b1 real numbers topic-1 euclid s division lemma … s division lemma and fundamental theorem of...

P-1S O L U T I O N S

REAL NUMBERS

SECTION

BSECTIONCHAPTER

1 TOPIC-1

Euclid’s Division Lemma and Fundamental Theorem of Arithmetic

WORKSHEET-1

Solutions

1. The given number ends in 5. Hence it is a multiple of 5. Therefore it is a composite number.

[CBSE Marking Scheme, 2016] 1

2.

HCF of

k

k

k

k

k

.

.

.

.

2

3

2

5

2

isk.

2

[CBSE Marking Scheme, 2015]

3. 90 = 2 × 32 × 5

and 144 = 24 × 32

HCF = 2 × 32 = 18 1

LCM = 24 × 32 × 5 = 720 1


4. 240 = 228 × 1 + 12 1

and 228 = 12 × 19 + 0

⇒ HCF of 240 and 228 = 12 1

5. HCF of 324 and 252 = 36 HCF of 36 and 180 = 36 \ HCF of 180, 252 and 324 is 36. [CBSE Marking Scheme, 2016] 3

6. LCM of 18, 24 and 36 is 72. 1½ 72 ) 999999 ( 13888 999936 ½ 63 → Remainder ½ \ Required number = 9,99,936. ½

7. Prime factors of : 378 = 2 × 33 × 7 1

180 = 22 × 32 × 5 1

420 = 22 × 3 × 7 × 5 1

\ HCF = 2 × 3 1

= 6 No, because HCF × LCM ≠ Product of three

numbers.

qqq

WORKSHEET-2

Solutions

1. a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then least prime factor of (a + b) is 2. 1

2. Given HCF (306, 1,314) = 18

LCM (306, 1,314) = ?

Let a = 306

b = 1,314 1

We know that

a × b = LCM (a, b) × HCF (a, b)

⇒ 306 × 1,314 = LCM (a, b) × 18

⇒

LCM (a, b) =

306 1 314

18

× ,

\

LCM (306, 1,314) = 22,338 1

3. y = 5 × 13 = 65 and x = 3 × 195 = 585 [CBSE Marking Scheme, 2015] 2

4.

2

3

5

7

371

1855

53

5565

11130 = 5565 × 2

= 1855 × 3

= 371 × 5

=371

7

= 53 × 7

\ x = 11,130 ½

½

½

½

P-2 M A T H E M A T I C S - X

5. Let n be any positive integer. By Eucild’s division lemma, n = 5q + r, 0 ≤ r < 5 n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4, where q∈ ω q is a whole number now n2 = (5q)2 = 25q2 = 5(5q2) = 5m n2 = (5q + 1)2 = 25q2 + 10q + 1 = 5m + 1 n2 = (5q + 2)2 = 25q2 + 20q + 4 = 5m + 4

Similarly n2 = (5q + 3)2 = 5m + 4 and n2 = (5q + 4)2 = 5m + 1 Thus square of any positive integer cannot be of

the form 5m + 2 or 5m + 3. [CBSE Marking Scheme, 2012] 3

6. Fundamental theorem of arithmetic : Every composite number can be expressed as the product of powers of primes and this factorization in unique.

2520 = 23 × 32 × 5 × 7 1

10530 = 2 × 34 × 5 × 13 1

\ LCM = 23 × 34 × 5 × 7 × 13 1

= 294840 1


qqq

WORKSHEET-3

Solutions

1. The smallest prime number is 2 and the smallest composite number is 22.

Hence, required HCF (22, 2) = 2 1

2.

a =

9 009

3 003

,

, = 3 ½

b =

1 001

143

,

= 7 ½

Since 143 = 11 × 13, so c = 11 or 13 ½

and d = 13 or 11. ½

3.

= 2 × 33816762

2 3381

3 1127

7 161

7 23

= 7 × 161

=1617

½

½

½

\ Composite number, x = 6762. ½

4. 3 × 12 × 101 + 4 = 4(3 × 3 × 101 + 1)

= 4(909 + 1)

= 4(910)

= a composite number

[QProduct of more than two factors] 2

[CBSE Marking Scheme, 2015] 5. By using Euclid’s Division Lemma, we have 92690 = 7378 × 12 + 4154 ½ Again we apply Euclid’s Division Lemma of divisor

7,378 and remainder 4154 7378 = 4154 × 1 + 3,224 4154 = 3224 ×1 + 930 3224 = 930 × 3 + 434 930 = 434 × 2 + 62 434 = 62 × 7 + 0 HCF (92690, 7378) = 62 1

Now, using Euclid’s Divison Lemma on 7161 and 62, we have

7161 = 62 × 115 + 31 ½

Again, applying Euclid’s Division Lemma on divisor 62 and remainder 31

62 = 31 × 2 + 0 Now, HCF (7161, 62) = HCF (62, 31) = 31 Hence, HCF of 92690, 7378 and 7161 is 31. 1

6. If 6n ends with 0, then it must have 5 as a factor.

But we know that only prime factor of 6n are 2 and 3.

\ 6n = (2 × 3)n = 2n × 3n

From the fundamental theorem of arithmetic, we know that the prime factorization of every composite numbers is unique.

\ 6n can never end with 0. 4


qqq

WORKSHEET-4

Solutions

1. HCF of 33 × 5 and 32 × 52

= 32 × 5

= 9×5

=45 1


2. (7 × 13 × 11) + 11 = 11 × (7 × 13 + 1) = 11 × (91 + 1) = 11 × 92 = 11 × 2 × 2 × 23 1 and (7 × 6 × 5 × 4 × 3 × 2 × 1) + 3

= 3 (7 × 6 × 5 × 4 × 2 × 1 + 1) = 3 × (1681) = 3 × 41 × 41 which is a composite number (more than one prime

factor). 1

3. 1656) 4025 (2

–3312

713)1656(2

–1426

230)713(3

– 690

23)230(10

–230

0

1

Hence HCF (1,656, 4,025) = 23 1


4. 1,200 = 4 × 3 × (2 × 5)2

= 24 × 3 × 52

The smallest natural number is 3.

[CBSE Marking Scheme, 2015] 2 5. To find LCM (9, 12, 15) 1

9 = 3 × 3 12 = 2 × 2 × 3

15 = 3 × 5

\ LCM (9, 12, 15) = 3 × 3 × 2 × 2 × 5

= 180 minutes 1

The bells will toll together after 180 minutes. 1

6. 16 = 2 × 2 × 2 × 2 = 24

36 = 2 × 2 × 3 × 3 = 22 × 32

HCF (16, 36) = 2 × 2

= 4

LCM (16, 36) = 24 × 32

= 16 × 9

= 144 1

We can check HCF and LCM are correct or wrong by using formula

HCF (a, b) × LCM (a, b) = Product of the numbers

= a × b 1

Product of the HCF and LCM should be equal to product of the numbers.

⇒ 4 × 144 = 16 × 36 ⇒ 576 = 576 ⇒ LHS = RHS 1

Hence our answer is correct.

7. Fundamental theorem of Arithmetic : Every composite number can be expressed as a product of primes and decomposition is unique, apart from the order in which the prime factors occur.

HCF = 24

LCM = 540

Let’s calculate

LCM

HCF =

540

24 = 22.5 not an integer

Since LCM is always a multiple of HCF, hence, two numbers cannot have HCF and LCM as 24 and 540 respectively. [CBSE Marking Scheme, 2015] 4

qqq

WORKSHEET-5

Solutions

1. We know that a × b = HCF (a, b) × LCM (a, b) ⇒ 1,800 = 12 × LCM (a, b)

⇒ LCM (a, b) = 1 800

12

,= 150 1

2. Let a be any positive integer ½

By division algorithm

a = 6q + r, where 0 ≤ r < 6

\ a = 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5 ½

But a is an even integer ½

\ a = 6q, 6q + 2, 6q+ 4 ½


3. By Euclid’s division algorithm,

a = bq + r

Take b = 4

\ Since 0 ≤ r < 4, r = 0, 1, 2, 3 1

So, a = 4q, 4q + 1, 4q + 2, 4q + 3 Clearly, a = 4q, 4q + 2 are even, as they are divisible by 2. Therefore A cannot be 4q, 4q + 2 as a is odd. But 4q + 1, 4q + 3 are odd, as they are not divisible by 2.

\ Any positive odd integer is of the form (4q + 1) or (4q + 3). 1

4. No.

15 does not divide 175.

LCM of two numbers should be exactly divisible by their HCF. 1

\ Two numbers cannot have their HCF as 15 and LCM as 175. 1

5. By Euclid’s division algorithm,

510 = 92 × 5 + 50

92 = 50 × 1 + 42

50 = 42 × 1 + 8

42 = 8 × 5 + 2 1


and 8 = 2 × 4 + 0

HCF (510, 92) = 2

92 = 22 × 23

510 = 2 × 3 × 5 × 17

LCM (510, 92) = 22 × 23 × 3 × 5 × 17

= 23460

HCF (510, 92) × LCM (510, 92)

= 2 × 23460 = 46920 1

Product of two numbers = 510 × 92 = 46920

⇒ HCF × LCM = Product of two numbers 1

6. We have 117 = 65 × 1 + 52

65 = 52 × 1 + 13

and 52 = 13 × 4 + 0 1

Hence, HCF = 13

\ 65m – 117 = 13

⇒ 65m = 117 + 13 = 130

\ m =

130

65= 2 1

Now, 65 = 13 × 5

117 = 32 × 13 LCM = 13 × 5 × 32 = 585 1

qqq

TOPIC-2Irrational Numbers, Terminating and Non-Terminating Recurring Decimals

WORKSHEET-6

Solutions

1. The decimal expansion of a rational number

terminates, if the denominator of rational no.

p

q,

when p and q are co-primes and q can be expressed as 2m5n. where m and n are non-negative integers.

e.g.

3

10 =

3

2 51 1

× = 0.3


2. Since

1

7

7

100×

=

1

100 = 0.01.

Thus smallest rational number is

7

100 1


3. Let 5 6 be a rational number, which can be put

in the form a

b, where b ≠0 ; a and b are co-prime.

\ 5 6 = a

b ½

6

=

a

b5

⇒ 6 = rational ½

But, we know that 6 is an irrational number.

Thus, our assumption is wrong.

Hence, 5 6 is an irrational number. 1

4. 15

4

5

40+ =

15

4

25

25

5

40

25

25× + × 1

=

375

100

125

1000+ 1

= 3.75 + 0.125 = 3.875 1

5. Let x = 0 3178.

⇒ x = .3178178178... 1

⇒ 10,000x = 3178.178178...

⇒ 10x = 3.178178...

Subtracting, 9990x = 3175 1

⇒

x = 3175

9990

635

1998= 1

6. If possible, let 3 be a rational number.

(i) \ a

b =

3 , where a and b are integers and co-primes

Squaring both sides, we have

a

b

2

2 = 3

⇒ a2 = 3b2

⇒ a2 is divisible by 3

\ a is divisible by 3. ....(i)

we can write a = 3c for some c (integer)

(3c)2 = 3b2

⇒ 9c2 = 3b2

⇒ b2 = 3c2

⇒ b2 is divisible by 3

⇒ b is divisible by 3 ....(ii)


From equation (i) and (ii), we have

3 is a factor of a and b which is contradicting the fact that a and b are co-primes.

Thus, our assumption that 3 is rational number is

wrong.

Hence, 3 is an irrational number. 2

(ii) let us assume to contrary that 7 + 2 3 is a rational

number.

\ 7 + 2 3 =

p

q, q ≠ 0 and p, q ∈ Integer

7 + 2 3 =

p

q

⇒

2 3 =

p

q− 7

⇒

2 3

=

p q

q

− 7

⇒

3

=

p q

q

− 7

2

p – 7q and 2q both are integers, hence 3 is a

rational number.

But this contradicts the fact that 3 is irrational

number. Hence 7 + 2 3 is an irrational number. 2

qqq

WORKSHEET-7

Solutions

1. A rational number is either terminating or non-terminating repeating. ½

An irrational number is non-terminating and non-repeating. [CBSE Marking Scheme, 2016] ½

2. 3

8 =

3

23

=3 5

2 5

3

3 3

×

×

= 375

103

= 375

1 000,

= 0.375 1

3. Denominator = 500

= 22 × 53 1

Decimal expansion,

257

500 =

257 2

2 2 5

514

102 3 3

×

× ×

=

= 0.514 1


4. Let 3 be a rational number.

3 = a

b.

(a and b are integers and co-primes)

On squaring both the sides,

3 =

a

b

2

2 1

⇒ 3b2 = a2

⇒ a2 is divisible by 3

\ a is divisible by 3 ...(i)

We can write a = 3c for some integer c

⇒ a2 = 9c2

⇒ 3b2 = 9c2

⇒ b2 = 3c2 1

⇒ b2 is divisible by 3

⇒ b is divisible by 3 ...(ii)

From equations (i) and (ii), we get 3 as a factor of ‘a’ and ‘b’,

which is contradicting the fact that a and b are co-

primes. Hence our assumption that 3 is a rational

number, is false. So 3 is an irrational number. 1

5. Let 2 be a rational number.

\

2 =

a

b,

(a, b are co-prime integers and b ≠ 0)

a = 2 b

Squaring, a2 = 2b2 1

⇒ 2 divides a2

⇒ 2 divides a.

So we can write a = 2c for some integer c, substitute for a, 2b2 = 4c2, b2 = 2c2

This means 2 divides b2, so 2 divides b.

\ a and b have ‘2’ as a common factor. 1

But this contradicts that a, b have no common factor other than 1.

\ Our assumption is wrong.


Hence, 2 is irrational.

Let

3

2 be rational

\ 3

2 =

a

b, where a and b are

integers, b ≠ 0

⇒ 3b = 2a

⇒

2 =

3b

a 1

3b

a is rational but 2 is not rational.

\ Our assumption is wrong.

\

3

2 is irrational. 1

qqq

WORKSHEET-8

Solutions

HOTS & Value Based Questions

1. Let x = 571

⇒ x = 571

Now 571 lies between the perfect squares of (23)2 and (24)2

Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23. 1

Since 571 is not divisible by any of the above numbers.

So, 571 is a prime number. 1

2. The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 1

\ LCM = 2 × 2 × 3 × 2 × 3 × 5 × 7

= 2520 1

3. Let the number of columns be x.

x is the largest number, which should divide both 104 and 96

104 = 96 × 1 + 8 1

96 = 8 × 12 + 0 1

\ HCF of 104 and 96 is 8 1

Hence, 8 columns are required. 1

4. Using Euclid’s algorithm, the HCF (30, 72)

72 = 30 × 2 + 12 ...(i)

30 = 12 × 2 + 6 ...(ii)

12 = 6 × 2 + 0 ...(iii)

HCF (30, 72) = 6 1

6 = 30 – 12 × 2 [From (ii)]

6 = 30 – (72 – 30 × 2) × 2

[From (i)]

6 = 30 – 2 × 72 + 30 × 4

6 = 30 (1 + 4) – 72 × 2

6 = 30 × 5 + 72 × (– 2)

\ x = 5, y = – 2 1

Also, 6 = 30 × 5 + 72(– 2) + 30 × 72 – 30 × 72 1

6 = 30 × (77) + 72 × (–32)

\ x = 77, y = – 32

Hence, x and y are not unique.

