

A kink (~) has been shown on the vertical axis showing that the marking on the vertical axisstarts from zero but has been shown to start from 400 as the data needs.

Examples 28.2 : The data below shows the number of students present in different classeson a particular day :

Classes VI VII VIII IX X

Number of 35 40 30 40 50students present

Represent the above data by a bar graph.

Solution : The bar graph for the above data is shown in Fig. 28.2.

304 Mathematics

Fig. 28.2

Example 28.3 : The data regarding causes of accidents in factories are given below:

Causes Percentage of Occurrence

Faulty Machinery 30%Electrical Disturbance 20%Delay in repairs 35%Mechanical Failure 10%Others 5%

Draw a bar graph to represent the above data.

Solution : The bar graph representing the above data is shown in Fig. 28.3 below:

Fig. 28.3


28.4.2 Interpretation of Bar graphs

After drawing a bar graph, we can draw some conclusions, which is called interpreting bargraphs.

Let us take some examples and do the same.

Example 28.4 : Read the bar graphs given in Fig. 28.1, and answer the followingquestions :

(i) In which year the maximum number of trees were planted ?

(ii) What trend the number of trees planted show ?

(iii) In which years the number of trees planted differ by 50 only ?

Solution : After reading the bar graph, the answers to the above questions are as follows :

(i) The maximum number (1500) of trees were planted in the year 2002, as in that year theheight of the bar is maximum.

(ii) The number of trees planted kept on increasing year after year

(iii) (a) The years 1997 and 1998

(b) The years 1999 and 2000

Examples 28.5 : Read the bar graph given in Fig. 28.3 and answer the followingquestions :

(i) Which cause is responsible for maximum accidents in factories ? Which is forminimum ?

(ii) Can you think of one of the “other” causes ?

(iii) How many percent of accidents could have been avoided by timely action?

Solution : (i) Delay in repairs is responsible for maximum (35%) of accidents. “Other causes”are responsible for minimum number of accidents

(ii) Carelessness of workers

(iii) (35 + 20)% or 55% accidents could have been avoided by taking steps for timely repairsand provision of equipment which can control electrical disturbances.


1. Enlist the possible forms of representing a data graphically.

2. What are the steps needed to represent a data by a Bar Graph ?

306 Mathematics

3. For the data on expenditure of a company over different heads, draw a bar graph :

Head Percentage of Expenditure

Salary of Employees 45%

Travelling Allowance 15%

Rent of Premises 20%

Machinery and materials 10%

Other expenditure 10%

4. Given below are data on causes of strikes in mills :

Causes Percentage

(i) Non fulfillment of economic demands 45

(ii) Overwork 20

(iii) Rivalry in unions 20

(iv) Non-congenial working conditions 10

(v) Others 5

Draw a bar graph depicting the above data.

5. From the bar graph given below, answer the following questions :

Fig. 28.4

(i) The names of two steel plants which produced maximum steel in the country duringthe time period.

(ii) What percentage of steel was produced in “other” plants ?


(iii) The steel plant at Durgapur produced how much less steel than at Bokaro?

(iv) What percentage of total steel under discussion was produced at Bhalai, Durgapurand Bokaro steel plants ?

28.5 HISTOGRAMS AND FREQUENCY POLYGONS

A histogram is a graphical representation of a continuous frequency distribution i.e. groupedfrequency distributions. It is a graph, including vertical rectangles, with no space between therectangles. The class-intervals are taken along the horizontal axis and the respective classfrequencies on the vertical axis using suitable scales on each axis. For each class, a rectangleis drawn with base as width of the class and height as the class frequency. The area of therectangles must be proportional to the frequencies of the respective classes.

A frequency polygon is the join of the mid-points of the tops of the adjoining rectangles. Themid-points of the first and the last classes are joined to the mid-points of the classes precedingand succeding respectively at zero frequency to complete the polygon.

Let us illustrate these with the help of examples.

Examples 28.6 : The following is the frequency distribution of weights of 30 students of classIX of a school. Draw a histogram to represent the data.

Classes : 45–50 50–55 55–60 60–65 65–70 Total

Frequency : 3 7 12 5 3 30

Solution : For drawing a histogram we go through the steps similar to those of a bar graph.They are given below :

Step 1 : On a paper, we draw two perpendicular lines and call them horizontal and verticalaxes.

Step 2 : Along the horizontal axis, we take classes of equal width :

45–50, 50–55, ...... As the axis starts from 45–50, we take one interval 40–45 beforeit and put a kink on axis before that

Step 3 : Choose a suitable scale on the vertical axis to represent the frequency. It can startfrom 0 to 12, with a step of 2, i.e., 0, 2, 4, 6, ...., 12, 14

Step 4 : Draw the rectangles as shown in Fig. 28.5.

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Fig. 28.5 shows the histogram required.

Note : A frequency polygon has been shown in dotted lines, as explained in the steps shownabove.

Example 28.7 : The daily earnings of 100 shopkeepers are given below :

Daily earnings 200-300 300-400 400-500 500-600 600-700 700-800 800-900(in Rs)

No. of shops 3 12 15 30 25 12 3

Draw a histogram and a frequency polygon to represent the above data.

Solution : Following the steps suggested in Example 28.6, the histogram and frequency polygonrepresenting the above data are given below in Fig. 28.6

Fig. 28.6


Example 28.8 : Draw a frequency polygon for the following data :

Pocket 0–50 50–100 100–150 150–200 200–250 250–300allowance(in rupees)

Number of 16 25 13 26 15 5students

Solution : To draw a frequency polygon without-drawing a histogram we go through thefollowing steps :

Step 1 : Draw two lines perpendicular to each other.

Step 2 : Find the class-marks of different classes. They are 25, 75, 125, 175, 225, 275

Step 3 : Plot the ordered pairs A (25, 16), B (75, 25), C(125, 13), D (175, 26), E(225, 15)and F(275, 5)

Step 4 : Join the points A, B, C, D, E and F and complete the polygon as explained before

The frequency polygon is given below :

Fig. 28.7

28.5.1 Reading a Histogram

Let us explain it with the help of an example

Example 28.9. The following histogram shows the monthly wages (in rupees) of workers ina factory

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Fig. 28.8

(i) Find the maximum number of workers getting a wage.

(ii) Find the least wage and highest wage with no. of workers earning them

(iii) How many workers get a monthly wage of Rs. 8000 or less ?

Solution :

(i) The maximum number of workers is 25 getting wages between Rs (7000 – 8000).

(ii) The least wage is between Rs (4000 – 5000) and 4 workers are getting that. Thecorresponding figures for highest wage are Rs (9000 – 10000) and four workers get that

(iii) 50 workers get a wage of Rs 8000 or less as

Rs (4000 – 5000) – 4 workers

Rs (5000 – 6000) – 10 workers

Rs (6000 – 7000) – 12 workers

Rs (7000 – 8000) – 24 workers

Total – 50


1. What is the difference between a bar graph and a histogram ?

2. Write various steps in the construction of a Histogram.


3. Draw a histogram for the following frequency distribution :

Height of students 135–140 140–145 145–150 150–155 155–160 160–165(in cm)

No. of students 3 5 12 7 5 3

Also draw a frequency polygon for the above data on the same sheet

4. Draw a frequency polygon for the data in Question 3 on a separate paper.

5. Draw a histogram and a frequency polygon for the following grouped data:

Annual income 4-6 6-8 8-10 10-12 12-14 14-16 16-18 18-20(in ten thousand rupees)

No. of families 25 20 15 15 13 7 3 2in a locality

6. Interpret the data represented by the following histogram by answering the followingquestions :

Fig. 28.9

Shirt sale in a week in a shop.

(i) The least number of shirts were sold in which class ?

(ii) The maximum number of shirts were sold in which class ?

(iii) How many shirts were sold upto the 42 shirt size ?

(iv) How many shirts of size 44–66 were sold ?

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28.6 GRAPHS RELATED TO DAY-TO-DAY ACTIVITIES

In addition to histograms and frequency polygons, we are sometimes faced with graphs of othertypes. When a patient is admitted in a hospital with fever the doctor/nurses prepare atemperature-time graph, which can be referred to any time for reference. Similarly, the velocitytime graph and pressure-volume graph are of day-to-day use. We shall learn to draw these graphsand interpret them in the sections below :

28.6.1 Temperature-Time Graph-Reading and Construction

Example 28.10 : The body temperature of a patient admitted in a hospital with typhoid feverat different times of a day are given below :

Time of the day 7 9 11 13 15 17 19 21 23hrs hrs hrs hrs hrs hrs hrs hrs hrs

Temperature 102 103 104 103 101 100 99 100 99(in °F)

Draw a graph to represent the above data.

Solution : The graph of the above data is given in Fig. 28.10. The graph has been obtainedby joining the points corresponding to pairs, like (7, 102), (9, 103), ........., (23, 99) in therectangular system of coordinates, by line-segments.

Fig. 28.10

Note : While drawing the graph it has been assumed that during the time interval in betweentimes, the same trend was present.


Example 28.11 : If the medicine was given to the patient at 9 hours, whose temperature-timegraph is shown in Fig. 28.10, answer the following questions :

(i) At what time of the day was the temperature highest ? At what time lowest?

(ii) After how much time, the action of medicine had started ?

(iii) What trend do you observe from the above graph ?

Solution :

(i) The temperature of the patient was highest at 11 hours and lowest at 19 hrs and 23 hrs.

(ii) The action of the medicine started 2 hours after the medicine was given as the temperaturestarted falling after that.

(iii) The administered medicine suited the patient as the temperature constantly fell after that,with the exception of period between 19 hrs and 21 hrs when it became slightly higherat 100°F but again fell after that

28.6.2 Velocity Time Graph

During a journey from one place to other, the speeds of vehicles keep on changing accordingto traffic congestions. This can be very well shown by a velocity-time graph. Let us illustrateit with the help of example :

Example 28.12. During a journey from city A to city B by car the following data regardingthe time and velocity of the car was recorded :

Time of the day 6 7 8 9 10 11 12 13 14 15 16 17(in hours)

Velocity 60 60 45 50 60 50 45 60 50 65 40 50(in km/hour)

Represent the above data by a velocity time graph.

Solution : As before the graph can be obtained by plotting the ordered pairs (6, 60), (7, 60),... (15, 65), ..., (17, 50) in the rectangular system of coordinates and then by joining them byline-segments.

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Fig. 28.11

Example 28.13 : Read the velocity-time graph given in Fig. 28.11 and reply the followingquestions :

1. At what time duration of the day, the velocity of the car

(i) was lowest ? was highest ?

(ii) constant

(iii) went on increasing

(iv) went on decreasing

2. What was the average speed of the car in the journey ?

Solution : 1 (i) At 16 hours; At 15 hours

(ii) The velocity was constant at 60 km/hour between 6 hours and 7 hours

(iii) Between 8 hours to 10 hours

(iv) Between 10 hours to 12 hours

2. The average speed of the car was

60 60 45 50 60 50 45 60 50 65 40 5012

+ + + + + + + + + + +FHG

IKJ km/hour

=63512 or 52.92 km/hour.


28.6.3 Pressure-Volume Graph

For a fixed quantity of a gas at a constant temperature, is there any relation between pressureand volume of the gas ? Let us see that from the following example :

Example 28.14 : The following data pertains to pressure and volume of a fixed quantity ofgas :

Pressure (p) 60 90 45 30 75(in Newton)

Volume (v) 90 60 120 180 72(in cm3)

Draw a graph to represent the above data.

Solution :

Fig. 28.12

The graph is obtained by joining the plot of the ordered pairs (60, 90), (90, 60), ..... (75, 72)by free hand curve.

Example 28.15 : Read the above graph, given in Fig. 28.12, and answer the followingquestions :

1. Full in the blank :

(i) As volume increases, the corresponding pressure ...

(ii) As pressure decreases, the volume ...

(iii) Pressure × Volume = ...

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2. What will be the pressure when volume is 100 cm3 ?

3. What will be the volume, when the pressure is 100 Newton ?

Solution : 1 (i) Decreases

(ii) Increases

(iii) Constant = 5400

2. We know that pv = 5400

∴ When volume = 100, p = 54 Newtons

as can be seen from the graph at point A

3. When p = 100 Newtons, v = 54 cm3

as can be seen from the graph at the point B.


Represent the data given in each of the questions below graphically :

1. For a town, the maximum temperature for the following months are given below :

Months March April May June July August September October

Maximumtemperature 35 38 38 42 45 40 38 35(in °C)

2. The body temperatures of a patient admitted in a hospital are given below:

Time of the day 8 9 10 11 12 13 14 15 16 17(in hours)

Temperature (in °F) 103 104 105 102 102 100 99 99 100 98

3. The speeds of a car going from station A to station B at different times of the day aregiven below :

Time of the day (in hours) 7 8 9 10 11 12 13 14 15

Speed (in km/hour) 45 45 50 60 60 75 60 60 50

4. The data on pressure and volume of a gas are given below :

Pressure (in Newtons) 60 80 50 30 40 20

Volume (in cm3) 40 30 48 80 60 120


5. For question No. 1, read the graph and reply the following questions :

(i) Find the range of temperature for given months

(ii) Which month had the least temperature ?

(iii) Which month had the highest temperature ?

(iv) In which month was the temperature less than 40°F ?

(v) Can you predict the temperature for the next two months ?

6. Read the graph of Question No. 2 and answer the following question :

(i) At what time of the day was the temperature of the patient maximum?

(ii) If the medicine takes at least two hours to show the effect, at what time of the daywas the medicine given ?

LET US SUM UP

Bar graphs are the graphical representation of ungrouped frequency data.

Histograms and frequency polygons are the graphical representation of continuousgrouped frequency data.