5. (i) Required number of minutes is the LCM of 18 and 12.

18 = 2 × 32

and 12 = 22 × 3 1

\ LCM of 18 and 12 = 22 × 32 = 36

Hence, Ravish and Priya will meet again at the starting point after 36 minutes. 1

(ii) LCM of numbers. 1

(iii) Healthy competition is necessary for personal development and progress. 1

qqq


POLYNOMIALS

SECTION

BSECTIONCHAPTER

2 TOPIC-1

Zeroes of a Polynomial and Relationship between Zeroes and Coefficients of Quadratic Polynomials

WORKSHEET-9

Solutions

1. Sum of the roots = –

Coefficient f

Coefficient f

o

o

x

x2

⇒

a

+ b = – -

b

a

=

b

a 1

2. Product of roots (zeroes)

=

2

1

a

=

α

α.2

1

⇒ 2a = 2 ⇒ a = 1


3. Q f (x) = x2 – 2x 1 = x (x – 2)

i.e. f (x) = 0 ⇒ x = 0 or x = 2 Hence, zeroes are 0 & 2. 1

4. Let p (x) = 3 x2 – 8x + 4 3

= 3 x2 – 6x – 2x + 4 3 1

= 3 x (x – 2 3 ) – 2(x – 2 3 )

= ( 3 x – 2) (x – 2 3 )

\ Zeroes are , x=

2

3, 2 3

1

5. Given, a and 2

αare the zeroes of x2 + 4x + 2a.

We know that,

Product of the zeroes =

Constant term

Coefficient of 2x

⇒

a ×

2

α =

2

1

a

⇒ 2 = 2a

\ a = 1 3

6. Factors of x2 + 7x + 12 :

x2 + 7x + 12 = 0

⇒ x2 + 4x + 3x + 12 = 0

⇒ x(x + 4) +3(x + 4) = 0

⇒ (x + 4) (x + 3) = 0

⇒ x = – 4, –3 ...(i) 1

Let p’(x) = x4 + 7x3 + 7x2 + px + q

If p(x) is exactly divisible by x2 + 7x + 12, then x = – 4 and x = – 3 are zeroes of p(x) [from eq (i)] 1

p(x) = x4 + 7x3 + 7x2 + px + q

p(– 4) = (– 4)4 + 7(– 4)3 + 7(– 4)2 + p(– 4) + q

but p(– 4) = 0

\ 0 = 256 – 448 + 112 – 4p + q

⇒ 0 = – 4p + q – 80

⇒ 4p – q = – 80 ...(ii)

and p(–3) = (–3)4 + 7(–3)3 + 7(–3)2 + p(–3) + q

but p(–3) = 0

\ 0 = 81 – 189 + 63 – 3p + q

⇒ 0 = –3 p + q – 45 1

⇒ 3p – q = – 45 ...(iii)

On solving eq. (ii) and eq. (iii) by elimination method, we get

4p – q = – 80

3p – q = – 45

– + +

p = – 35

On putting the value of p in eq. (i),

4(–35) – q = – 80

⇒ –140 – q = – 80

⇒ – q = 140 – 80

⇒ – q = 60

\ q = – 60

Hence, p = – 35, q = – 60 1

qqq


WORKSHEET-10

Solutions

1. The number of zeroes of p(x) is 1. 1

2. Sum of zeroes = 6, Product of zeroes = 9

\ Quadratic polynomial is x2 – 6x + 9 1

Also x2 – 6x + 9 = 0

⇒ (x – 3) (x – 3) = 0

⇒ x = 3, 3

Hence zeroes are 3, 3. 1

3. According to the question,

Sum of zeroes =

21

8

and Product of zeroes =

5

16 1

So, quadratic polynomial

= x2 – (Sum of zeroes) x

+ Product of zeroes

= x2 –

21

8

x + 5

16

=

1

16 (16x2 – 42x + 5)

= (16x2 – 42x + 5)

1

16. 1

4. Given, polynomial, f(x) = ax2 – 5x + c

Let the zeroes of f (x) are a and b, then according to the question

Sum of zeroes, (a + b) = Product of zeroes, (ab) = 10

Now a + b = –

Coeff. of

Coeff. of 2

x

x a= −

−5 1

⇒ 10 = +5

a

\ a =

1

2

and ab = Constant term

Coeff. of 2x 1

⇒ 10 = 2c

\ c = 5

Hence a =

1

2 and c = 5 1

5. Let a and b be the zeroes of the polynomial, then as per the question

b = 7a

\ a + 7a = 8a = – -8

3

½

⇒ a =

1

3 ½

and a × 7a =

2 1

3

k +

⇒ 7a2 =

2 1

3

k +

⇒ 7 1

3

2

=

2 1

3

k +

1

⇒ 7 ×

1

9 =

2 1

3

k +

⇒

7

3 – 1 = 2k

\

2

3 = k 1

qqq

WORKSHEET-11

Solutions

1. Let p(x) = 4x2 – 12x + 9

= 4x2 – 6x – 6x + 9

⇒ p(x) = 2x(2x – 3) – 3(2x – 3)

0 = (2x – 3) (2x – 3)

The zeros are

3

2,

3

2 ⇒x =

3

2,

3

2

Hence, zeroes of the polynomial are

3

2, 3

2. 1

2. x – 3 ) x3 – 3x2 + 6x – 15 ( x2 + 6

x3 – 3x2

– +

6x – 15 1

+ 6x – 18 – +

+ 3

Remainder = 3

Hence – 3 must be added 1



3. Let p(x) = 3x2 + 11x – 4

= 3x2 + 12x – x – 4

= 3x(x + 4) – 1(x + 4) ½

= (3x – 1) (x + 4)

So, zeroes are, m =

1

3 and n = – 4 ½

Now, m

n+n

m =

1

3

4

4

1

3

−+

−

= 1

1212

-- ½

=

-145

12 ½

4. f (x) = 2x2 – 7x + 3

Sum of roots = p + q = –

Coefficient of

Coefficient of

x

x2

= –

-7

2

7

2

= ½

Product of roots = pq = Constant term

Coefficient of 2x =

3

2 ½

We know that

(p + q)2 = p2 + q2 + 2pq

⇒ p2 + q2 = (p + q) 2 – 2pq ½

=

7

23

49

4

3

1

37

4

2

= =- - ½


5. p(y) = 6y2 – 7y + 2

a + b = – -

7 7

66

=

and

ab =

2

6 =

1

3

Now 1 1

α β+ =

α + β

αβ =

7 6

2 6

7

2

/

/=

1

and

1 1

α β× =

1

αβ = 3 1

The required polynomial is

y2 –7

2y + 3 =

1

2 [2y2 – 7y + 6]. 1

qqq

WORKSHEET-12

Solutions

1. p(x) = ax2 + bx + c

Let a and

1

α be the zeroes of p(x), then

Product of zeroes,

a ×

1

α =

c

a 1

So, required condition is, c = a 1 2. Since, – 1 is a zero of the polynomial p(x) = kx2 – 4x + k, then p (–1) = 0 1 \ k (–1)2 – 4 (–1) + k = 0 ⇒ k + 4 + k = 0 ⇒ 2k + 4 = 0

⇒ 2k = – 4

Hence, k = – 2 1

3. x2 – 4 3x + 3 = 0

If a and b are the zeroes of x2 – 4 3 x + 3.

then a + b = –

b

a

⇒ a + b = –

( )−4 3

1

⇒ a + b = 4 3

and

ab = c

a

⇒ ab = 3

1

⇒ ab = 3

\ a + b – ab= 4 3 – 3. 2

4. Given, p(x) = 3x2 – 4x – 7 and a and b are its zeroes.

Sum of zeroes = a + b = – Coefficient of

Coefficient of 2

x

x

= – -4

3

4

3

=

Product of zeroes = ab =

Constant term

Coefficient of 2x

=

−

= −7

3

7

3 1

For the new polynomial,

Sum of zeroes = 1 1

α β+ =

α β

αβ

+=

−

=−

4

37

3

4

7


Product of zeroes= 1 1

α β× =

1 1

7

3

3

7αβ=

−=

− 1

\ Required quadratic polynomial = x2 – (Sum of zeroes) x + Product of zeroes

= x2 –

- -4

7

3

7

+

x

= 1

7 (7x2 + 4x – 3)

= (7x2 + 4x – 3)

1

7. 1

5. From the given polynomial we will find the value, the sum of the zeroes, and the multiple of the zeroes.

a

+ b =

−ba

= -8

1 = – 8

a × b =

c

a =

6

1 = 6 1

Sum of zeroes =

1 1

α β+

=

α β

αβ

+

=-8

6=

-4

3 ½

Product of the zeroes =

1 1 1 1

6α β αβ× = =

½

Now for making a polynomial

p(x) = x2 – (a + b)x + ab

\The polynomial is : p(x) =

1

6 (6x2 + 8x + 1) 1

qqq

WORKSHEET-13

Solutions

1. f(x) = x2 – 7x – 8

Let other zero be k, then

Sum of zeroes – 1 + k = –

-7

1

= 7

⇒

k = 8 1

2. Sum of zeroes = – Coefficient of

Coefficient of

x

x2

\ a + b = – a ⇒ 2a + b = 0 1

Product of zeroes =

Constant term

Coefficient of x2

\ ab = b ⇒ a = 1

then b = – 2 1

3. Given, f(x) = 14x2 – 42k2x – 9 Let one zero be a \ The other = – a, \ Sum of zeroes = a + (– a) = 0

Sum of zeroes = – Coefficient of

Coefficient of

x

x2 1

According to the question,

Sum of zeroes = 42

14

2k = 3k2

\ 3k2 = 0 ⇒ k = 0. 1

4. f(x) = 4x2 + 4x – 3

⇒

f1

2

= 41

44

1

23

+

-

= 1 + 2 – 3 = 0

and

f -

3

2

= 49

44

3

23

+

- -

= 9 – 6 – 3 = 0

\

1

2

3

2,- are zeroes of polynomial 4x2 + 4x – 3. 1

Sum of zeroes =

1

2

3

2- = – 1 =

-4

4

= – Coefficient of

Coefficient of 2

x

x 1

Product of zeroes =

1

2

3

2

3

4

−

=−

=

Constant term

Coefficient of 2x

\ Relation between zeroes and coeff. of polynomial is verified. 1

5. p (x) x2 –2 2 x = x(x – 2 2 ), p (x) = 0

⇒ x (x – 2 2 ) = 0

⇒ zeroes are 0 and 2 2

Sum of zeroes = 2 2 = –

Coefficient of

Coefficient of 2

x

x

and

Product of zeroes = 0 =

Constant term

Coefficient of 2x 3


qqq


TOPIC-2Problems on Polynomials

WORKSHEET-14

Solutions

1. x3 – 5x2 + 6x + 4 = g(x) (x – 3) + 4

⇒ g(x) = x x x

x

3 25 6

3

− +

− 1

x – 3 ) x3 – 5x2 + 6x ( x2 – 2x

+ x3 – 3x2

– +

– 2x2 + 6x

– 2x2 + 6x

+ –

×

Hence g(x) = x2 – 2x. 1


2. 2x + 5

2x2 – x + 1) 4x3 + 8x2 + 8x + 7

4x3 – 2x2 + 2x – + –

+ 10x2 + 6x + 7

+ 10x2 – 5x + 5

– + –

+ 11x + 2

Thus, Quotient = 2x + 5 1

and Remainder = 11x + 2 1


3. x2 + 2x ) x3 + 5x2 + 7x + 3 ( x+ 3 1

x3 + 2x2

3x2 + 7x + 3

3x2 + 6x

Remainder = x + 3 1

Hence – (x + 3) must be added. 1

[CBSE Marking Scheme, 2016] 4. 2x + 2

3x2 – 2x + 1) 6x3 + 2x2 – 4x + 3

6x3 – 4x2 + 2x

– + –

6x2 – 6x + 3

6x2 – 4x + 2 – + – – 2x + 1

Quotient = 2x + 2; Remainder = – 2x + 1 1

p(x) = g(x) q(x) + r(x)

= (3x2 – 2x + 1) (2x + 2) + (–2x + 1)

= 6x3 – 4x2 + 2x + 6x2 – 4x + 2 – 2x + 1 1

= 6x3 + 2x2 – 4x + 3 Verified 1

5. x = – 1 and x = – 3 are zeroes. 1

⇒ (x + 1 ) (x + 3) = x2 + 4x + 3

x2 + 4x + 3 ) x3 + 5x2 + 7x + 3 ( x + 1

x3 + 4x2 + 3x 1

– –

x2 + 4x + 3

x2 + 4x + 3

– – 1

0

\ x + 1 = 0

⇒ x = – 1

\ The third zero is – 1. 1


qqq

WORKSHEET-15

Solutions

1. g(x) = x2 + 3x + 1, f(x) = 3x4 + 5x3 – 7x2 + 2x + 4

½

3x2 – 4x + 2

x2 + 3x + 1) 3x4 + 5x3 – 7x2 + 2x + 4

3x4 + 9x3 + 3x2

– – –

– 4x3 – 10x2 + 2x

– 4x3 – 12x2 – 4x

+ + +

2x2 + 6x + 4

2x2 + 6x + 2

– – –

2 \ Remainder, r(x) = 2 1½ \ r(x) ≠ 0, g(x) is not a factor of p(x).