The graphical representation of data from day-to-day life is the join of points correspondingto ordered pairs represented by the data.

The graphical representations show the trends readily and at a glance only.

TERMINAL EXERCISE

Draw the bar graph for the following data in each case :

1. Height of samplings (in m) 0.5 0.75 1.0 1.25 1.50 1.75 2.00

No. of samplings 15 18 25 40 12 8 7

2. Weight (in kg) 7 8 9 10 11 12 15

No. of baskets of apples 4 5 7 8 5 4 3

3. Number of parcels 120 150 80 60 40 50received in a post officeWeight of parcels (in kg) 1 2 3 4 5 6

4. Interpret the data given in Question 1 and 2.

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Draw a histogram and frequency polygon for the data in each case below:

5. Weight (in kg) 40–45 45–50 50–55 55–60 60–65 65–70

No. of students 4 7 8 9 6 3in the class

6. Daily earning 100-120 120-140 140-160 160-180 180-200(in rupees)

No. of workers 5 7 8 3 2

7. Read the graphs for Question Nos. 5 and 6 and interpret them.8. The minimum temperatures of a town for a year are given below :

Month Jan Feb March April May June July Aug Sep Oct Nov Dec

Min. temp. 12 14 16 20 20 24 25 24 22 18 16 12(in °C)

Draw a graph to represent the above data and interpret it.9. A man left New Delhi for Lucknow by car at 7 AM. The speed of the car at different

times of the day is given below :

Time of the day 7 9 11 13 15 17 19 21(in hour)Velocity (in km/hour) 45 50 60 65 70 60 55 40

Represent the above data by a velocity time graph and answer the following questionsAt what time was the velocity of the car(i) Maximum

(ii) Minimum(iii) Between (50–60) km/hour(iv) Can you give hypothesis regarding the places where speed is extreme ?

10. The following data pertains to a gas in a container :

Pressure (in Newton) 100 80 50 40 125 200

Volume (in cm3) 40 50 80 100 32 20

Represent the above data by a pressure volume graph. What relation do you find betweenpressure and volume from the data ?


ANSWERS


5. (i) Bhilai and Rourkela (ii) 8.33%

(iii) 50 (in ten thousand tonnes) (iv) 62.5%


6. (i) (44–46) size (ii) (40–42) size

(iii) 700 (iv) 50

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29

Measures of Central Tendency

29.1 INTRODUCTION

Sometimes we are required to describe the data arithmetically. The measures with which wedo that are called Arithmetical Descriptors of Data. The name itself suggests that these measuresdescribe the data arithmetically— for example, average age of a group, average height of aclass, median score of the group or modal collar size of a team. Because of the fact that thesemeasures are representative of the group they represent, they are called Measures of CentralTendency –central because these are the measures around which the measures of all themembers of the group gather around. Represented graphically, the graph of the observationsof the group will be around and close to the measure of central tendency.

In this lesson, we will study about different measures of central tendency for ungrouped dataand some of them for grouped data also. We will learn to calculate them by the formulae andwill also learn properties of some of them.

29.2 OBJECTIVES


define mean of raw, ungrouped and grouped data

calculate mean of raw, ungrouped and grouped data by ordinary and short-cut-methods

define mode and median of raw data

calculate mode and median of raw data

cite properties of mean and median

29.3 MEAN OF RAW DATA

If x1, x2, x3, ..., xn are n observations, then their mean x is defined as

x = Sum of all observationsNumber of observations

or x =x x x x

nn1 2 3+ + + + ...

...(i)

Measures of Central Tendency 321

In abbreviated form, (i) can be written as

x =x

n

ii

n

=∑

1


Example 29.1 : Find the mean of first 10 natural numbers.

Solution : We are required to find the mean of first ten natural numbers

Mean = Sum of first 10 natural numbers10

or x = 1 2 3 1010

+ + + ... +

= 5510 = 5.5

Thus, the mean of first 10 natural numbers is 5.5

Examples 29.2 : Find the mean of first five prime numbers

Solution : First five prime numbers are 2, 3, 5, 7 and 11

Their mean x = 2 3 5 7 115

+ + + +

= 285 = 5.6

Thus, the mean of first five prime numbers is 5.6.

Example 29.3 If the mean of 5, 7, 9, x, 11 and 12 is 9, find x.

Solution : We know that the mean x of observations is given by

x = Sum of observationsNumber of observations

As, x is given to be 9

∴ 9 = 5 7 9 11 126

+ + + + +x

or 54 = 44+ x

⇒ x = 54 – 44 = 10

∴ The value x is 10

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29.4 MEAN OF UNGROUPED DATA

Let the frequency distribution for ungrouped data be given as below :

Observations : x1 x2 x3 ... x10 ... xn

Frequency : f1 f2 f3 ... f10 ... fn

The above data shows that observation x1 is repeated f1 times, x2, f2 times and so on.

∴ The sum of all these observations is given by

f1x1 + f2x2 + ..... + f10x10 + ...... + fnxn

and the total number of observation is given by f1 + f2 + ...... + f10 + ..... + fn

∴ x , the mean of observations, is given by

x =f x f x f x f x

f f f fn n

n

1 1 2 2 10 10

1 2 10

+ + + + ++ + + + +

... ...... ... ...(ii)

In abbreviated form (ii) is written as

x =f x

f

i ii

n

ii

n=

=

∑

∑1

1

or x =f x

n

i ii

n

=∑

1 , where n = fii

n

=∑

1

Let us take some examples to illustrate

Example 29.4 : Find the mean of the following data :

xi : 2 5 8 10 12 15

fi : 4 5 3 5 2 1

Solution : We make the tabular form of data as follows :

xi fi fi . xi

2 4 85 5 258 3 24

10 5 5012 2 2415 1 15

∑ fi = 20 146 ←∑ f xi i


We know Mean xb g is given by

x =∑∑f xfi i

i

=14620 = 7.3

Example 29.5 : If the mean of the following data is 7, find x

xi : 4 x 6 7 9 11

fi : 2 4 6 10 6 2

Solution : Let us put the data, as usual, in tabular form. We get the following:

xi fi fi . xi

4 2 8x 4 4x6 6 367 10 709 6 54

11 2 22

∑ fi = 30 (190 + 4x)←∑ f xi i

We know that x = 7

∴ 7 = 190 430+ x

or 210 – 190 = 4xor x = 5

∴ The value x is 5.

29.5 MEAN OF GROUPED DATA

We know that for grouped data, the frequency in a class is centered at its class mark. Therefore,the first step in finding mean of grouped data is to find the class marks, say, x1, x2, x3, ....xk for k classes with class frequencies f1, f2, f3, ....., fk. Further, the steps are the same as formean for ungrouped data.

Thus, the mean x for grouped data is given by

x =f x f x f x

f f fk k

k

1 1 2 2

1 2

+ + ++ + +

.......... ....(iii)

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where, x1, x2, ....., xk are class marks of k classes and f1, f2, ....., fk are their frequencies.

In abbreviated form (iii) can be written as

x =f x

f

i ii

k

ii

k=

=

∑

∑1

1


Example 29.6 : Find the mean of the following grouped data :

Classes 0–10 10–20 20–30 30–40 40–50 50–60 60–70

Frequencies 1 3 5 7 5 3 1

Solution : Let us put the data in tabular form

Classes Frequencies (Class marks)(fi) xi fixi

0–10 1 5 005

10–20 3 15 045

20–30 5 25 125

30–40 7 35 245

40–50 5 45 225

50–60 3 55 165

60–70 1 65 065

∑ fi = 25 875 ←∑ f xi i

∴ Mean x =∑∑f xfi i

i = 875

25 = 35

You can very well see that if the class marks and frequencies are large, it is difficult to findthe mean as it involves lots of calculations. To simplify the procedure, we follow the followingmethod, called ‘Assumed Mean Method” or “Short cut Method”. The method involves thefollowing steps :

Step 1 : Find the class-marks

Step 2 : Take any convenient class mark as assumed mean A (usually the central oneor with maximum frequency is taken)


Step 3 : Find the deviations (xi – A)

Step 4 : Find the step deviations x A

ci − , where c is the interval and call them di

Step 5 : Find ∑ f di i and use the following formula to find

x = A f df ci i

i+ ∑∑ ×

Let us try to apply this method on Example 6.

Example 29.7 : Find the mean of the following grouped data using short cut method :

Classes : 0–10 10–20 20–30 30–40 40–50 50–60 60–70

Frequencies : 1 3 5 7 5 3 1

Solution : Let the assumed mean be 35.

Classes fi xi di = x Ac

i − fidi

0–10 1 5 –3 –3

10–20 3 15 –2 –6

20–30 5 25 –1 –5

A← 30–40 7 35 0 0

40–50 5 45 1 5

50–60 3 55 2 6

60–70 1 65 3 3

0 ←Σfidi

∴ x = 35+ ∑∑f df ci i

i

=35 + 0 = 35

which is same as found in Example 29.6.

Example 29.8 : Find the mean of the following data

Classes : 0–100 100–200 200–300 300–400 400–500 500–600

Frequencies : 15 25 40 30 10 5

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Solution : Let the assumed mean be 250.

Classes fi xi di = xi − 250100 fidi

0–100 15 50 –2 –30100–200 25 150 –1 –25200–300 40 250 0 0300–400 30 350 1 30400–500 10 450 2 20500–600 5 550 3 15

∑ fi = 125 10 ←∑ f xi i

∴ x = A f df ci i

i+ ∑∑ ×

= 250 10125 100+ × = 258


1. Write the formulae for calculating mean of(i) Raw data

(ii) Ungrouped data(iii) Grouped Data–ordinary method(iv) Grouped Data–Shortcut method.

2. Find the mean of first 6 multiples of 53. Find the mean of all multiples of 7 which are less than 100.4. Find the mean of following data :

xi : 7 9 12 15 21 24

fi : 3 4 5 8 3 2

5. Find the mean of following grouped data (1) by ordinary method, (ii) by short cut methodand confirm your results.

(a) Classes : 20-40 40-60 60-80 80-100 100-120 120-140 140-160

Frequencies : 3 6 9 12 9 6 5

(b) Classes : 0-500 500-1000 1000-1500 1500-2000 2000-2500 2500-3000

Frequencies : 14 16 20 30 12 8


29.6. MEDIAN OF RAW DATA

Median is that value of the variate which divides the raw data into two equal halves, i.e., equalnumber of observations on its both sides, when arranged in order (ascending or descending)

For finding the median of raw data, we first arrange the observations in ascending (ordescending) order of magnitude and then follow the following :

(i) If the number of observations n is odd, the median is n +FH IK12 th observation

(ii) If the number of observations n is even, the median is the average of n2FH IK th and n

2 1+FH IK thobservations

Let us illustrate it with examples.

Example 29.9 : Find the median of the following data :

(a) 5, 6 , 9 , 3, 18, 16, 10

(b) 16, 4, 18, 25, 17, 19, 12, 20, 15, 7

Solution : (a) Here the number of observations is odd (7 here). Arranging the data in ascendingorder, we have

3, 5, 6, 9, 10, 16, 18

The median is 7 12+FH IK th or 4th observation

∴ Median = 9

(b) Here n is even (10)

Arranging the data is ascending order, we have

4, 7, 12, 15, 16, 17, 18, 19, 20, 25

The median is average of 102FH IK th and 10

2 1+FH IK th observations i.e. 5th and 6th observations

∴ Median = 16 172+ = 16.5

Example 29.10 : If the median of the following data is 20, find x :

19, x, 16, 18, 22, 17, 24, 23, 21

Solution : As the median is 20, and there are four observations more than 20 and fourobservations less than 20, x must be 20

Arranging the data in ascending order, we have

16, 17, 18, 19, x, 21, 22, 23, 24

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1. Fill in the blanks suitably :

(i) Median is that value of the variable ...............

(ii) When the number of observations n is odd, median is given by ............... observation

(iii) When the number of observations n is even, median is given by ............... observation.

2. Find the median of the following :

(a) 5, 12, 16, 18, 20, 25, 10

(b) 18, 80, 42, 28, 15, 72, 69, 52, 46

(c) 15, 18, 7, 12, 19, 25, 17, 20

(d) 6, 12, 9, 10, 16, 28, 25, 13, 15, 17

3. If the median of the following data arranged in ascending order is 17, find x

12, 15, 16, x, 18, 19, 20, 22

29.7 MODE OF RAW DATA

Mode is that value of the variable which occurs most frequently in the data. The methodinvolved is to find the distinct values of observations in the data and find the frequency ofoccurrence of each. The observation with the maximum frequency is called mode.


Example 29.11 : Find the mode of the following data :

15, 11, 13, 11, 15, 12, 16, 18, 15, 19, 11, 15, 16, 8, 12

Solution : Here the distinct observation are

8 11 12 13 15 16 18 19

with their respectively frequencies as

1 3 2 1 4 2 1 1

Thus, the observation 15 has maximum frequency (4 here)

∴ Mode of the data is 15

Example 29.12. Find the value of x so that the mode of the data is 11

12, 11, 18, 20, 25, 11, 13, x, 13, 18, 13, 11, 19, 20

Solution : The distinct values of observations are

11, 12, 13, 18, 19, 20, 25, x

with frequencies

3 1 3 2 1 2 1

The mode of data is 11, then 11 has to have maximum frequency, that is more than 3. Therefore,x = 11.


Example 29.13 : If two of the 20’s in Example 29.12, where x = 11, are changed to 13 eachfind the mode of the new data.

Solution : Now the data becomes

xi : 11 12 13 18 19 25

fi : 4 1 5 2 1 1(* As x = 11)

∴ The new mode is 13.