2. x – 3

x2 – 3x + 2) x3 – 6x2 + 11x + 8 x3 – 3x2 + 2x – + – – 3x2 + 9x + 8 – 3x2 + 9x – 6 + – + 14 Remainder = 14 So – 14 should be added. 2


3. 2x2 + 2x – 1

4x2 + 3x – 2) 8x4 + 14x3 – 2x2 + ax + b

8x4 + 6x3 – 4x2

– – +

8x3 + 2x2 + ax

8x3 + 6x2 – 4x

– – +

– 4x2 + (a + 4)x + b

– 4x2 – 3x + 2

+ + – (a + 7)x + b – 2

For exact division, remainder is zero, then

(a + 7) x + b – 2 = 0 2

⇒ a + 7 = 0, b – 2 = 0

⇒ a = – 7, b = 2. 1

4. x2 – 4x + (8 – k)

x2 – 2x + k) x4 – 6x3 + 16x2– 25x + 10

x4 – 2x3 + kx2

– + –

– 4x3 + (16 – k)x2 – 25x + 10

– 4x3 + 8x2 – 4kx 1

+ – +

(8 – k)x2 – (25 – 4k)x + 10

(8 – k)x2 – (16 – 2k)x + (8k – k2)

– + –

(2k – 9)x + (10 – 8k + k2) 1

Given, remainder = x + a

⇒ 2k – 9 = 1 ⇒ k =

10

2 = 5 1

and a = 10 – 8k + k2 = 10 – 40 + 25

= – 5 1


qqq

WORKSHEET-16

Solutions

1. 3x2 + 4x + 1) 6x4 + 8x3 + 17x2 + 21x + 7 ( 2x2 + 5

6x4 + 8x3 + 2x2

– – –

15x2 + 21x + 7

15x2 + 20x + 5

– – –

x + 2

\ ax + b = x + 2

On comparing, we get

a = 1 and b = 2. [CBSE Marking Scheme, 2015] 2

2. x2 + x + 3

x2 – 2 ) x4 + x3 + x2 – 2x – 3

x4 – 2x2

– +

x3 + 3x2 – 2x

x3 – 2x 1

– +

3x2 – 3

3x2 – 6

– +

+ 3 1

Remainder ≠ 0

⇒ x2 – 2 is not a factor of the given polynomial. 1

3. p(x) = 2x3 – x2 – 13x – 6 p(3) = 2(3)3 – (3)2 – 13(3) – 6 ½ = 2(27) – 9 – 39 – 6 = 54 – 54 = 0 ½ ⇒ x – 3 is a factor of p(x). ½

p x

x

( )

− 3 =

2 13 6

3

3 2x x x

x

− − −

− 1

By long division,

2x2 + 5x + 2 x – 3 ) 2x3 – x2 – 13x – 6 2x3 – 6x2

– + 5x2 – 13x 5x2 – 15x – + 2x – 6 2x – 6 0 0


p x

x

( )

− 3 = 2x2 + 5x + 2

= 2x2 + 4x + x + 2 ½

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

⇒

−1

2, – 2 are the other zeroes of p(x)

½

All the zeroes of p(x) are

−1

2, – 2, 3. ½

qqq

WORKSHEET-17

Solutions

1. –x

–x2 + 2) x3 – 4x + 6

x3 – 2x

– +

– 2x + 6

Thus, Quotient = – x

and Remainder = – 2x + 6. 2

2. x – 3

x – 2) x2 – 5x + 16

x2 – 2x

– +

– 3x + 16

– 3x + 6

+ –

10

Quotient = x – 3, Remainder = 10 2


3. On dividing x3 + 2x2 – 9x + 1 by x + 4, remainder should be zero.

x + 4) x3 + 2x2 – 9x + 1 (x2 – 2x – 1

x3 + 4x2

– –

– 2x2 – 9x + 1

– 2x2 – 8x

+ +

– x + 1

– x – 4

+ +

5 2

Hence –5 should be added to x3 + 2x2 – 9x + 1 to make it completely divisible by x + 4. 1

4. As x = 3 2 and –3 2 are the zeroes of 4x4 + x3

– 72x2 – 18x, So (x – 3 2 ) and (x + 3 2 ) are the

factors of 4x4 + x3 – 72x2 – 18x

Now, (x – 3 2 ) (x + 3 2 ) = 0

⇒ x2 – 18 = 0

Dividing 4x4 + x3 – 72x2 – 18x by x2 – 18

x2 – 18) 4x4 + x3 – 72x2 – 18x (4x2 + x 1

4x4 – 72x2

– +

x3 – 18x

x3 – 18x

– +

0

\ 4x4 + x3 – 72x2 – 18x = (x2 – 18) (4x2 + x)

= (x2 – 18) x(4x + 1)

= (x – 3 2 ) (x + 3 2 )(x) (4x + 1) 2

Hence other two zeroes are 0 and -1

4. 1

qqq

WORKSHEET-18

Solutions


1. If x + 7 is a factor then (–7) is a root.

So f (–7) = (–7)4 + 10 (–7)3 + 25(–7)2 + 15(–7) + k = 0 (when it is a root the polynomial should be equal to zero when value is substituted) 1

2401 – 3430 + 1225 – 105 + k = 0 ⇒ 3626 – 3535 + k = 0 ⇒ 91 + k = 0 \ k = (–91) 1

2. As x = 2 ±

3

are the zeroes of p(x), so x –

2 ±( )3 are the factors of p(x). 1

Now, x x− +( ){ } − −( ){ }2 23 3

= ( ) ( )x x− −{ } − +{ }2 3 2 3

= ( )x − − ( )2 32 2

= x2 – 4x + 1 1

Dividing p(x) by x2 – 4x + 1


x2 – 2x – 35

x2 – 4x +1) x4 – 6x3 – 26x2 + 138x – 35

x4 – 4x3 + x2

– + –

– 2x3 – 27x2 + 138x

– 2x3 + 8x2 – 2x

+ – +

–35x2 + 140 x – 35

–35x2 + 140 x – 35

+ – +

0

\ p(x) = x − +( ){ }2 3 x − −( ){ }2 3

x x2

2 35− −( ) As, x2 –2x – 35 = (x + 5)(x – 7) Hence, other two zeroes of p(x) are – 5 and 7. 1 3. Dividend = (Divisor × Quotient) + Remainder

Hence, we get 4x4 – 5x3 – 39x2 – 46x – 2 = g(x) (x2 – 3x – 5) + (– 5x + 8)

1

⇒ 4x4 – 5x3 – 39x2 – 46x – 2 + 5x – 8

= g(x) (x2 – 3x – 5) ⇒ 4x4 – 5x3 – 39x2 – 41x – 10

= g(x) (x2 – 3x – 5)

⇒ g(x) =

4 5 39 41 10

3 5

4 3 2

2

x x x x

x x

− − − −

− −( )

1

4x2 + 7x + 2

x2 – 3x – 5 ) 4x4 – 5x3 – 39x2 – 41x – 10

4x4 – 12x3 – 20x2

– + +

7x3 – 19x2 – 41x

7x3 – 21x2 – 35x – + +

2x2 – 6x – 10

2x2 – 6x – 10 – + +

0

Therefore, g(x) = 4x2 + 7x + 2 1

qqq


PAIR OF LINEAR EQUATIONS IN

TWO VARIABLES

SECTION

BSECTIONCHAPTER

3TOPIC-1Graphical Solution of Linear Equations in Two Variables Consistency/Inconsistency

WORKSHEET-19

Solutions

1. Here a1 = 2, b1 = 1, c1 = 3

and a2 = 4, b2 = 2, c2 = 6

Clearly,

a

a1

2

=

b

b1

2

=

c

c1

2

1

i.e.

2

4 =

1

2 =

3

6 1

Hence lines are coincident.[CBSE Marking Scheme, 2016]

2. The given equations can be re-written as :

3x + 2y – 5 = 0 2x – 3y – 7 = 0 On comparing with ax + by + c = 0, we have

a1 = 3, b1 = 2, c1 = – 5 a2 = 2, b2 = – 3, c2 = – 7

a

a1

2 =

3

2,

b

b1

2

=

2

3- 1

Thus, 3

2 ≠

2

3-

i.e. a

a1

2

≠ b

b1

2

Hence, the pair of linear equations is consistent. 1

3. The given equations can be re-written as

2x – 3y – 8 = 0

4x – 6y – 9 = 0

On comparing with ax + by + c = 0, we have

a1 = 2, b1 = – 3, c1 = – 8

a2 = 4, b2 = – 6, c2 = – 9

a

a1

2

= 2

4 =

1

2,

b

b1

2

= -

-

3

6 =

1

2

and c

c1

2

= -

-

8

9 =

8

9 1

Thus, 1

2 =

1

2 ≠

8

9

i.e. a

a1

2

= b

b1

2

≠ c

c1

2

Hence, the pair of linear equations is inconsistent. 1

4. Let amounts contributed by two sections X-A and X-B be ` x and ` y.

x + y = 1,500 ...(i)

1

y – x = 100 ....(ii) 1

1500

800 (700, 800)

700 15,00

Point of intersection = (700, 800) 1 Hence x = 700 and y = 800. 1


qqq

WORKSHEET-20

Solutions

1. Since, am = bl ⇒

a

l =

b

m

c

n≠

So, ax + by = c and lx + my = n has no solution. 1


2. The given equations can be re-written as :

3

2x +

5

3y – 7 = 0

9x – 10y – 14 = 0

On comparing with ax + by + c = 0, we have

a1 =

3

2, b1 =

5

3, c1 = – 7

a2 = 9, b2 = – 10, c2 = – 14

a

a1

2

= 3

18 =

1

6

b

b1

2

= 5 3

10- =

-1

6 1

Thus, 1

6 ≠

-1

6

i.e., a

a1

2

≠ b

b1

2 1

Hence, the pair of linear equations is consistent. 3. The pair of linear equations is : x – y = 8 ⇒ x – y – 8 = 0 …(i) and 3x – 3y = 16 ⇒ 3x – 3y – 16 = 0 …(ii) 1 On comparing with ax + by + c = 0, we have a1 = 1, b1 = – 1, c1 = – 8 ½ a2 = 3, b2 = – 3, c2 = – 16

\

a

a1

2

= 1

3,

b

b1

2

= 1

3,

c

c1

2

= 1

2 ½

As a

a

b

b

c

c

1

2

1

2

1

2

= ≠

the lines are parallel having no solution. ½ So, the pair of linear equations is inconsistent. ½

qqq

WORKSHEET-21

Solutions

1. ad ≠ bc ⇒

a

c ≠

b

d

Hence, the pair of given linear equations has unique solution. 1

2. 3

6 ≠

2

4−

i.e., a

a1

2

≠

b

b1

2

1

Hence, the pair of linear eqations is consistent. 1[CBSE Marking Scheme, 2016]

3. Yes, it is consistent. We have, for the equation

2x + 3y – 9 = 0

a1 = 2, b1 = 3 and c1 = – 9 ½

and for the equation, 4x + 6y – 18 = 0

a2 = 4, b2 = 6 and c2 = – 18 ½

Here

a

a1

2

= 2

4 =

1

2

b

b1

2

= 3

6 =

1

2

and c

c1

2

= -

-

9

18 =

1

2 ½

It is clear that

a

a1

2

= b

b1

2

=

c

c1

2

½

Hence, system is consistent and dependent.

4. Given linear equation is 3x + 4y = 9 (i) Intersecting line is 3x – 5y = 10 1 (ii) Coincident line is 6x + 8y = 18. 1


5. x + 2y = 5

⇒ y = 5

2

- x

x 1 3 5

y 2 1 0

2x – 3y = – 4

⇒ y = 2 4

3

x +

x 1 4 –2

y 2 4 0

Graph of two equations is :

5

4

3

2

1

0x

y

1 2 3 4 5Marks

– 5 – 4 – 3 – 2 – 1

– 1

– 2

– 3

– 4

– 5

y'

x'

(1, 2)

(3, 1)

(5, 0)

(4, 4) 2– 3

= – 4

x

y

(–2, 0)

x y+ 2 = 5

Lines meet x-axis at (5, 0) and (– 2, 0) respectively.



WORKSHEET-22

Solutions

1. Given equations are :

4x + py + 8 = 0 ...(1)

2x + 2y + 2 = 0 ...(2)

The condition of unique solution,

a

a1

2

≠b

b1

2

1

Hence,

4

2 ≠

p

2 or

2

1 ≠

p

2

\ p ≠ 4

1

2. Pair of linear equations kx – 4y – 3 = 0

and 6x – 12y – 9 = 0

Condition for infinite solutions :

a

a1

2 =

b

b1

2

=

c

c1

2 1

k

6 =

-

-

4

12 =

3

9

⇒ k = 2 1

3. For equation, 2x + 3y – 4 = 0

a1 = 2, b1 = 3, c1 = – 4 ½

For equation, (k + 2) x + 6y – (3k + 2)=0

a2 = k + 2, b2 = 6, c2 = – (3k + 2) ½

For infinitely many solutions

a

a1

2

= b

b1

2

= c

c1

2

⇒

2

2k +

= 3

6 =

4

3 2k + ⇒12 = 3k + 6

⇒ 6 = 3k ⇒ k = 2 1

4. 3x – 4y + 3 = 0

⇒ y = 3 3

4

x +

x 3 7 –1

y 3 6 0 and 3x + 4y – 21 = 0

⇒ y = 21 3

4

- x

x 3 7 11

y 3 0 –3

5

4

3

2

1

0

x

y

1 2 3 4 5– 5 – 4 – 3 – 2 – 1

y'

x'

(3, 3)

(3, 1)

(7, 0)

(7, 6)

3– 4

+ 3 =

0

x

y

(–1, 0)

6

7

8

– 4

– 5

– 6

– 7

– 8

– 3

– 2

– 1– 6– 7 6 7 8 9 10 11

– 9

9

(11, –3)

3x + 4– 21 = 0

y

(i) These lines intersect each other at point (3, 3). Hence x = 3 and y = 3.

(ii) The vertices of triangular region are (3, 3), (–1, 0) and (7, 0).

(iii) Area of ∆ = 1

2 × 8 × 3

⇒ Area = 12 sq unit. 4

qqq

TOPIC-2Algebraic Methods to Solve Pair of Linear Equations and Equations Reducible to Linear Equations

WORKSHEET-23

Solutions

1. x + 2y – 2 = 0

x – 3y – 7 = 0

x

− −14 6 =

y

− +2 7 =

1

3 2− − 1

⇒

x

−20 =

y

5 =

−1

5

⇒ x = 4

and y = – 1 1[CBSE Marking Scheme, 2016]

2. Let age of father and son be x and y respectively.

x + y = 40 ...(i)

x = 3y ...(ii)

By solving eqns. (i) and (ii), we get

x = 30 and y = 10

Ages are 30 years and 10 years.



Alternative Method :

Let age of father and son be x and y years respectively. x + y = 40 ...(i) x = 3y ...(ii) Substitute the value of x from eqn. (ii) in, (i) \ 3y + y = 40 ⇒ 4y = 40 \ y = 10 yrs. From (i), x + 10 = 40 \ x = 30 yrs. \ Age of father is 30 years and Age of son is 10 years. 3

3. Let cost of 1 chair = ` x and cost of 1 table = ` y

According to the question, 4x +3y = 2100 ...(i) 5x +2y = 1750 ...(ii) Multiplying eqn. (i) by 2 and eqn. (ii) by 3, 8x +6y = 4200 ...(iii) 15x + 6y = 5250 ...(iv) eqn. (iv) – eqn. (iii) ⇒ 7x = 1050 ⇒ x = 150 Substituting the value of x in (i), y = 500 Cost of chair and table = ` 150, ` 500 respectively. [CBSE Marking Scheme, 2015] 4

4. Substitute 1

x

= X and

1

y = Y

⇒ 2X + 3Y = 2 ....(i)

4X – 9Y = – 1 ....(ii)

Multiply eqn. (i) by 3,

6X + 9Y = 6 ....(iii)

Adding eqn. (iii) and eqn. (ii),

4X – 9Y = – 1

6X + 9Y = 6

+ + +

10X = 5

10X = 5 ⇒ X =

5

10

1

2=

⇒

1 1

2x= ⇒x = 4

Now 3Y = 2 – 2X

= 2 – 2

1

2

= 1

⇒ Y =

1

3

1 1

3⇒ =

y

⇒ y = 9

Thus x = 4, y = 9 is the solution. 4

qqq

WORKSHEET-24

Solutions

1. 3x + 2y – 7 = 0 ....(i)

4x + y – 6 = 0 ....(ii)

From eqn. (ii).

4x + y – 6 = 0

⇒ y = 6 – 4x ½

Value of y put in eqn. (i)

3x + 2y – 7 = 0

⇒ 3x + 2(6 – 4x) – 7 = 0

⇒ 3x + 12 – 8x – 7 = 0

⇒ 5 – 5x = 0 ½

⇒ 5x = 5

\ x = 1 ½

Substitute the value of x in eq (ii) to get value of y,

4x + y – 6 = 0

⇒ 4(1) + y – 6 = 0

⇒ 4 + y – 6 = 0

⇒ y – 2 = 0

\ y = 2 ½

Hence, values of x and y are 1 and 2 respectively.