1. Define mode of raw data.

2. Does the mode depend on number of observations ?

3. Find the mode of the following data :

(a) 15, 25, 18, 16, 25, 19, 18, 25, 16, 19, 25

(b) 1, 9, 22, 16, 15, 28, 9, 14, 16, 9, 28

(c) 28, 7, 28, 17, 7, 19, 15, 7, 28, 7, 15, 18, 7

4. (a) Find the value of x so that the mode of the following data is 37 :

12, 33, 37, 18, 19, 37, x, 33, 12, 33, 18, 37

(b) What will be the new mode if one of the 37 after x is replaced by 33.


28, 22, 21, 29, 12, 13, 18, 22, 14, 16, 28, 29,

13, 28, 22, 14, 22, 16, 19, 16, 15, 18, 22, 29

29.8 PROPERTIES OF MEAN

1. If two groups with observations n1 and n2 have their means x1 and x2 , then the combinedmean x of these is given by

x =n x n x

n n1 1 2 2

1 2

++

2. If x1, x2, x3, ..., xn observations have a mean x ,

then x x x x x x x xn1 2 3− + − + − + + −b g b g b g b g... = 0

i.e. x xii

n−

=∑ b g

1= 0

3. If each observation in the data is increased by a, then the mean is increased by a.4. If each observation in the data is decreased by b, then the mean is decreased by b.

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5. If each observation in the data is multiplied by some constant, then the mean getsmultiplied by the same constant.

6. If each observation in the data is divided by some non-zero constant, then the mean getsdivided by the same non-zero constant.

29.9 PROPERTIES OF MEDIAN

1. Median is the middlemost observation of the data i.e., it divides the data exactly in twohalves, when arranged in order (ascending or descending).

2. Median, unlike mean, is not affected by extreme values.

3. Median may not lie in the data itself.

4. Median can be determined graphically also while mean can not.

LET US SUM UP

The three measures of central tendency are mean, median and mode.

Mean is that value in the group of observations which describes it arithmetically.

Mean x of raw data is given by xx

n

ii

n

= =∑

1

Mean x of grouped data is given by

(i) x = f x

f

i ii

k

ii

k=

=

∑

∑1

1

, where k is the number of classes and xi′s are class marks

(ii) x = Af d

fc

i ii

k

ii

k+ ×=

=

∑

∑1

1

where A is the assumed mean di = x A

ci −

and c is the length of interval.

Median of raw data is given by

(i) n2FH IK th observation if the number of observation (n) is odd.

(ii) average of n2FH IK th and n

2 1+FH IK th observations of n is even.

Mode is that value of variate which occurs most frequently.


TERMINAL EXERCISE

1. Find the mean of

(i) 2, 9, 19, 12, 15 and 9

(ii) 3, 6, 9, 12, 15, 18, 19, 21

(iii) 5, 2, 19, 15, 13, 18, 13, 11

2. If the mean of 8, 12, x , 14, 18 is 12, find x.

3. If the mean of 13, 12, 18, 14, p, 19, 17, 14 is 14 find the value of p.

4. Find the mean of the following :

(a) xi : 5 9 13 17 22 25

fi : 3 5 12 8 7 5

(b) xi : 16 18 20 22 24 26

fi : 1 3 5 7 5 3

5. Find the mean of the following data :

(a) Classes : 10-20 20-30 30-40 40-50 50-60 60-70

Frequencies : 2 3 5 7 5 3

(b) Classes : 100-200 200-300 300-400 400-500 500-600 600-700

Frequencies : 3 5 8 6 5 3

6. Find the mean of data given in 5(a) and 5(b) by assumed mean method.

7. Find the median of the following data :

(a) 3, 19, 17, 15, 8, 5, 9

(b) 15, 13, 8, 22, 29, 12, 14, 17, 6

8. If the mode of the following observation is 12, find x

(a) 16, 13, 8, 12, 19, 17, 12, 16, x, 19

(b) 15, 16, 19, 12, 13, 16, 12, 12, 12, 16, x


(a) 8, 5, 2, 5, 3, 5, 3

(b) 19, 18, 17, 16, 17, 15, 14, 17, 15

10. If the median of the follows data is 25, find x

16, 18, 19, 23, x, 29, 31, 35.

332 Mathematics

ANSWERS


1. (i) xx

n

ii

n

= =∑

1 (ii) xf x

f

i ii

n

ii

n= =

=

∑

∑1

1

(iii) xf x

f

i ii

k

ii

k= =

=

∑

∑1

1

(iv) x A f df ci i

i= + ×Σ

Σ

2. 17.5 3. 52.5 4. 13.92

5. (a) 92.4 (b) 1420


1. (i) Which divided the data into two equal halves.

(ii) n th+12

(iii) Average of n2 th and n

2 1+FH IK th2. (a) 16 (b) 46 (c) 17.5 (d) 143. 16


1. The value of the variable which occur most frequency is defined mode of raw data2. No3. (a) 25 (b) 9 (c) 74. (a) x = 37 (b) New mode = 335. 22

Terminal Exercise

1. (i) 11 (ii) 12.875 (iii) 122. 8 3. 54. (a) 15.775 (b) 21.755. (a) 42.6 (b) 396.76. (a) 42.6 (b) 396.77. (a) 9 (b) 148. (a) 12 (b) 129. (a) 5 (b) 17 10. 27

Introduction to Probability 333

30

Introduction to Probability

30.1 INTRODUCTION

In our day-to-day life, we sometimes make the following statements :

It may be foggy today.

The train may reach in time.

Ram is likely to get good marks in his examination.

Probably, he will come today.

The usage of terms of the type “may”, “likely” or “probably” convey the meaning that theevent, we are talking about, is not certain to occur. Under the head of probability, we will studythe measure of this uncertainty under given conditions.

If we go by the dictionary meaning of the word “probable” it says “likely though not certain”.The word probability has been derived from the same and is intended to be used to have ameasure of uncertainty in making statements or taking decisions in such situations. Thus,probability measures the degree of uncertainty involved and hence measures the degree ofcertainty of occurrence of events.

Interestingly, the origin of the theory of probability has been the study of games of chance andis connected with events like tossing a coin, throwing a die, selecting a card at random froma well shuffled pack of cards.

30.2 OBJECTIVES


define an experiment, an outcome and an event

define different types of events

define probability of occurrence of an event

solve problems based on tossing a coin, throwing a die, and drawing a card from a wellshuffled pack of cards.

334 Mathematics


We assume that the learner is already familiar with :

the terms connected with coin-head and tail

the terms connected with a dice the dots on six faces as (•, ••, •••, , , )

four fundamental operations on numbers

terminology connected with “playing cards”

30.4 SOME DEFINITIONS

Before saying something about probability, it will be desirable to have knowledge of basic termsbeing used in the text. We will take them one by one below

30.4.1 Experiment

An activity which ends in some well defined result is called an experiment.

Drawing a card from a pack of cards, throwing a die and tossing a coin are examples ofexperiments.

30.4.2 Trial and Outcome

Performing an experiment once is called a trial and the result of the trial is called as outcome.

30.4.3 Random Experiment

An experiment in which all possible outcomes are known before but will be the result of atrial can not be surely predicted, is called a random experiment.

For example, tossing a fair coin or throwing a fair die

Note : By the word fair, we mean that no trick has been played with the object being tossed,otherwise we can always predict the result. If both the faces of a coin have head, the resultwill always be head.

30.4.4 Event

All the possible outcomes of a trial are called events. Any specific of these is called an event

30.5 TYPES OF EVENTS

Let us now study about some types of events and sample space :

30.5.1 Sample space

The collection of all possible outcomes of an experiment constitute its Sample Space.

For example, when a coin is tossed, the sample space consists of Head (H) and Tail (T).

When a die is thrown, the sample space consists of •, ••, •••, , , , (sometimescalled 1, 2, 3, 4, 5 and 6)

• •• • • ••• •• • •• • •

• •• •• ••• •

• • •• • •


30.5.2 Favourable Event

The cases which ensure the occurrence of an event, are called favourable to that event.For example, the favourable cases to the event “occurrence of an odd number, when a die isthrown are 1, 3, 5.

30.5.3 Mutually Exclusive Events

If two events cannot occur together, they are said to be mutually exclusive events. For example,

In a throw of a coin, head and tail are mutually exclusive events.

30.5.4 Equally Likely Events

Two events are said to be equally likely events if they have equal chance of occurrence

For example, in a fair toss of a coin, occurrence of head and tail are equally likely events.

30.6 PROBABILITY OF OCCURRENCE OF AN EVENT

The probability P(E) of occurrence of an event E is defined as

P(E) =Number of outcomes favourable to the event

Total number of possible outcomes

Thus, the probability of occurrence of a head (H) when a coin is tossed is 12 .

Suppose there are m cases (outcomes) which are favourable to an event capital X and n is thetotal number of outcomes, then the probability of occurrence of the event X is given by

P(X) =mn ...(i)

We know that m ≤ n ...(ii)

From (i) and (ii), we can say that

0 ≤ P(X) ≤ 1

When do we say that the probability of occurrence of event x is zero ? When there is no outcomefavourable to the event. Such an event is called an Impossible event. For example getting anumber greater than 6 when a die is thrown once is an impossible event.

In the case, when P(X) = 1, all the possible out comes are favourable to the event. In that case,the event is said to be a Sure event. For example, getting a number less than 7, when a dieis thrown once is a sure event.

Let us now introduce another concept, called probability of non-occurrence of an event.

336 Mathematics

If P(X) = mn , then there are (n – m) outcomes which are not favourable to the occurrence of

X. Then the probability of non-occurrence of the event X, called P X( ) is given by

P X( ) = n mn

nn

mn

− = −

= 1− mn = 1 – P(X)

i.e., the probability of non-occurrence of an event equals [1 – Probability of occurrence of theevent]

Note : X is called an event complementary to event X.


Example 30.1 : Write the sample space if a die is thrown once.

Solution : We know that when a die is thrown once, the likely outcomes are

•, ••, •••, , , ,

or 1, 2, 3, 4, 5, 6

This is the required sample space.

Example 30.2 : 3 males and 4 females appear for an interview, of which one candidate is tobe selected. Write the sample space.

Solution : Let the three male candidates be denoted by M1, M2, M3 and four female candidatesbe denoted by F1, F2, F3 and F4. The sample space consists of M1, M2, M3, F1, F2, F3, F4.

Examples 30.3 : Find the probability of occurrence of a tail (T) when an unbiased coin is tossedonce.

Solution : When a coin is tossed once, the sample space is H, T

There is only one outcome favourable to a tail (T).

∴ The probability of occurrence of a tail = 1 2

Example 30.4 : A die is tossed once, find the probability of occurrence of

(i) 2

(ii) an even number

Solution : When a die is tossed, the sample space is

1, 2, 3, 4, 5, 6

• •• •• ••• •

• • •• • •


(i) Probability of occurrence of 2, i.e.,

P(2) = 16

because only one outcome is favourable to the event out of 6 possible outcomes.

(ii) Probability of occurrence of an even number

The even numbers are 2, 4, 6

The number of favourable outcomes = 3

∴ P (even number) = 36

12=

Example 30.5 : Find the probability of getting an odd number when a die is tossed once.

Solution : The number of odd numbers is 3

(1, 3, 5 are odd numbers)

If E is the event, called, occurrence of an odd number, then

P(E) = 36

12=

You could have found P(E) as

P(E) = 1 – P (even number) = 1 12

− = 12

Example 30.6 : In case of Example 2; find the probability of selection of a

(i) male candidate

(ii) female candidate

Solution : Total number of candidates = 7

Numbers of males = 3

∴ (i) P (selection of a male candidate) = 37

(ii) P (selection of a female candidate) = 47

Alternatively P (selection of female candidate) = 1 – P (selection of a male candidate)

= 1 37

47− =

338 Mathematics

Example 30.7 : From a well shuffled pack of cards, if a card is drawn, find the probabilityof occurrence of a

(i) red card

(ii) black king

(iii) a queen

Solution : (i) Since there are 26 red cards in a pack of 52 cards, the probability of occurrenceof a red card, P (R), is given by

P(R) = 2652

12=

(ii) Out of 52 cards, there are two black kings

∴ The probability of occurrence of a black king P(B) is given by

P(B) = 252

126=

(iii) Similarly, the probability of occurrence of a queen P (Q) is given by

P(Q) = 452

113=

Examples 30.8 : A bag contains 4 red, 3 black and 2 white balls. A ball is drawn from thebag. Find the probability (i) that it is a red ball (ii) it is not black.

Solution : (i) No. of red balls = 4

Total no. of balls = 9

∴ P (a red ball) = 49

(ii) No. of favourable cases = 6 (4 red + 2 white)

or P (getting a black ball) = 39

13=

∴ P (not getting a black ball) = 1 13

23− =

∴ P (not getting a black ball) = 23


1. Define

(i) Sample space

(ii) Experiment and a random experiment


(iii) Event and outcome

(iv) Types of events

(v) Probability fo occurrence of an event

2. Write the sample space for the following :

(i) when a fair coin is tossed once

(ii) when a fair die is tossed once

(iii) when 6 boys and 5 girls appear for a selection.

3. Find the probability of occurrence of

(a) a multiple of 2 when a die is tossed

(b) a multiple of 3 when a die is tossed

(c) a number greater than 4 when a die is tossed

(d) a number greater than 6 when a die is tossed

(e) a multiple of 4 which is less than 3

4. Find the probability of selection of

(a) a boy

(b) a girl

for Question 2(iii)

5. An urn contains 6 black and 4 red balls. One ball is selected from the urn at random.Find the probability of selecting a

(i) black ball

(ii) red ball

6. A group of people have following distribution of members according to age

Number Age

5 30 years and below

10 (31 – 40) years

15 (41 – 60) years

A person is selected at random from the group. Find the probability of getting a personwith age

(i) less than or equal to 40 yrs

(ii) beyond 40 years

(iii) less than or equal to 30 years.