2. x

- - 80 48 =

y

- - - 4 100

1

60 4+=

⇒ x = 2

and y =

- 3

2 3


Alternative Method : Given : 5x + 4y – 4 = 0 ...(i)

x – 12y – 20 = 0 ...(ii)

By cross-multiplication method,

x

4 4

12 20

-

- -

=

y

-

-

4 5

20 1

=

1

5 4

1 12-

⇒

x

- -80 48 =

y

- - -4 100

1

60 4+=

x

-128 =

1

64- and

y

96 =

1

64-

⇒ x = 2 and y =

-3

2 3

3. 2x – y + 3 = 0 ...(i)

3x – 5y + 1 = 0 ...(ii)

Multiplying eqn. (i) by 5, we get

5(2x – y + 3) = 0

⇒ 10x – 5y + 15 = 0

Subtracting eqn. (ii) from eqn. (i), we get


10x – 5y + 15 = 0 3x – 5y + 1 = 0 (–) (+) (–) 7x + 14 = 0 ⇒ 7x = – 14

\ x =

-14

7 = – 2

Substituting the value of x in eqn. (i),

2x – y + 3 = 0 ⇒ 2 (–2) – y + 3 = 0 ⇒ – 4 – y + 3 = 0 ⇒ – y – 1 = 0 ⇒ – y = 1

\ y = – 1

Hence,

x = – 2

and y = – 1 4

qqq

WORKSHEET-25

Solutions

1. By elimination method : Given, 3x + 4y = 10 …(i) and 2x – 2y = 2 …(ii) On multiplying eqn. (i) by 1 and eqn. (ii) by 2 and

adding, 1(3x + 4y) + 2(2x – 2y) = 1 × 10 + 2 × 2 1 ⇒ 3x + 4y + 4x – 4y = 10 + 4 ⇒ 7x = 14 \ x = 2 On substituting x = 2 in eqn. (i), 3 × 2 + 4y = 10 ⇒ 4y = 4 \ y = 1 Hence, x = 2 and y = 1 1 By substitution method : Given, 3x + 4y = 10 …(i) and 2x – 2y = 2 …(ii) From eqn. (ii), 2y = 2x – 2 ⇒ y = x – 1 …(iii) On substituting y = x – 1 from eqn. (iii) in eqn. (i), 1 3x + 4(x – 1) = 10 ⇒ 7x = 14 \ x = 2 From eqn. (iii), y = 2 – 1 = 1 \ y = 1 Hence, x = 2 and y = 1 1 2. By elimination method : Given, 3x – 5y = 4 …(i) and 9x = 2y + 7 …(ii) On multiplying eqn. (i) by 3 and eqn. (ii) by 1, 9x – 15y = 12 …(iii) 9x – 2y = 7 …(iv) On subtracting eqn. (iv) from eqn. (iii), (9x – 15y) – (9x – 2y) = 12 – 7 ⇒ – 15y + 2y = 5 ⇒ –13y = 5

\ y = – 5

13 1

On substituting y =

-5

13 in eqn. (i),

3x – 5

-5

13

= 4

⇒ 39x + 25 = 52

⇒ 39x = 27

\ x =

9

13

Hence x =

9

13 and y = –

5

13 1

By substitution method : Given, 3x – 5y = 4 …(i) and 9x = 2y + 7 …(ii) From eqn. (ii), 2y = 9x – 7

⇒ y =

9 7

2

x - …(iii)

On substituting y from eqn. (iii) in eqn. (i),

3x – 5 ×

9 7

2

x -

= 4

⇒ 6x – 45x + 35 = 8 ⇒ – 39x = – 27

\ x = 9

13

On substituting x = 9

13 in eqn. (iii), 1

y =

99

137

2

× -

= 81 91

2 13

-

×

= –

10

26

\ y = –5

13

Hence, x = 9

13 and y = –

5

13 1

qqq


WORKSHEET-26

Solutions

1. Given, a pair of linear equations is :

s – t = 3

⇒ s = t + 3 ...(i)

and

s

3 2+t

= 6 ...(ii)

On substituting s = t + 3, from eqn. (i) in eqn. (ii), we get

t t++

3

3 2 = 6

⇒ 2(t + 3) + 3t = 36 ½

⇒ 5t + 6 = 36

⇒ 5t = 30

⇒ t = 6 ½

From eqn., (i), s = 6 + 3 = 9 ½

Hence, s = 9, t = 6 ½

2. Let the actual speed of the train be x km/hr and actual time taken be y hr.

Distance = Speed × Time ½ Q = xy km According to the given condition, xy = (x + 10)(y – 2) ⇒ xy = xy – 2x + 10y – 20 ⇒ 2x – 10y + 20 = 0 ⇒ x – 5y + 10 = 0 [divide by 2] …(i) ½ and xy = (x – 10)(y + 3) ⇒ xy = xy + 3x – 10y – 30 ⇒ 3x – 10y – 30 = 0 …(ii) 1 On multiplying eqn. (i) by 3 and subtracting eqn. (ii)

from eqn. (i), 3 × (x – 5y + 10) – (3x – 10y – 30) = 0

⇒ – 5y = – 60 \ y = 12 1 On substituting y = 12 in eqn. (i), x – 5 × 12 + 10 = 0 ⇒ x – 60 + 10 = 0 ⇒ x = 50 Hence, the distance covered by the train = 50 × 12 = 600 km. 1

3. Let the incomes of two persons be 11x and 7x.

Also the expenditures of two persons be 9y and 5y.

\ 11x – 9y = 400 ...(i)

and 7x – 5y = 400 ...(ii)

Multiplying eqn. (i) by 5 and eqn. (ii) by 9 and subtracting,

55x – 45y = 2,000 ...(iii)

63x – 45y = 3,600 ...(iv) 1

– + –

On subtracting, – 8x = –1600

⇒ x =

-

-

1 600

8

, = 200 1

Substituting this value of x in eqn. (i),

2200 – 9y = 400

⇒ 9y = 2200 – 400 = 1,800

⇒ y =

1 800

9

, = 200 1

Their monthly incomes are 11 × 200 = ` 2200 and 7 × 200 = ` 1400. [CBSE Marking Scheme, 2012] 1

qqq

WORKSHEET-27

Solutions

1. Given, a pair of linear equations is :

3x – y = 3 or y = 3x – 3 ...(i) ½

and 9x – 3y = 9 ...(ii) ½

On substituting y from eqn. (i) in eqn. (ii),

9x – 3(3x – 3) = 9

i.e., 9 = 9 ½

It is a true statement. Hence, eqn. (i) and (ii) have infinitely many solutions. ½

2. Let the two numbers be x and y (x > y)

We are given that,

x – y = 26 …(i) 1

and x = 3y …(ii)

On substituting x from eqn. (ii) in eqn. (i), we get

3y – y = 26

⇒ 2y = 26

\ y = 13 1

On substituting y = 13 in eqn. (ii), we get

x = 3 × 13 = 39

\ x = 39 1

Hence, the two numbers are 39 and 13.

3. Let the speed of the car I from A = x km/hr. Speed of the car II from B = y km/hr. Same dirrection :

Distance covered by car I = 150 + (distance covered by car II)

⇒ 15x = 150 + 15y ⇒ 15x – 15y = 150 ⇒ x – y = 10 ...(i) 1


Opposite direction :

Distance covered by car I + distance covered by car II

= 150 km x + y = 150 ...(ii) 1

Adding eqns. (i) and (ii),

2x = 160

\ x = 80

Substituting x = 80 in eqn. (i),

y = 70 1

\ Speed of the car I from A = 80 km/hr

and speed of the car II from B = 70 km/hr. 1


4. Let the fraction be

x

y.

x

y

- 2

1+ =

1

2

⇒ 2x – 4 = y + 1

⇒ 2x – y = 5 ...(i) 1

Also,

x

y

+ 4

3- =

3

2

⇒ 2x + 8 = 3y – 9

⇒ 2x – 3y = –17 ...(ii) 1 Subtracting eqn. (ii) from eqn. (i), 2y = 22 ⇒ y = 11 1

Substituting this value of y in eqn. (i), 2x – 11 = 5 \ x = 8 Hence, x = 8, y = 11

\

Fraction =

8

11 1

qqq

WORKSHEET-28

Solutions

1. Given, a pair of linear equations is : 0.2 x + 0.3y = 1.3 ...(i) and 0.4 x + 0.5y = 2.3 ...(ii)

From eqn. (i), 2x + 3y = 13 [Q Multiply by 10] ⇒ 3y = 13 – 2x

⇒ y = 13 2

3

- x

...(iii)½

On substituting y from eqn. (iii) in eqn. (ii),

4

10

5

10

13 2

3x

x+ ×

( )- =

23

10

⇒ 4x + 5

3 (13 – 2x) = 23

⇒ 12x + 5 (13 – 2x) = 3 × 23 ⇒ 12x + 65 – 10x = 69 ⇒ 2x = 69 – 65 = 4 \ x = 2 ½

On substituting x = 2 in eqn. (iii), we get

y =

13 2 2

3

9

3

- ×=

i.e., y = 3 ½

Hence, x = 2, y = 3 ½

2. Let the supplementary angles be x° and y° (x° > y°)

Now, x° + y° = 180° …(i) and x° – y° = 18° …(ii) From eqn. (ii), y° = x° – 18 …(iii) 1 On substituting y from eqn. (iii) in eqn. (i), x° + x° – 18° = 180°

⇒ 2x° = 198° \ x° = 99° 1

On substituting x° = 99° in eqn. (iii), y° = 99° – 18° = 81° \ y° = 81° 1

Hence, the angles are 99° and 81°

3. Let the number of red balls be x and white balls be y.


1

2y

=

1

3x or 2x – 3y = 0 ...(i)

and 3(x + y) – 7y = 6 1

or 3x – 4y = 6 ...(ii)

Multiplying eqn. (i) by 3 and eqn. (ii) by 2 and then subtracting, we get

6x – 9y = 0 1

6x – 8y = 12

– + –

– y = – 12

\ y =12

2x – 36 = 0

Substituting 12 in eqn. (i),

\ x = 18 1

\ x = 18, y = 12

Hence, number of red balls = 18

and number of white balls = 12 1


qqq


WORKSHEET-29

Solutions

1. Let length = x and breadth = y Then according to the first condition, (x – 5)(y + 3) = xy – 9 ⇒ 3x – 5y = 6 ...(i) 1 According to the second condition, (x +3)(y +2) = xy + 67 ⇒ 2x + 3y = 61 ...(ii) 1 Multiplying eqn. (i) by 3 and eqn. (ii) by 5 and then

adding, 9x – 15y = 18 10x + 15y = 305

\ x = 323

19 = 17 1

Substituting this value of x in eqn. (i), 3(17) – 5y = 6 ⇒ 5y = 51 – 6 ⇒ y = 9 1 Hence, perimeter = 2(x + y) = 2(17 + 9) = 52 units. 2. 2(3x – y) = 5xy ...(i) 2(x + 3y) = 5xy ...(ii) Divide eqns. (i) and (ii) by xy,

6 2

y x- = 5 ...(iii)

and

2 6

y x+ = 5 ...(iv) 1

Let

1

y = a and

1

x = b,

then equations (iii) and (iv) become

6a – 2b = 5 ...(v) 2a + 6b = 5 ....(vi) 1 Multiplying eqn. (v) by 3 and then adding with eqn.

(vi), 20a = 20 \ a = 1 Substituting this value of a in eqn. (v),

b =

1

2 1

Now

1

y

= a = 1

⇒ y = 1

and

1

x = b =

1

2 ⇒ x = 2. 1

qqq

WORKSHEET-30

Solutions

1. Let the sum of the ages of the 2 children be x and the age of the father be y years.

\ y = 2x

i.e. 2x – y = 0 ...(i) 1

and 20 + y = x + 40

x – y = – 20 ...(ii) 1

Subtracting (ii) from (i),

x = 20

From (i), y = 2x = 2 × 20 = 40

y = 40 1

Hence, the age of the father = 40 years. 1

[CBSE Marking Scheme, 2012] 2. Let the number of students in the class be x and the

number of rows be y.

\ The number of students in each row = x

y

According to the given condition,

x = x

y+

3 (y – 1)

⇒ x = x –

x

y+ 3y – 3

⇒

x

y – 3y + 3 = 0 …(i) 1

and x =

x

y- 3

(y + 2)

⇒ x = x +

2x

y – 3y – 6

⇒

2x

y – 3y – 6 = 0 …(ii) 1

On putting

x

y = u in eqn. (i) and (ii),

u – 3y + 3 = 0 …(iii)

and 2u – 3y – 6 = 0 …(iv) 1

On subtracting eqn. (iii) from eqn. (iv),

u – 9 = 0

⇒ u = 9

On substituting u = 9 in eqn. (iii),

9 – 3y + 3 = 0


⇒ 3y = 12

\ y = 4 …(v)

Now, u = 9

⇒

x

y

= 9

⇒ x = 36 [from eqn. (v)]

\ The number of students in the class = 36. 1

qqq

WORKSHEET-31

Solutions

1. Let the ages of Ani and Biju be x yr. and y yr, respectively.


x – y = ± 3 …(i)

Also, age of Ani’s father Dharam = 2x years

And age of Biju’s sister =

y

2 years


2x –

y

2 = 30 1

⇒ 4x – y = 60 …(ii)

Case I : When x – y = 3 …(iii)

On subtracting eqn. (iii) from eqn. (ii),

3x = 57

\ x = 19 years

On putting x = 19 in eqn. (iii),

19 – y = 3

\ y = 16 years 1

Case II : When x – y = – 3 …(iv)

On subtracting eqn. (iv) from eqn. (ii),

3x = 60 + 3

⇒ 3x = 63

\ x = 21 years

On putting x = 21 in eqn. (iv), we get 21 – y = – 3 \ y = 24 years 1

Hence, age of Ani is 19 year and age of Biju is 16 years or age of Ani is 21 years and age of Biju is 24 years. 1

2. Let the amount of their respective capitals be ` x and ` y.

\ According to the given condition, x + 100 = 2(y – 100) 1

⇒ x – 2y = – 300 …(i) and 6(x – 10) = y + 10 ⇒ 6x – y = 70 …(ii) 1 On multiplying eqn. (ii) by 2 and subtracting from

eqn. (1), x – 2y = – 300 ⇒ + 12x – 2y = + 140 – + – – 11x = – 440 ⇒ x = 40 1

On substituting x = 40 in eqn. (i), 40 – 2y = – 300 ⇒ 2y = 340 \ y = 170 1

Hence, the amount of their respective capitals are ` 40 and ` 170.

qqq


TOPIC-1Solution of Quadratic Equations

WORKSHEET-32

Solutions

1. 3 62x + = 9

⇒ 3x2 + 6 = 81 ⇒ 3x2 = 81 – 6 = 75

⇒ x2 =

75

3 = 25

⇒ x = ± 5 \ Positive root = 5 1 [CBSE Marking Scheme, 2015]

2. 6x2 – x – 2 = 0

⇒ 6x2 + 3x – 4x – 2 = 0 ½

⇒ 3x(2x + 1) –2(2x + 1) = 0 ½

⇒ (2x + 1)(3x – 2) = 0 ½

⇒ 3x – 2 = 0 or 2x + 1 = 0

⇒ x =

2

3 or x = –

1

2

\ Roots of equation are

2

3 and –

1

2.