340 Mathematics

LET US SUM UP

Probability is a measure of index of uncertainty of occurrence of an event.

An activity that ends in some well defined result is called an experiment. An experimentin which the sample space of outcomes is known but which one will occur cannot bepredicted is called a random experiment.

Performing an experiment once is called a trial and its result is called an outcome.

The collection of all possible outcomes is called sample space.

Outcomes which ensure the occurrence of an event are called favourable cases to theoccurrence of that event

The events which cannot occur together are called mutually exclusive events

Events which have equal chances of occurrence are called equally likely events.

Probability of occurrence of an event.

=Number of outcomes favourable to the event

Total Number of outcomes

TERMINAL EXERCISE

1. Define probability of occurrence of an event.

2. Define sample space. Write the sample space for

(i) selection of a candidate having 8 male and 6 female candidates

(ii) when a die is tossed

(iii) An urn with 4 red and 3 black balls

3. Find the probability of the following events when a die is tossed once:

(i) occurrence of 4

(ii) occurrence of a number greater than 2

(iii) occurrence of a number less than 3

(iv) occurrence of multiple of 4

(v) occurrence of a number greater than 6

(vi) occurrence of a number greater than or equal to 1

(vii) occurrence of a number which is prime.


4. An urn contains 6 red, 4 black and 5 green balls. A ball is selected at random. Find theprobability of selection of a

(i) red ball

(ii) black ball

(iii) green ball

(iv) non-red ball

(v) non black ball

5. Find the probability of a non-leap year (365 days) containing 53 Sundays.

6. In an interview for the selection of a candidate, 6 boys and 2 girls have been called. Findthe probability of selection of a

(i) boy

(ii) girl (using the concept of complementary events)

342 Mathematics

ANSWERS


2. (i) H, T

(ii) 1, 2, 3, 4, 6

(iii) B1, B2, ....., B6, G1, G2, ....., G5

3. (a) 12 (b) 1

3 (c) 13 (d) zero (e) zero.

4. (a) 611 (b) 5

11

5. (i) 35 (ii) 2

5

6. (i) 12 (ii) 1

2 (iii) 13

Terminal Exercise

3. (i) 16 (ii) 2

3 (iii) 13 (iv) 1

6 (v) zero

(vi) 1 (vii) 12

4. (i) 25 (ii) 4

15 (iii) 13 (iv) 3

5 (v) 1115

5. 17

6. (i) 34 (ii) 1

4


APPENDIX

A DIE

A die is a stable cube with its six faces marked with dots from1 to 6. (See Fig. 30.1). The plural of the word “die” is “dice”.Whenever, we roll a die and ask for a number, if that numbercomes on the top, the attempt is described as “win”. If youcarefully see the figure on the side, the number 6 ( ) hascome on the top and if the player had asked for that number, hewould have won in that trial.

Remmeber, for an unbiassed die, each number 1 to 6 is equally likely to come on the top (occur)

A PACK OF CARDS

A pack of card consists of 52 cards divided into 4 suits (colours) called

(i) Spades ♠ — Black colour

(ii) Hearts ♥ — Red colour

(iii) Diamonds ♦ — Red colour

(iv) Clubs ♣ — Black colour

Each suit consists of 13 cards Ace, 2, 3, ..., 10, Jack, Queen and King. Jack, Queen and Kingare called face cards

If we draw a card from a well shuffled pack of cards, each of the 52 cards have equal chanceof being drawn.

∴ Probability (of selection of a card) = 152

Particularly, probability (Jack of spades) = 152

Fig. 30.1

Practice Work 205

Secondary CourseMathematics

Practice Work–Algebra

Maximum Marks : 25 Time : 45 Minutes

Instructions :1. Answer all the questions on a separate sheet of paper.

2. Give the following informations on your answer sheet.

Name

Enrolment number

Subject

Topic of practice work

Address

3. Get your practice work checked by the subject teacher at your study centre so that youget positive feedback about your performance.

Do not send your practice work to National Institute of Open Schooling.

1. Fractions are called : 1

(A) positive integers

(B) positive rationals

(C) negative rationals

(D) negative integers

2. Which of the following is not an integer ? 1

(A) – 4

(B) – 25

(C) 36

(D)45

206 Mathematics

3. Which of the following algebraic expressions is not a polynomial ? 1

(A) x2(x + 1)

(B) 5

(C) 3 3x x+

(D) 3 2x x+

4. The reciprocal of 23

3FH IK−

is : 1

(A) −FH IK−2

33

(B) 23

3FH IK

(C) 32

3FH IK

(D) 32

3FH IK−

5. The tenth term of the A.P. 1

2, 4, 6, ... is :

(A) –20

(B) 18

(C) 20

(D) 22

6. If * is an operation defined on ‘a’ and ‘b’ as follows : 2

a * b = – a + b – (–3), then find (–4) * 3

7. Find two numbers nearest to 1000 which are exactly divisible by 4, 5, 6, 7 and 10 each.2

8. Express 0 67. as a rational number. 2

Practice Work 207

9. Simplify : 3 2 22 2

11 2

. x xx x

−−

−

− − . 2

10. The sum of the first four and the first five terms of an A.P. are 26 and 40 respectively.If the first term is 2, find the common difference. 2

11. If mth, nth and rth term of an A.P. are x, y and z respectively, prove that

x(n – r) + y(r – m) + z(m – n) = 0. 4

12. The area of a rectangle gets increased by 4 m2 if its length is increased by 4 m and breadthdecreased by 2 m. If the length is increased by 2 m and breadth decreased by 2 m, thearea is decreased by 8 m2. Find the dimensions of the rectangle. 6

Practice Work 289


Practice Work–Commercial Mathematics




Name

Enrolment number

Subject


Address



1. A saree is available for Rs 450 whose marked price is Rs 600. Rate of discount givenby the shop is : 1

(A) 15%

(B) 20%

(C) 25%

(D) 30%

2. A customer has to pay Rs 20 more, if the discount is reduced from 20% to 15%. Themarked price of the item is : 1

(A) Rs 420

(B) Rs 400

(C) Rs 380

(D) Rs 385

290 Mathematics

3. A man sold an article for Rs 187.50. If he had spent Rs 12.50 on transportation and hada gain of 25%, the cost price of the article was : 1

(A) Rs 175.00

(B) Rs 137.50

(C) Rs 150.00

(D) Rs 140.00

4. A sum of money becomes Rs 2000 in 2 years and Rs 2250 in 4 years at the same rateof simple interest. The sum is : 1

(A) Rs 2000

(B) Rs 1850

(C) Rs 1750

(D) Rs 1600

5. Which one is a better investment ? 1

(A) 12% per annum interest compounded yearly.

(B) 6% per half year interest compounded half yearly.

(C) 3% per quarter interest compounded quarterly.

(D) 1% per month interest compounded monthly.