½


3. 15x2 – 10 6 x + 10 = 0

⇒ 3x2 – 2 6 x +2 = 0 ½

⇒ 3x2 – 6 x – 6 x + 2 = 0 ½

⇒ 3x( 3x – 2) – 2 ( 3x – 2) = 0 ½

⇒ ( 3 x – 2 )( 3 x – 2 ) = 0

⇒ x =

2

3

2

3,

½


4. x x x x

x x

2 2

2

3 2 3 2

2

+ + + +

+

-

- =

4 8 2 3

2

x x

x

- - -

- 1

(2x2 + 4) (x – 2) = (2x – 11) (x2 + x – 2) ⇒ 5x2 + 19x – 30 = 0 1

⇒ (5x – 6) (x + 5) = 0

⇒ x = – 5, 6

5 1


5.

2

55

2

524

2x

x

x

x- --

+

= 0

Let

2

5

x

x( )- = y 1

\ y2 + 5y – 24 = 0 1

(y + 8) (y – 3) = 0 1

y = 3, – 8

⇒ y = 3

⇒ x = 15

or y = – 8

⇒ x = 4 1


qqq

WORKSHEET-33

Solutions

1. Putting x = -1

2

in 3x2 + 2kx – 3 = 0

3 -

1

2

2

+ 2k -

1

2

– 3 = 0

⇒

3

4 – k – 3 = 0

⇒ k =

3

4 – 3

⇒ k =

3 12

4

-

QUADRATIC EQUATIONS

SECTION

BSECTIONCHAPTER

4


k =

-9

4 1


2. 4 3 x2 + 5x – 2 3 = 0

⇒ 4 3 x2 + 8x – 3x –2 3 = 0 ½

⇒ 4x( 3 x + 2)– 3 ( 3 x + 2) = 0 ½

⇒ ( 3 x + 2)(4x – 3 ) = 0 ½

⇒ x = –

2

3

3

4,

½

[CBSE Marking Scheme, 2013, 2012]

3. x x2 ( 3+1) + 3-

= 0

⇒ x x x2- -3 1 + 3 = 0 ½

⇒ x x x( )- - -3) 1 ( 3 = 0

⇒ ( )x x- -3) ( 1

= 0 1

⇒ x = 3 , x = 1 ½


4. 2x(2x + 3) + (x – 3) + (3x + 9) = 0 1

⇒ 2x2 + 5x + 3 = 0 1

⇒ (x + 1) (2x + 3) = 0 ½

⇒ x = – 1, x = −3

2 ½


5. 1

1

2

2x x++

+ =

4

4x +

⇒ x x

x x

+ + +

+ +

2 2 1

1 2

( )

( )( )

=

4

4x + 1

⇒ 3 4

3 22

x

x x

+

+ +

= 4

4x +

⇒ (3x + 4)(x + 4) = 4(x2 + 3x + 2) 1 ⇒ 3x2+ 16x + 16 = 4x2 + 12x + 8 ⇒ x2 – 4x – 8 = 0 1

⇒ x = - - - - -( ) ( ) ( )( )4 4 4 1 8

2 1

2±

×

⇒ x = 4 48

2

4 4 3

2

±=

±

⇒ x = 2 ± 2 3

⇒ x = 2 + 2 3 or 2 – 2 3 1

qqq

WORKSHEET-34

Solutions

1. 3 x2 – 2x – 3 = 0

⇒ 3 x2 – 3x + x – 3 = 0

⇒ 3 x (x – 3 ) +1(x– 3 ) = 0

⇒ (x – 3 )( 3 x +1) = 0

\ x = 3 ,

-1

3 1


2. (x + 3)(x – 1) = 3(x –

1

3)

⇒ x2 + 2x – 3 = 3x – 1 1

⇒ x2 – x – 2 = 0

⇒ x2 – 2x + x – 2 = 0

⇒ x(x – 2) + 1(x – 2) = 0

⇒ (x – 2)(x + 1) = 0

\ x = 2, – 1 1


3.

2

5

3

5

2x x- -

= 0

⇒

2 5 3

5

2x x- -

= 0 ½

⇒ 2x2 – 5x – 3 = 0 ½

⇒ 2x2 – 6x + x – 3 = 0

⇒ 2x(x – 3) +1(x – 3) = 0 ½

⇒ (2x + 1)(x – 3) = 0

\ x = –

1

2, 3 ½


4. (x – 2) (2x – 3 + 2x) = 2x2 – 3x 1

or (x – 2) (4x – 3) = 2x2 – 3x 1

4x2 – 11x + 6 = 2x2 – 3x

or 2x2 – 8x + 6 = 0

or x2 – 4x + 3 = 0 ½

(x – 1) (x – 3) = 0 ½

x = 1, 3



5. a(x – b)(x – c) + b(x – a)(x – c) = 2c(x – a)(x – b) 1½

x2(a + b – 2c) + x(–ab –ac –ab – bc + 2ac + 2bc) = 0

x2(a + b – 2c) + x(–2ab + ac + bc) = 0 1½

x = ac bc ab

a b c

+

+

-

-

2

2


qqq

WORKSHEET-35

Solutions

1. Here, A = 4, B = –4a2, C = (a4 – b4)

x =

4 4 4 4

2 4

2 2 2 4 4a a a b± ×

×

( ) ( )- - -

=

4 16 16 16

8

2 4 4 4a a a b± +-

=

4 16

8

2 4a b±

1

⇒

x =

4 4

8 2

2 2 2 2a b a b±

=±

\ x1 =

a b2 2

2

+

and

x2 =

8 2

2 2a b- -=

1


Alternative Method : (4x2 – 4a2x + a4) – b4 = 0

⇒ (2x – a2)2 – (b2)2 = 0 1

\ (2x – a2 + b2) (2x – a2 – b2) = 0

⇒ x =

a b a b2 2 2 2

2 2

-,

+

1

2. Here, a = 9, b = – 6b2, c = – (a4 – b4)

x =

6 6 4 9

2 9

2 2 2 4 4b b a b± × × ( ){ }×

( )- - - -

x =

6 36 36 36

18

2 4 4 4b b a b± + -

1

=

6 36

18

2 4b a±

=

6 6

18

2 2b a±

\

x =

a b b a2 2 2 2

3 3

+,

-

1


Alternate Method : (9x2 – 6b2x + b4) – a4 = 0

⇒ (3x – b2)2 – (a2)2 = 0

\ (3x – b2 + a2) (3x – b2 – a2) = 0

⇒ x =

b a b a2 2 2 2

3 3

-,

+

1

3. xa

a bx

a b

a

2+

++

+

x + 1 = 0

x x

a

a b

a b

ax

a

a b+

+

++

++

= 0 1

x

a

a bx

a b

a+

+

++

= 0 1

⇒

x =

- -a

a b

a b

a++

,( )

1


4.

3

1

4

1x x++

- =

29

4 1x -

⇒

3 3 4 4

12

x x

x

-

-

+ +

=

29

4 1x -

⇒

7 1

12

x

x

+

- =

29

4 1x - 1

⇒ (7x + 1)(4x – 1) = 29x2 – 29

⇒ 28x2 – 7x + 4x – 1 = 29x2 – 29

⇒ 28x2 – 3x – 1 = 29x2 – 29

⇒ x2 + 3x – 28 = 0 1

⇒ x2 + 7x – 4x – 28 = 0

⇒ x(x + 7) – 4(x + 7) = 0 1

⇒ (x + 7)(x – 4) = 0

\ x = 4, –7 1

qqq


WORKSHEET-36

Solutions

1. 4x2 + 4bx + b2 – a2 = 0 ⇒ (2x + b)2 – a2 = 0 ½

⇒ (2x + b + a) (2x + b – a) = 0 ½

⇒ x =

- -a bx

a b+( )=

2 2,

[CBSE Marking Scheme, 2015] ½ + ½

2. x2 – 2ax – (4b2 – a2) = 0 Given equation can be written as x2 – 2ax + a2 – 4b2 = 0 or (x – a)2 – (2b)2 = 0 1

\ (x – a + 2b) (x – a – 2b) = 0 ½

⇒ x = a – 2b, x = a + 2b ½


3. 1

( 1)( 2)+

1

( 2)( 3)x x x x- - - - =

2

3

⇒

x x- -

- - -

3+ 1

( )( )( )x x x1 2 3 =

2

3

⇒ 2 4

1 2 3

x

x x x

-

- - -( )( )( ) =

2

3 1

⇒

2 2

1 2 3

( )

( )( )( )

x

x x x

-

- - - =

2

3 ½

⇒

2

1 3( )( )x x- -

=

2

3

⇒ 3 = (x – 1) (x – 3) ½

⇒ x2 – 4x + 3 = 3

⇒ x2 – 4x = 0

⇒ x(x – 4) = 0

⇒ x = 0 or x = 4 1

4. Let time taken by pipe A be x minutes. and time taken by pipe B be x + 5 minutes.

In one minute pipe A will fill 1

x, tank

In one minute pipe B will fill 1

5x +, tank

Both pipes A + B will fill

1 1

5x x+

+ tank in one minute

Then according to the question.

1 1

5x x+

+ =

9

100 2

⇒ x x

x x

+ +

+

5

5( ) =

9

100

⇒ 100(2x + 5) = 9x(x + 5)

⇒ 200x + 500 = 9x2 + 45x

⇒ 9x2 – 155x – 500 = 0 1

⇒ 9x2 – 180x + 25 x – 500 = 0

⇒ 9x(x – 20) + 25(x – 20) = 0

⇒ (x – 20)(9x + 25) = 0

⇒ x = 20, -25

9

rejecting negative value, x = 20 minutes

and x + 5 = 25 minutes

Hence pipe A will fill the tank in 20 minutes and pipe B will fill in 25 minutes. 1

qqq

WORKSHEET-37

Solutions

1. Here, a = 4, b = – 4p, c = (p2 – q2) The roots are given by the quadratic formula

x =

- -b b ac

a

±2 4

2

\

x =

4 16 4 4

2 4

2 2 2p p p q± − × × −×

( )

1

=

4 16 16 16

8

2 2 2p p p q± +-

=

4 4

8

p q±

\ The roots are

p q+

2,

p q-

2.

1


2. Let the side of the smaller square be y and the side of the longer square by x, then. ½

4x – 4y = 24 ⇒ x – y = 6 ⇒ x = y + 6 ½

According to the question, x2 + y2 = 468 ⇒ (y + 6)2 + y2 = 468 ⇒ 2y2 + 12y + 36 = 468 ⇒ 2y2 + 12y – 432 = 0


⇒ y2 + 6y – 216 = 0 ⇒ (y + 18)(y – 12) = 0 \ y = – 18, 12 Rejecting y = – 18, as side can not be negative \ y = 12 and x = 18 1


3. a(x – a) + b(x – b) = 2[x2 – (a + b)x + ab] ½

ax – a2 + bx – b2 = 2x2 – 2(a + b)x + 2ab

2x2 – 3(a + b)x + (a + b)2 = 0

2x2 – 2(a + b) x – (a + b) x + (a + b)2 = 0 ½

[2x – (a + b)] [x – (a + b)] = 0

x = a + b,

a b+

2 1


4. 3 2 2 2 32x x- - = 0

⇒ 3 3 2 2 32x x x- - 2+ =0 1

⇒ 3 6 6x x x( ) ( )- -+ 2 =0

⇒ ( )( )x x- 6 3 2+ =0 1

x =

6

2

3, x = -

½ +½



3 2 2 2 32x x- - = 0

We know that 2 2 can be written as 3 2 2-

\ 3 3 2 2 2 32x x- - -[ ] = 0

⇒ 3 3 2 2 2 32x x x- -+ = 0 1

⇒ 3 3 3 2 2 2 2 32x x x- -+ = 0

⇒3 3 2 2 2 3x x x- -.

+

= 0

⇒

3 6 2 6x x x- -

+

= 0 1

⇒ ( )( )x x- 6 3 2+ = 0

⇒ x = 6

or x =

-

2

3 =

-

2

3 1

5. Let the speed while going be x km/h

Therefore 150 150

10x x-

+ =

5

2 2

⇒ x2 + 10x – 600 = 0

⇒ (x + 30)(x – 20) = 0

⇒ x = 20 1

\ Speed while going =20 km/h ½

and speed while returning = 30 km/h ½


qqq

WORKSHEET-38

Solutions

1. 3 3 2 2 2 32x x x− + − = 0

3 6 2 6x x x[ ] [ ]− + − = 0 ½

or ( )( )x x− +6 3 2 = 0 ½

⇒ x = 62

3,−

1


2. Substituting x =

2

3 in ax2 + 7x + b = 0

4

9

14

3a b+ + = 0

⇒ 4a + 42 + 9b = 0 ⇒ 4a + 9b = – 42 ...(i) ½ x = – 3 9a – 21 + b = 0 ⇒ 9a + b = 21 ...(ii) ½ Solving (i) and (ii), a = 3 and b = – 6 1

3. Given, 6 7x + – (2x – 7) = 0

⇒ 6 7x + = (2x – 7)

Squaring both sides 6x + 7 = (2x – 7)2 1 ⇒ 6x + 7 = 4x2 – 28x + 49 ⇒ 4x2 – 34x + 42 = 0 ⇒ 2x2 – 17x + 21 = 0 ⇒ 2x2 – 14x – 3x + 21 = 0 ⇒ 2x(x – 7) – 3(x – 7) = 0 ⇒ (x – 7)(2x – 3) = 0

⇒

x = 7 or x = 3

2 1

4. x x23 3 30+ - = 0 ½

⇒ x x x2 5 3 2 3 30+ - - = 0 1

⇒ x x x( ) ( )+ +5 3 2 3 5 3-

= 0 ½

⇒ ( ) ( )x x+ 5 3 2 3 - = 0

⇒ x = -5 3 2 3,

½ + ½



5. Let the fraction be

x

x2 1+ ½

\

x

x

x

x2 1

2 1

++

+

=

58

21 1

21[x2 + (2x + 1)2] = 58 (2x2 + x) ⇒ 11x2 – 26x – 21 = 0 1

x = 3, −7

11( )rejected

\ Fraction = 3

7 ½+1


qqq

WORKSHEET-39

Solutions

1. On completing the square, ⇒ (x – 2)2 – 8 – 4 = 0 ½

⇒ (x – 2)2 – 12 = 0 ⇒ (x – 2)2 = 12

⇒ (x – 2)2 = (2 3 )2 ½

⇒ x – 2 = ± 2 3

⇒ x = 2 ±2 3 ½

\ x = 2 + 2 3 , 2 – 2 3 ½


2. Let units digit and tens digit of the two digit number be x and y

\ Number is 10y + x

According to question,

10y + x = 4(y + x) ½

⇒ 10y + x = 4y + 4x

⇒ 10y – 4y = 4x – x

⇒ 6y = 3x

⇒ 2y = x ...(i)

Also,

10y + x = 3xy ...(ii)

⇒ 10y + 2y = 3(2y)y

⇒ 12y = 6y2

⇒ 6y2 – 12y = 0

⇒ 6y (y – 2) = 0

⇒ y = 0 or y = 2 ½

Rejecting y = 0 as the number can not be zero.