6. The sides of a triangle are in the ratio 1 : 1.5 : 2. If the perimeter is 13.5 cm, find thelength of each side. 2

7. A watch is sold for Rs 405 at a loss of 10%. Another watch is sold for Rs 540 at a gainof 8%. Find the total gain or loss in both transactions. 2

8. The cost price of 15 pens is equal to selling price of 12 pens. Find the profit percent.2

9. A shopkeeper allows 25% discount on his articles in a sale. However he has to charge10% sales tax on goods sold. Find the marked price of an article, if a customer has topay Rs 330 inclusive of sales tax, to buy it. 2

10. The population of a village in the year 2000 was 8000. Its population increased by 5%during the year 2001 and 4% during the year 2002. Find the total population at the endof the year 2002. 2

11. The difference between the compound interest and simple interest as a certain sum ofmoney at 10% per annum for 2 years is Rs 155. Find the sum, if the interest is compoundedyearly. 4

12. A T.V. set is available for Rs 7500 cash or Rs 2000 as cash down payment followed by6 monthly instalments of Rs 1000 each. Find the rate of interest charged under instalmentplan. 6

190 Mathematics


Practice Work–Geometry




Name

Enrolment number

Subject


Address



1. Lines AB and CD intersect each other at O as shown in the adjacent figure. A pair ofvertically opposite angles is : 1

(A) 1, 2

(B) 2, 3

(C) 3, 4

(D) 2, 4

2. Which of the following statements is true for a ∆ABC ? 1

(A) AB + BC = AC

(B) AB + BC < AC

(C) AB + BC > AC

(D) AB + BC + AC = 0

Practice Work 191

3. The quadrilateral formed by joining the mid points of the pair of adjacent sides of arectangle is a : 1

(A) rectangle

(B) square

(C) rhombus

(D) trapezium

4. In the adjacent figure, PT is a tangent to the circle at T. If ∠BTA = 45° and ∠PTB = 70°,Then ∠ABT is : 1

(A) 110°

(B) 70°

(C) 45°

(D) 25°

5. Two points A, B have co-ordinates (2, 3) and (4, x) respectively. If AB2 = 13, the possiblevalue of x is : 1

(A) – 6

(B) 0

(C) 9

(D) 12

6. In ∆ABC, AB = 10 cm and DE is parallel to BC such that AE = 14 AC. Find AD. 2

7. If ABCD is a rhombus, then prove that 4AB2 = AC2 + BD2. 2

192 Mathematics

8. Find the co-ordinates of the point on x-axis which is equidistant from the points whoseco-ordinates are (3, 8) and (9, 5). 2

9. The co-ordinates of the mid-point of a line segment are (2, 3). If co-ordinates of one ofthe end points of the segment are (6, 5), then find the co-ordinates of the other end point.

2

10. The co-ordinates of the vertices of a triangle are (3, –1), (10, 7) and (5, 3). Find theco-ordinates of its centroid. 2

11. In an acute angled triangle ABC, AD⊥BC. Prove that

AC2 = AB2 + BC2 – 2BC. BD 4

12. Prove that parallelograms on equal (or same) bases and between the same parallels areequal in area. 6


Practice Work–Mensuration




Name

Enrolment number

Subject


Address



1. If ‘a’ is the side of an equilateral triangle, then its altitude is : 1

(A)3

22a

(B)3

2 2a

(C)3

2a

(D)3

2a

238 Mathematics

2. The sides of a triangle are 30 cm, 40 cm and 50 cm. Its area is : 1

(A) 120 cm2

(B) 600 cm2

(C) 750 cm2

(D) 1200 cm2

3. The perimeter of a square of side ‘l’ is given by : 1

(A) l2

(B) 4l

(C) l 2

(D) 2l

4. The base of an isosceles triangle is 8 cm and one of the equal sides is 5 cm. The heightof the triangle is : 1

(A) 5 cm

(B) 4 cm

(C) 3 cm

(D) 2 cm

5. The edge of a cube, whose volume equals that of a cuboid of dimensions63 cm × 56 m × 21 cm is : 1

(A) 21 cm

(B) 28 cm

(C) 36 cm

(D) 42 cm

6. How long will a man take to walk round the boundary of a square field of area90000 sq. m at the rate of 6 km an hour ? 2

7. One of the diagonals and a side of a rhombus are 8 cm and 5 cm respectively. Find thearea of the rhombus. 2

8. The radii of two right circular cylinder are in the ratio 3 : 4 and their heights are in theratio 5 : 2. Find the ratio of their curved surface areas. 2

9. Three equal cubes are placed end-to-end in a row. Find the ratio of the total surface areaof the new cuboid to that of the sum of the surface areas of the three cubes. 2

Practice Work 239

10. A room is 7 m long and 4 cm wide. It has two doors of 2 m × 1 m each and two windows1 m × 1m each. The cost of painting the walls at the rate of Rs 5 per square meter isRs 300. Find the height of the room. 2

11. Spherical marbles of 1.4 cm diameter are dropped into a cylindrical vessel of diameter7 cm containing water. Find the number of marbles that should be dropped, if the waterlevel in the cylinder rises by 5.6 cm. 4

12. A cubic meter of iron is melted to form a wire of diameter 3.5 mm Find the length of

the wire. Useπ =FHG

IKJ

227 6

284 Mathematics


Practice Work–Trigonometry




Name

Enrolment number

Subject


Address



1. If sin θ = pq , then cos θ is : 1

(A)qp

(B) 1− pq

(C)q p

q

2 2−

(D)p q

q

2 2−

Practice Work 285

2. In ∆ABC, right angled at B, AB = 5 cm, AC = 13 cm and ∠C = θ, then cos θ is equalto : 1

(A)5

13

(B)135

(C)5

12

(D)1213

3. If sinθ = 35 , then the value of 1+ cosecθb g is : 1

(A) 85

(B) 83

(C) 95

(D) 59

4. If 2 3tanθ = , then the value of 6 54sin cossin cos

θ θθ θ−+ is : 1

(A) 2

(B)45

(C)137

(D)47

286 Mathematics

5. Representation of sinθ interms of tanθ is : 1

(A)tan

tanθ

θ1 2+

(B)tan

tanθ

θ1 2−

(C)1

1 2tan tanθ θ+

(D)1

1 2tan tanθ θ−

6. Show that 2 21 2sin cos tan

tanA A A

A=

+. 2

7. Prove that sin4A – cos4A = sin2A – cos2A. 2

8. Eliminate θ from x = a sec tanθ θ+b g 2

y = b sec tanθ θ−b g9. For what value of θ, cos θ and sec θ are equal ? 2

10. If x = a sin cosθ θ+b g 2

y = b sin cosθ θ−b g , prove that

xa

yb

2

2

2

2 2+ =

11. Prove that tan sintan sin

secsec

θ θθ θ

θθ

+−

= +−

11 . 4

12. The angles of elevation of the top of a light house from two boats, in the opposite directionof light house, in the sea are 45° and 60°. If the distance between the boats is 60 m, findthe height of the light house. 6

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