⇒ x = 4

\ Required number is 24. 1

3. Let the number of wickets taken by Zahir be x.

The number of wickets taken by Harbhajan = 2x – 3 ½

According to question, x(2x – 3) = 20 ½

⇒ 2x2 – 3x = 20

\Required quadratic equation,

2x2 – 3x – 20 = 0 1


4. x2 + 5x – (a2 + a – 6) = 0

x =

- -5 25 4 6

2

2± + +( )a a

1

[ ]Qx

b b ac

a=

±- -2 4

2

=

-5

2

± ( )2 1a +

1

=

2 4

2

2 6

2

a a- - -,

x = a – 2, x = – (a + 3) [CBSE Marking Scheme, 2015] ½ + ½

5.

c = 25 cm

B

a

bC A

Here a + b + c = 60, c = 25 \ a + b = 35 1

Using Pythagorus theorem

a2 + b2 = 625

Using identity (a + b)2 = a2 + b2 + 2ab 1

352 = 625 + 2ab

⇒ ab = 300 1

Area of ∆ABC = 1

2 ab =150 cm2. 1


qqq


TOPIC-2Discriminant and Nature of Roots

WORKSHEET-40

Solutions

1. (k + 1)x2 – 2(k + 1)x + 1 = 0 has equal roots D = 0

b2 = 4ac 1

4(k + 1)2 = 4(k + 1)

k + 1 = 1

k = 0 1


2. 2 is the root of x2 + kx + 12 = 0 ⇒ (2)2 + 2k + 12 = 0 ⇒ 2k + 16 = 0 k = – 8 ½ Put k = – 8 in x2 + kx + q = 0 ⇒ x2 –8x + q = 0 ½ For equal roots (–8)2 – 4(1)q = 0 ½ 64 – 4q = 0 4q = 64 q = 16 ½ [CBSE Marking Scheme, 2016]

3. Given, 2 is a root of the equation, 3x2 + px – 8 = 0

Putting x = 2 in 3x2 + px – 8 = 0

12 + 2p – 8 = 0

⇒ p = – 2

Given, 4x2 – 2px + k = 0 has equal roots

i.e., 4x2 + 4x + k = 0 has equal roots 1

\ D = b2 – 4ac = 0

⇒ (4)2 – 4(4)(k) = 0

⇒ 16 – 16k = 0 1

⇒ 16k = 16

\ k = 1 1

4. kx(x – 2 5 ) + 10 = 0

⇒ kx2 – 2 5 kx + 10 = 0 ½

Here, a = k, b = – 2 5 k, c = 10

Given, roots are equal, D = b2 – 4ac = 0 ½

⇒ (–2 5 k)2 – 4 × k × 10 = 0 ½

⇒ 20k2 – 40k = 0 ½

⇒ 20k(k – 2) = 0

⇒ k(k – 2) = 0

Q k ≠ 0 ½

\ k = 2 ½

5. For equal roots of x2 + 2px + mn = 0, 4p2 – 4mn = 0

⇒ p2 = mn ...(i) 1

For equal roots of

x2 – 2 (m + n)x + (m2 + n2 + 2p2) = 0

4(m + n)2 – 4(m2 + n2 + 2p2) = 0 1

m2 + n2 + 2mn – m2 – n2 – 2(mn) = 0 (From (i)) 1

\ If roots of x2 + 2px + mn = 0 are equal then those

of x2 – 2a(m + n)x + (m2 + n2 + 2p2) = 0 are also equal. [CBSE Marking Scheme, 2016] 1

qqq

WORKSHEET-41

Solutions

1. a = 9, b = – 3k , c = k Since roots of the equation are equal

b2 – 4ac = 0

(–3k)2 – (4 × 9 × k) = 0

9k2 – 36 k = 0

k2 – 4k = 0

k (k – 4) = 0

k = 0 or k = 4 Since k = 0 is not possible for the equation

k = 4.

2. Since the roots are equal, then ½

D = 0

D = b2 – 4ac = 0

⇒ (–2k)2 – 4 (k)(6) = 0

⇒ 4k2 – 24k = 0

⇒ 4k(k – 6) = 0

⇒ k = 0, 6 1

But k ≠ 0, as coefficient of x2 can not be zero

k = 6 ½



3. b2 – 4ac = (– 4 3 )2 – 4(3)(4) 1

= 48 – 48 = 0 ½

\ Roots are real and equal.

\ Roots are

- -b

a

b

a2 2

,

1

or

2 3

3

2 3

3,

½


4. For equation x2 + kx + 64 = 0

b2 – 4ac = 0

⇒ k2 – 4 × 1 × 64 = 0

⇒ k2 – 256 = 0

⇒ k = ± 16 ...(i) 1

and for equation x2 – 8x + k = 0

b2 – 4ac = 0

⇒ (– 8)2 – 4 × 1 × k = 0

⇒ 64 = 4k

⇒ k =

64

4 = 16 ...(ii) 1

From (i) and (ii), we get k = 16 ½

For k = 16, given equations will have equal solution. [CBSE Marking Scheme, 2014] ½

5. (i) For x2 + kx + 64 = 0 to have real roots

k2 – 256 ≥ 0 ...(i) 1½

(ii) For x2 – 8x + k = 0 to have real roots

64 – 4k ≥ 0 ...(ii) 1½

For (i) and (ii) to hold simultaneously

k = 16 1


qqq

WORKSHEET-42

Solutions

1. As the equation has equal roots

i.e., D = 0

D = b2 – 4ac = 0 1

⇒ p2 – 4 × 4 × 3 = 0

⇒ p2 – 48 = 0

⇒ p2 = 48

⇒ p = ± 4 3 1

2. a = 13 3 , b = 10, c = 3 ½

b2 – 4ac = (10)2 – 4(13 3 )( 3 ) ½

= 100 – 156

= – 56 ½

As b2 –4ac < 0 ½

So, the equation has no real roots.


3. Given, kx2 + 1 –2(k – 1)x + x2 = 0

i.e., (k + 1)x2 – 2(k – 1)x + 1 = 0 1

For equal roots D = b2 – 4ac = 0

Here, a = k + 1, b = –2(k – 1), c = 1

4(k – 1)2 – 4(k + 1) × 1 = 0

⇒ 4k2 – 8k + 4 – 4k – 4 = 0 1

⇒ 4k2 – 12k = 0

⇒ 4k(k – 3) = 0

\ k = 0, 3 1


4. Since equation has equal roots, \ D = 0 i.e., b2 – 4ac = 0 {2(2k – 3)}2 – 4(k – 2)(5k – 6) = 0 1

⇒ 4(4k2 – 12k + 9) – 4(k – 2)(5k – 6) = 0

⇒ 4k2 – 12k + 9 – 5k2 + 6k + 10k – 12 = 0 1

⇒ k2 – 4k + 3 = 0

⇒ k2 – 3k – k + 3 = 0

⇒ k(k – 3) –1(k – 3) = 0 ⇒ (k – 3)(k – 1) = 0

\ k = 1, 3 1


5. Since the given equation has real roots,

i.e., D = b2 – 4ac = 0 1

Here, a = 1, b = – 8, c = k. 1

\ (–8)2 – 4(1)(k) = 0 1

⇒ 64 – 4k = 0

4k = 64 ⇒ k =

64

4

⇒ k = 16 1

qqq


WORKSHEET-43

Solutions


1. 3 9 2

2 2- -( ) + ( )k

= 10 1

QDist = +

( ) ( )x x y y1 22

1 22

- -

⇒ (– 6)2 + k2 – 4k + 4 = 100 ⇒ k2 – 4k + 40 = 100 ⇒ k2 – 4k – 60 = 0 ⇒ k2 – 10k +6k – 60 = 0 ⇒ k(k – 10) + 6 (k – 10) = 0 ⇒ (k – 10)(k + 6) = 0 \ k = 10, – 6 1


2.

B

x

PA

O

x + 7 ½

Let P be the location of the pole such that its distance from gate B, x metres.

\ AP = x + 7 ½

AB is diameter ⇒ ∠APB = 90° and AB = 17 m ½ \ x2 + (x + 7)2 = (17)2

x2 + x2 + 14x – 240 = 0 or x2 + 7x – 120 = 0 1½

x = -7 49 480

2

± + = 8, –15

\ x = 8 m, x + 7 = 15 m 1 [CBSE Marking Scheme, 2016]

3. (i) Let the cost price of the toy be ` x, then gain = x%

⇒ Gain = ` xx

×

100

= `x2

100

½

\ S.P. = C.P. + Gain

= x + x2

100

But S.P. = ` 24 ½

\ x + x2

100 = 24

⇒ 100x + x2 = 2400 ⇒ x2 + 100x – 2400 = 0 ⇒ x2 + 120x – 20x – 2400 = 0 ⇒ x(x + 120) – 20(x + 120) = 0 ⇒ (x – 20)(x + 120) = 0 ⇒ x = 20 or x = – 120 1

⇒ x = 20, [Q x = – 120 is not possible]

Hence the cost price of the toy is ` 20.

(ii) Quadratic equation. ½

(iii) Genuine profit. ½

4. Let the usual speed of plane be x km/h.

\ 1500 1500

250x x-

+ =

1

2 2

⇒ x2 + 250x – 750000 = 0 (x + 1000) (x – 750) = 0 ⇒ x = 750 \ Speed of plane = 750 km/h. 1 Values depicted here are : (i) Helping the needy. (ii) Quick help to the injured. 1 [CBSE Marking Scheme, 2016]

qqq

WORKSHEET-44

Solutions


1. Let the number of books bought = x.

\1200 1200

10x x-

+ = 20 1

⇒ x2 + 10x – 600 = 0 1

⇒ (x + 30)(x – 20) = 0

⇒ x = – 30 or x = 20 1

Since number of books cannot be negative,

\ x = 20

\ Number of books bought = 20. 1


2. Let the number of books be x and cost of each book be ` y.

\ xy = 80 ½

Also, (x + 4)(y – 1) = 80 ½


⇒ xy – x + 4y – 4 = 80 ½

⇒ 80 480

4- -xx

+

= 80 ½

⇒ - -xx

+320

4

= 0

⇒ – x2 + 320 – 4x = 0

⇒ x2 + 4x – 320 = 0

⇒ x2 + 20x – 16x – 320 = 0 ½ + ½

⇒ x(x + 20) – 16(x + 20) = 0

⇒ (x – 16)(x + 20) = 0

⇒ x = 16

(as it cannot be negative)

\ No. of books bought = 16 ½ + ½


3.

A B192 km

Let speed of passenger train be x km/h ½ \ speed of superfast train = (x + 16) km/h ½

By question, Tpassenger = 192

x

⇒

192 192

16x x−

+

= 2 ½

⇒ 192(x + 16) – 192x = 2(x2 + 16x) ½

⇒ 192x + 192 × 16 – 192x = 2(x2 + 16x) ½

⇒ x2 + 16x – 1536 = 0

⇒ x2 + 48x – 32x – 1536 = 0

⇒ x(x + 48) – 32(x + 48) = 0 ½

⇒ (x – 32)(x + 48) = 0

⇒ x = 32 or – 48

Since speed can’t be negative, therefore – 48 is not possible.

\ Speed of passenger train = 32 km/h ½

and Speed of fast train = 48 km/h. ½


4. (i) Suppose Arjun had x arrows.

Number of arrows used to cut arrows of Bheeshm

=

3

x

Number of arrows used to kill the rath driver = 6

Number of other arrows used = 3

Remaining arrows = 4 x + 1

So,

xx

26 3 4 1+ + + +

= x 1

⇒ x + 20 + 8 x = 2x

⇒ x = 20 + 8 x

Put x = y2, then y2 = 20 + 8y ⇒ y2 – 8y – 20 = 0 ⇒ y2 – 10y + 2y – 20 = 0 ⇒ y(y – 10) + 2(y – 10) = 0 ⇒ (y – 10) (y + 2) = 0 ⇒ y = 10 or y = – 2 1 ⇒ y = 10, [Qy cannot negative] ⇒ x = y2

⇒ x = 100. Hence the number of arrows which Arjun had is

100. (ii) Quadratic equations. ½

(iii) Bravery, fight for truth. ½

qqq


ARITHMETIC PROGRESSION

SECTION

BSECTIONCHAPTER

5

TOPIC-1To Find nth Term of the Arithmetic Progression

WORKSHEET-45

1. a = 11, d = – 3, an = – 150

an = a + (n – 1)d

– 150 = 11 + (n – 1)(– 3)

– 150 = 11 – 3n + 3

3n = 164

⇒ n = 164

3 = 54·66

Hence – 150 is not a term of the given A.P. 1

2. Here, a = 5, d = – 3 Q l = a + (n – 1)d \ – 49 = 5 + (n – 1)(– 3) ⇒ – 49 = 5 – 3n + 3 ⇒ 3n = 49 + 5 + 3

⇒ n =

57

3 = 19th term. 1


3. Here, a = 3, d = 7 – 3 = 11 – 7 = 4 ½

an = a + (n – 1)d, let an = 184 ½

⇒ 184 = 3 + (n – 1) 4 ½

⇒

181

4 = n – 1

⇒ 45.25 = n – 1

⇒ 46.25 = n

⇒ 184 is not a term of given A.P. ½


4. Given, a3 = a + 2d = 7

⇒ a7 – 3 × 7 = 2 1

⇒ a + 6d – 21 = 2

a + 6d = 23

a + 2d = 7

– – – 1

⇒ 4d = 16

⇒ d = 4

a + 8 = 7

a = – 1 ½

a20 = a + 19d

= – 1 + 76 = 75

\ an = a + (n – 1) d

= – 1 + 4n – 4 ½

= 4n – 5.

5. Given, t7 =

1

9

t9 =

1

7

i.e. a + 6d =

1

9 ... (i) 1

and

a + 8d =

1

7 ... (ii) 1

From, equation (ii) – equation (i), we get

a + 8d – a – 6d =

1

7

1

9-

⇒ 2d =

2

63

⇒ d =

1

63

\ from (ii),

a + 8 ×

1

63 =

1

7

⇒ a =

1

7

8

63-

⇒ a =

9 8

63=

1

63

-

\ t63 =

1

6362

1

63+ ×

=

1 62

63

+

⇒ t63 =

63

63 = 1

1

6. Let the three numbers in A.P. be a – d, a, a + d.

3a = 12

⇒ a = 4.


Also, (4 – d)3 + 43 + (4 + d)3 = 288 1

⇒ 64 – 48d + 12d2 – d3 + 64 + 64 + 48d + 12d2 + d3

= 288

⇒ 24d2 + 192 = 288 1

⇒ d2 = 4

d = ± 2 1

The numbers are 2, 4, 6, or 6, 4, 2. 1


qqq

WORKSHEET-46

1. Here given sequence is an A.P.

2 8 18, , , ...

= 2 2 2 3 2, , ...

Where, a = 2 , d = 2 , n = 10

\ an = a + (n – 1) d

⇒ a10 = 2 10 1 2+ ( )-

= 2 9 2+

= 10 2

= 200 1

2. Let us write A.P. in reverse order

i.e., 184, ........ 13, 10, 7. ½

d = 7 – 10 = – 3 ½

a = 184, n = 7

a7 = a + 6d

a7 = 184 + 6 (– 3)

= 184 – 18 = 166. ½

Hence, 166 is the 7th term from the end. ½


3. Let the nth term be zero.

then an = 0

⇒ a + (n – 1)d = 0

⇒ 150 + (n – 1)(–3) = 0 1

⇒ 150 – 3n + 3 = 0

⇒ –3n = –153

\ n = 51. 1

Therefore, the first negative term is 52nd term.

4. Let the first term of A.P. be a and common difference be d.

a9 = 7a2

⇒ a + 8d = 7(a + d) ...(i) ½

a12 = 5a3 + 2

⇒ a + 11d = 5(a + 2d) + 2 ...(ii) 1

From (i), a + 8d = 7a + 7d

– 6a + d = 0 ...(iii)

From (ii), a + 11d = 5a + 10d + 2

– 4a + d = 2 ...(iv)

Subtracting (iv) from (iii),

– 2a = – 2

⇒ a = 1 1

From (iii),

– 6 + d = 0

d = 6 ½


5. Here given, a3 = 9 ⇒ a + 2d = 9 ....(i) a8 – a5 = 6 ⇒ (a + 7d) – (a + 4d) =6 ⇒ 3d = 6 ⇒ d = 2 ....(ii) 1 Substituting (ii) in (i), we get ⇒ a + 2(2) = 9 ⇒ a = 5 1 So, A.P. is 5, 7, 9, 11, ..... 1 [CBSE Marking Scheme, 2015]

6. a, 7, b, 23 and c are in A.P.

Let the common difference be d

\ a + d = 7 .....(i) ½

a + 3d = 23 .....(ii) ½

From (i) and (ii), we get

a = – 1, d = 8 1

b = a + 2d

b = – 1 + 2 × 8

⇒ b = – 1 + 16

⇒ b = 15 ½

\ c = a + 4d

= – 1 + 4 × 8

= – 1 + 32

c = 31 ½

\ a = – 1, b = 15, c = 31 1


qqq


WORKSHEET-47

1. Here, a = 2 2 2, = 8a d+ =

d = 2 2 2 2- =

\ Next term = 32 2+

= 4 2 2+

= 2 × 5

= 50 1 [CBSE Marking Scheme, 2012]

2. Common difference,

d = 6 3-

=

3 2 1-( )

= 9 6-

= 3 6-

= 12 9-

= 2 3 3-

The given series is not in A.P. as common difference does not exist. [CBSE Marking Scheme, 2015] 1

3. Here, a = 7 , = 28a + d

\ d = 28 7- = 2 7 7−

= 7

⇒ Next term = 7 + ( )4 1 7-

⇒ = 7(4)

⇒ = 7 16× 1

= 112 .

4. Let the 1st term be a and common difference be d.

According to the question, a32 = 2a12

\ a + 31d = 2(a + 11d)

a + 31d = 2a + 22d

a = 9d 1

a70 = a + 69d

= 9d + 69d = 78d

Q a31 = a + 30d

= 9d + 30d = 39d

a70 = 2a31 Hence Proved. 1


5. Given, a8 = 0 ⇒ a + 7d = 0 ⇒ a = – 7d ½

⇒ a38 = a + 37d

⇒ a38 = – 7d + 37d = 30d ½

⇒ a18 = a + 17d

= – 7d + 17d = 10d ½

⇒ a38 = 30d = 3 × 10d = 3 × a18

\ a38 = 3a18. Hence Proved. ½


6. Let the four numbers be

a – 3d, a – d, a + d, a + 3d

\ a –3d + a – d + a + d + a + 3d = 56

⇒ 4a = 56

a = 14 1

Hence numbers are 14 – 3d, 14 – d, 14 + d, 14 + 3d

Now, according to question,

( )( )

( )( )

14 3 14 3

14 14

-

-

d d

d d

+

+ =

5

6

⇒ 196 9

196

2

2

-

-

d

d =

5

6

⇒ 6(196 – 9d2) = 5(196 – d2)

⇒ 6 × 196 – 54d2 = 5 × 196 – 5d2

⇒ 6 × 196 – 5 × 196 = 54d2 – 5d2

⇒ (6 – 5) × 196 = 49d2

⇒ d2 = 196

49 = 4

⇒ d = ±2 1

\ The numbers are

14 – 3 × 2, 14 – 2, 14 + 2, 14 + 3 × 2

8, 12, 16, 20 1

7. a = a’ + (p – 1)d, b = a’ + (q – 1)d, c = a’ + (r – 1)d

1½

a(q – r) = [a’ + (p – 1)d][q – r]

b(r – p) = [a’ + (q – 1)d][r – p]

and c[p – q] = [a’ + (r – 1)d][p – q] ½

\ a(q – r) + b(r – p) + c(p – q) = a’ [q – r + r – p + p –

q] + d [p(q – r) – q + r + (q – 1) (r – p) + (r – 1) (p – q)]

½

= a’ × 0 + d[pq – pr + qr – pq + pr – qr + (– q + r – r + p – p + q)] = 0 ½


qqq


WORKSHEET-48

1. If 2k + 1, 3k + 3, 5k – 1 are in A.P.

then (5k – 1) – (3k+ 3) = (3k + 3) – (2k + 1) ½

⇒ 5k – 1 – 3k – 3 = 3k + 3 – 2k – 1

⇒ 2k – 4 = k + 2

⇒ 2k – k = 4 + 2

or k = 6 ½

2. Here, a = – 5, d =

- - -5

2( 5)=

5

2

nth term = a +(n – 1)d

25th term = – 5 + (25 – 1) ×

5

2

= – 5 + 60 = 55 1 [CBSE Marking Scheme, 2015]

3. Given, 5a5 = 8a8 ⇒ 5(a + 4d) = 8(a + 7d) 1 ⇒ 5a + 20d = 8a + 56d ⇒ 3a + 36d = 0 ⇒ 3(a + 12d) = 0

⇒ a + 12d = 0 \ a13 = 0. 1 [CBSE Marking Scheme, 2012]

4. Let the first term be a and common difference be d.

a +4d = 20 ...(i) ½

a + 6d + a + 10d = 64

a + 8d = 32 ...(ii)1

Solving equations (i) and (ii),

common difference d = 3 ½


5. Sn = 3n2 + 5n Sn – 1 = 3(n – 1)2 + 5(n – 1) = 3(n2 + 1 – 2n) + 5n – 5 1 = 3n2 + 3 – 6n + 5n – 5 = 3n2 – n – 2 an = Sn – Sn – 1 = 3n2 + 5n – (3n2 – n – 2) 1 = 6n + 2 ½ A.P. is 8, 14, 20, .................... ½ a15 = a + 14d = 8 + 14(6) = 92. [CBSE Marking Scheme, 2012]

7.

[Topper Answer, 2016]

qqq


WORKSHEET-49

TOPIC-2Sum of nth Terms of an Arithmetic Progression

1. Here, a = 5, n = 10, d = 5

Q S =

na n d

22 1[ ( ) ]+ -

\ S10 =

10

22 5 10 1[ ( ) ]× + - 5

= 5 [10 + 9 × 5]

= 5 [10 + 45]

= 5 × 55 = 275 1

2. Here, a = 2, d = 2, n = 5

Q S =

na n d

22 1[ ( ) ]+ -

\ S5 =

5

22 2 5 1[ ( ) ]× + - 2

=

5

2[ ]4 + 4 × 2

=

5

2[ ]4 + 8

=

5

212 5 6× = × = 30 1


3. Here a = 65, d = – 5, Sn = 0 ½

n

2 [130 + (n – 1) (– 5)] = 0 1

⇒ n = 27 ½ [CBSE Marking Scheme, 2016]

4. Here a = 18, d = – 2, Sn = 0 ½

Therefore

n

2[36 + (n – 1) (– 2)] = 0 1

⇒ n = 19 [CBSE Marking Scheme, 2016] ½

5. Sn =

3 13

2

2n n+

an = Sn – Sn–1 ⇒ a25 = S25 – S24

=

3 25 13 25

2

3 24 13 24

2

2 2( ) ( ) ( ) ( )+ +-

1

=

1

2{3(252 – 242) + 13(25 – 24)} 1

=

1

2(3 × 49 + 13) = 80 1


6. S1 = 1 + 2 + 3 + ....n.

S2 = 1 + 3 + 5 + ...upto n terms

S3 = 1 + 4 + 7 + ...upto n terms

⇒ S1 = n n( )+ 1

2 ½

Also, S2 = n

2 [2 × 1 + (n – 1)2]

= n

2 [2n] = n2 ½

and S3 = n

2 [2 × 1 + (n – 1)3]

= n n( )3 1

2

- ½

Now, S1 + S3 = n n n n( ) ( )+

+1

2

3 1

2

- ½

= n n n[ ]+ +1 3 1

2

-

= n n[ ]4

2

= 2n2 = 2S2 Hence Proved. 1

7. a = 8, d = 1/3 years, Sn = 168 ½

Sn =

na n d

22 1[ ( ) ]+ -

⇒ 168 =

nn

22 8 1

1

3( ) ( )+

-

½

n2 + 47n – 1008 = 0 1

⇒ n2 + 63n – 16n – 1008 = 0

⇒ (n – 16)(n + 63) = 0

⇒ n = 16 or n = – 63

n = 16

(n cannot be negative) 1

Age of the eldest participant = a + 15d = 13 years


qqq


WORKSHEET-50

1. Here, a = 10, d = 6 – 10 = – 4, n = 16

Q S =

na n d

22 1[ ( ) ]+ -

\ S16 =

16

22 10 16 1[ ( ) ]× + - -( 4)

= 8[20 + 15 × (– 4)] = 8[20 – 60] = 8 × (– 40) = – 320 1


2. Here, a = 6, d = 6, n = 5

Q S =

na n d

22 1[ ( ) ]+ -

\ S5 =

5

22 6 5 1[ ( ) ]× + - (6)

=

5

212[ ] + 4 × 6

=

5

212[ ] + 24

=

5

236[ ]

= 5 × 18 = 90 1 [CBSE Marking Scheme, 2012]

3. Here a = 27, d = – 3, Sn = 0 ½

\

n

2[54 + (n – 1)(–3)]

= 0 1

⇒ n = 19 ½ [CBSE Marking Scheme, 2016]

4. S5 + S7 = 167

⇒

5

22 4

7

22 6( ) ( )a d a d+ + + = 167

⇒ 24a + 62d = 334 or 12a + 31d = 167 ....(i) ½ S10 = 235 ⇒ 5(2a + 9d) = 235 or 2a + 9d = 47 ...(ii) ½ Solving (i) and (ii), wet get a = 1, d = 5 ½

Here A.P. = 1, 6, 11, .... ½ [CBSE Marking Scheme, 2015]

5. S1 =

n

2[10 + (n – 1)2] ½

S2 =

n

2[10 + (n – 1)4] ½

S3 =

n

2[10 + (n – 1)6] ½

S1 + S3 =

n

2[20 + 2n – 2 + 6n – 6]

=

n

2[20 + 8(n – 1)]

= 2 ×

n

2[10 + 4(n – 1)] 1

= 2S2 ½


6. Let a be the first term and the common difference be d.

S12 = 6[2a + 11d]

= 12a + 66d 1

S8 = 4[2a + 7d]= 8a + 28d ½

S4 = 2[2a + 3d] = 4a + 6d ½

3(S8 – S4) = 3[(8a + 28d) – (4a + 6d)]

= 3[4a + 22d]

= 12a + 66d

= 6[2a + 11d]

= S12 1


7. Let total time be n minutes

Total distance convered by thief = (100n) metres ½

Total distance covered by policeman = 100 + 110 + 120 + ... + (n – 1) terms ½

\ 100n = n - 1

2[200 + (n – 2)10] 1

n2 – 3n – 18 = 0 ½

(n – 6)(n + 3) = 0 ½

⇒ n = 6 ½

Policeman took 5 minutes to catch the thief. ½


qqq

WORKSHEET-51

1. Let the sum of n terms of A.P. = Sn. Given, Sn = 2n2 + 5n Now, nth term of A.P. = Sn – Sn – 1

⇒ an = (2n2 + 5n) – [2 (n – 1)2

+ 5 (n – 1)]

= 2n2 + 5n – [2 n2 – 4n

+ 2 + 5n – 5]

= 2n2 + 5n – 2n2 – n + 3


an = 4n + 3 ½

4th term a4 = 4 × 4 + 3 = 19 ½


2. Let the sum of k terms of A.P. is Sn = 3k2 – k

Now kth term of A.P.

= Sn – Sn –1

ak = (3k2 – k) – [3 (k – 1)2 – (k – 1)]

= 3k2 – k – [3k2 – 6k + 3 – k + 1]

= 3k2 – k – 3k2 + 7k – 4

= 6k – 4 ½

first term a = 6 × 1 – 4 = 2 ½


3. Here, a1 = – 1, a2 = – 5 and d = – 4

Q Sn =

na n d

22 1[ ( ) ]+ -

½

\ S16 =

16

22 1 16 1[ ( ) ( ) ]× +- - - ( 4)

½

= 8 [– 2 – 60] = 8 (– 62) = – 496 1


4. Here n = 25, tn = 7 – 3n Taking n = 1, 2, 3, ..... t1 = 7 – 3 × 1 = 4 t2 = 7 – 3 × 2 = 1 t3 = 7 – 3 × 3 = – 2 ½ \ Given A.P. is 4, 1, – 2, ..... . Here, a = 4, d = 1 – 4 = – 3 ½

Now, Sn =

na n d

22 1[ ( ) ]+ -

½

=

25

22 4 25 1[ ( ) ]× + - - ( 3)

=

25

28 24[ ]+ ( 3)-

=

25

28 72( )-

= – 800 ½ [CBSE Marking Scheme, 2012]

5. Here, a14 = 2a8 Let first term be a and common difference be d. ⇒ a = – d. ...(i) 1 a6 = – 8 ⇒ a + 5d = – 8 ...(ii) ½

Solving (i) and (ii), we get

a = 2, d = – 2 ½

S20 =

20

22 2 20 1 2[ ( )( )]× + - -

= 10[4 + 19 × (– 2)]

= 10(4 – 38)

= 10 × (– 34) = – 340 1


6. Let a, A be the first terms and d, D be the common difference of two A.P.’s

Then according to question,

S

S

n

n'

=

na n d

nA n D

n

n

22 1

22 1

7 1

4 27

[ ( ) ]

[ ( ) ]

+

+

=+

+

-

-

1

⇒ 2 1

2 1

a n d

A n D

+

+

( )

( )

-

-

=

7 1

4 27

n

n

++

⇒ a

nd

An

D

+

+

( )

( )

-

-

1

21

2

=

7 1

4 27

n

n

++ ½

Putting n - 1

2 = m – 1

⇒ n = 2m – 1 1

a m d

A m D

+

+

( )

( )

-

-

1

1

= 7 2 1 1

4 2 1 27

( )

( )

m

m

-

-

+

+

⇒ a

A

m

m

= 14 6

8 23

m

m

-

+ ½

qqq

WORKSHEET-52

1. Here, an = 3 – 2n

Taking n = 1, tn = 3 – 2 = 1

n =15, t15 = 3 – 2 × 15

= 3 – 30 = – 27 1

Sn =

na l

2( )+

S15

=

15

21 27[ ( )]+ -

=

15

226[ ]-

= 15 × (– 13) = – 195 1



2. Tn = Sn – Sn – 1 ½

Tn – 1 = Sn – 1 – Sn – 2 ½

Sn – 2Sn – 1 + Sn – 2 = Sn – Sn – 1 – Sn – 1 + Sn – 2

= (Sn – Sn – 1)

– (Sn – 1 – Sn –2) ½

= Tn – Tn – 1 = d. ½


3. Let the sum of first n terms of A.P. = Sn

Given, Sn = 5n – n2

Now, nth term of A.P. = Sn – Sn – 1 1

⇒ an = (5n – n2) – [5(n – 1) – (n – 1)2] = 5n – n2 – [5n – 5 – (n2 + 1 – 2n)] = 5n – n2 – (5n – 5 – n2 – 1 + 2n) = 5n – n2 – 7n + 6 + n2

= – 2n + 6 ⇒ an = – 2(n – 3) \ nth term an = – 2(n – 3) 1

4. a = 5, l = 45 (given) \ 45 = 5 + (n – 1)d ⇒ (n – 1)d = 40 ....(i) Given, Sn = 400 1

\ 400 =

n

2(5+45)

800 = 50 n ⇒ n = 16 \ from (i), (n – 1)d = 40 \ 15d = 40

⇒ d =

40

15 =

8

3 1


5. Let the first term be a and the common difference be d.

a12 = a + 11d = – 13 ...(i) ½ S4 = 2[2a + 3d] = 24 2a + 3d = 12 ...(ii) 1 Solving (i) and (ii), we get a = 9 and d = – 2 1

Thus, S10 =

10

22 9 10 1 2[ ( )( )]× + - -

= 5[18 – 18] = 0 ½ [CBSE Marking Scheme, 2015]

6. Let the first 20 terms

S20 =

20

22 19( )a d+

⇒ 400 =

20

22 19( )a d+

⇒ 400 = 10 [2a + 19d] or 2a + 19d = 40 ...(i) 1

Also, S40 =

40

22 39( )a d+

⇒ 1600 = 20 [2a + 39d] or 2a + 39d = 80 ...(ii) 1 from (i) and (ii), we get a = 1 and d = 2

S10 =

10

22 1 10 1 2[ ( )( )]× + -

1

= 5 [2 + 9 × 2] = 5 [2 + 18] = 5 × 20 = 100 1 [CBSE Marking Scheme, 2015]

qqq

WORKSHEET-53

HOTS & Value based Answers

1. The sequence goes like this,

110, 120, 130,................. 990

Since they have a common difference of 10, they form an A.P. a = 110, an = 990, d = 10 ½

an = a+ (n – 1) × d

990 = 110 + (n – 1) × 10

990 – 110 = (n – 1) × 10 ½

880 = (n – 1) × 10

n – 1 = 88

n = 89 ½

These are 89 terms between 101 and 999 divisible by

2 and 5. ½

2. Let A.P. is 105, 112, 119, ............., 994 which is divisible by 7.

Here, a = 105, d = 112 – 105 = 7, Tn = 994, then

Tn = a + (n – 1)d ½

⇒ 994 = 105 + (n – 1) × 7

⇒ 889 = (n – 1) × 7 ½

⇒ n – 1 =

889

7=127

½

⇒ n = 127 + 1 = 128. ½

Therefore, the first negative term is 52nd term.

3. Two digit numbers which are divisible by 7 are

14, 21, 28, ......98. ½

It forms an A.P.

a = 14, d = 7, an = 98 ½

an = a + (n – 1)d

98 = 14 + (n – 1)7 ½

98 – 14 = 7n – 7

84 + 7 = 7n

⇒ 7n = 91

⇒ n = 13 ½



4. Here a = 101, d = 7, an = 997 1

⇒ an = a + (n – 1)d ½

997 = 101 + (n – 1) × 7 ½

\ n = 129. 1


5. Here, a = 12, l = 264, d = 4

n = l a- -

d+1=

264 12

4+1

1

= 252

4 + 1 = 63 + 1 = 64 1

There are 64 multiples of 4 that lie between 11 and

266. 1

6. Here, required money is ` 2500

a = saving in 1st week = ` 100

d = difference in weekly saving = ` 20

A.P. formed by saving, 1


Sequence is 100, 120, 140, ..... upto 12 terms

Q Sn =

n

2[2a + (n – 1)d]

\ S12 =

12

2[2 × 100 + (12 – 1) × 20]

⇒ = 6[200 + 11 × 20]

⇒ = 6[200 + 220]

⇒ = 6 × 420 = 2520 ½

She will be able to send her daughter to school after 12 weeks. ½

Value : small saving can fulfill your big desires or anyone else’s. [CBSE Marking Scheme, 2015] 1

qqq

WORKSHEET-54

HOTS & Value based Answers

1. Here an = n2 + 1

d = an – an – 1 1

⇒ d = (n2 + 1) – [(n – 1)2 + 1]

⇒ = n2 + 1 – n2 + 2n – 1 – 1

⇒ d = 2n – 1 1

Since common difference depends upon n and not constant, n2 + 1 can not be the nth term of an A.P. 1


2. S7 = 49 ⇒ 2a + 6d = 14 ½ S17 = 289 ⇒ 2a + 16d = 34 ½ Solving equations to get a = l and d = 2 1

Hence Sn = n

2[2 + (n – 1)2] = n2. 1


3. Let the three digits be a – d, a, a + d. ½

Sum = a – d + a + a + d = 3a = 15 given ½

\ the three digits are 5 – d, 5, 5 + d.

\ Original number = 100(5 – d) + 10 × 5 + 1(5 + d)

= 555 – 99d

Revered number = 100(5 + d) + 10 × 5 + 1(5 – d)

= 555 + 99d

According to question,

(555 – 99d) – (555 + 99d) = 594

– 198d = 594

⇒ d = 594

198- = – 3 1

\ The three digits are

5 – (–3), 5, 5 + (–3)

i,e., 8, 5 and 2 ½

\ Original number is 8 × 100 + 5 × 10 + 2 × 1 = 852. ½

4. The list of 2-digit odd positive numbers are 11, 13 ...... 99. 1

This is an A.P. with a = 11, d = 2

We have, tn = a + (n – 1)d

So, 99 = 11 + (n – 1)2

⇒ n = 45 1

\ Sn =

na n d

2[2 +( 1) ]-

=

45

2[2(11)+(44)2]

⇒ Sn =

45

2[22+88]

= 2475. 1

5. (i) Since, each section of class plants the same number of trees as the class number and there are three sections of each class.

\Total number of trees planted by the students

= 3[1 + 2 + ....... + 12]

= 312

2[2×1+(12 1)×1]-

1

= 3[6(2 + 11)]

= 18 × 13 = 234

\ Students planted 234 trees. 1

(ii) Arithmetic Progression. 1

(iii) Our duty towards earth is to save our environment. 1

qqq


LINES (IN TWO DIMESIONS)

SECTION

BSECTIONCHAPTER

6 TOPIC-1

Distance Between Two Points and Section Formula

WORKSHEET-55

Solutions

1. Let P(x, y) is equidistant from A(–5, 3) and B (7, 2)

AP = BP

⇒ (( ) ( ) )x y+ +5 32 2-

=

(( ) ( ) )x y− +7 2

2 2-

½

⇒ x2 + 10x + 25 + y2 – 6y + 9

= x2 – 14x + 49 + y2 – 4y + 4 ½

10x – 6y + 34 = – 14x – 4y + 53

10x + 14x – 6y + 4y = 53 – 34

24x – 2y = 19

24x – 2y – 19 = 0 is the required relation. 1


2.

(–1, 2)(2, 5) ( , )x y

3 : 4CA B

Given that, AC

CB =

3

4

Applying section formula for x co-ordinate,

– 1 = 3 4 2

3 4

x +

+

( )

⇒ –7 = 3x + 8

or x = – 5 1

Similarly for y co-ordinate,

2 = 3 4 5

3 4

y +

+

( )

14 = 3y + 20

or y = – 2

\ (x, y) is (–5, –2) 1

Hence x2 + y2 = (–5)2 + (–2)2

= 25 + 4 = 29 1

3. AP =

3

7AB ⇒ AP : PB = 3 : 4 1

A (–2, –2)

P

B (2, –4)3 : 4

(x, y)

\ x =

6 8

7

-

=

-2

7 1

\ y =

- -12 8

7 =

-20

7 ½

P =

− −

2

7

20

7,

½


4. By section formula

9a – 2 =

3 8 1 3 1

3 1

( ) ( )a a+ +

+ ...(i) 1

– b =

3 5 1 3

3 1

( ) ( )+

+

-

...(ii) 1

From (ii),

– b =

15 3

4

-

= 3

b = – 3 1

From (i),

9a – 2 =

24 3 1

4

a a+ +

4(9a – 2) = 27a + 1

36a – 8 = 27a + 1

9a = 9

a = 1 1


qqq


WORKSHEET-56

Solutions

1. The point on the y-axis is (0, 12) \ Distance between (5, 12) and (0, 12)

d = 0 5 12 12

2 2- -( ) + ( )

= 25 0+ = 5 units. 1


2. Let the point P be (2y, y) 1

PQ = PR

⇒

( ) ( )2 2 52 2y y− + +

=

( ) ( )2 3 6

2 2y y+ + − ½

Solving to get y = 8

Hence coordinates of point P are (16, 8). ½


3. Let the point on y-axis be (0, y) and AP : PB = K 1

½

Therefore

5

1

−+

k

k = 0 gives k = 5

Hence required ratio is 5 : 1 ½

y =

− −4 5 6

6

( )

=

−13

3 ½

Hence point on y- axis is 0

13

3, .−

½


4.

A (1, 2)

P

B (6, 7)2 : 3

(x, y)

AP =

2

5AB ⇒ AP : PB = 2 : 3 1

\ x =

12 3

5

+

= 3; y =

14 6

5

+

= 4 1+½

P(x, y) = (3, 4) ½


5. P(x, y), A(6, 2), B(– 2, 6)

PA = PB ⇒ PA2 = PB2 ½

(x – 6)2 + (y – 2)2 = (x + 2)2 + (y – 6)2 ½

⇒ x2 – 12x + 36 + y2 – 4y + 4

= x2 + 4 x + 4 + y2 – 12y + 36

⇒ – 12x – 4y = 4 x – 12y

⇒ 12y – 4y = 4x + 12x

⇒ 8y = 16x 2

⇒ y = 2x Hence proved.


6. Let P(x1, y1) and Q(x2, y2) are two points which divide AB in three equal parts.

By section formula

P(x1, y1) =

1 4 2 2

1 2

1 6 2 3

1 2

× + ×

+

× + ×

+

( ) ( ),

( ) ( )- - -

1

=

- - -4 4

3

6 6

3

+ +

,( )

= (0, – 4) 1

Q(x2, y2) =

2 4 1 2

2 1

2 6 1 3

2 1

× + ×

+

× + ×

+

( ) ( ),

( ) ( )- - -

1

=

- - -8 2

3

12 3

3

+ +

,( )

= (– 2, – 5) 1


qqq

WORKSHEET-57

Solutions

1.

P

(– 5, – 4) (– 2, 3)

A B

(– 3, )k

Let the ratio in which P divides AB be n : 1

Considering x co-ordinate for section formula

–3 = ( ) ( )- -2 1 5

1

n

n

+

+

⇒ –3(n + 1) = – 2n – 5

⇒ –3n –3 = – 2n – 5

5 – 3 = 3n – 2n

2 = n ½

\ Ratio n

1 =

2

1 or 2 : 1 ½

Now, Considering y co-ordinate

k = 2 3 1 4

2 1

( ) ( )+

+

-


k = 6 4

3

2

3

-=

\ k =

2

3 1

2. |PQ| = |PR|

[ ( )] [ ( )]x a b y b a- - -+ +

2 2

=

[ ( )] [ ( )]x a b y b a- - -

2 2+ +

P ( )x, y

R

( )a – b, a + b

Q

( )a + b, b – a

Squaring, we get

[x – (a + b)]2 + [y – (b – a)]2

= [x – (a – b)]2 + [y – (a + b)]2 1

⇒ [x – (a + b)]2 – [x – a + b]2

= (y – a – b)2 – (y – b + a)2

⇒ (x – a – b + x – a + b) (x – a – b – x + a – b)

= (y – a – b + y – b + a)(y – a – b – y + b – a)

⇒ (2x – 2a) (– 2b) = (2y – 2b) (– 2a)

⇒ (x – a) b = (y – b) a

⇒ bx = ay. Hence proved. 1


3.

A (7, 2)

B (9, 10)

E(x , y )1 1

F (x , y )2 2

C (1, 4)

\Co-ordinates of point E

=

9 7

2

10 2

2

+ +

,

= (8, 6) ½

Co-ordinates of point F

=

7 1

2

2 4

2

+ +

,

= (4, 3) ½

Length of EF =

8 4 6 32 2−( ) + ( )-

=

4 32 2

( ) + ( )

= 5 units ...(i) 1

Length of BC = 9 1 10 4

2 2- -( ) + ( )

= 8 6

2 2( ) + ( )

=10 units ...(ii) From equation (i) and (ii), we get

EF = 1

2BC. Hence proved. 1


qqq

WORKSHEET-58

Solutions

1. Given A(3, 0), B(6, 4) and C(–1, 3)

\ AB2 = (3 – 6)2 + (0 – 4)2 = 9 + 16 = 25 ½

BC2 = (6 + 1)2 + (4 –3)2 = 49 + 1 = 50 ½

CA2 = (–1 –3)2 + (3 – 0)2 = 16 + 9 = 25 ½

AB2 = CA2 ⇒ AB = CA

\ Triangle is isosceles

Also, 25 + 25 = 50

⇒ AB2 + CA2 = BC2 ½

Since pythagoras theorem is verified, therefore triangle is a right angled triangle.

2.

A (5, 2)

B (2, –2) C (–2, t)


AB2 = (2 – 5)2 + (– 2 –2)2 =9 + 16 = 25 BC2 = (– 2 – 2)2 + (t + 2)2 = 16 + (t + 2)2

AC2 = (5 + 2)2 + (2 – t)2 = 49 + (2 – t)2 1 Since∆ABC is a right angled triangle \ AC2 = AB2 + BC2

⇒ 49 + (2 – t)2 = 25 + 16 + (t + 2)2

⇒ 49 + 4 – 4t + t2 = 41 + t2 + 4t + 4 ⇒ 53 – 4t = 45 + 4t ⇒ 8t = 8 \ t =1 1


3. Co-ordinates of point R = (4, 0) 1 \ QR = 8 units ½ Let the co-ordinates of point P be (0, y) 1 Since PQ = QR ⇒ (– 4 – 0)2 + (0 – y)2 = 64 ⇒ 16 + y2 = 64

x′ x

y′

P (0,y)

Q

(– 4,0) (4,0)

R

y

⇒ y = ±4 3 1

Coordinates of P are (0, 4 3 )

or (0, − 4 3 ) ½ [CBSE Marking Scheme, 2015]

qqq

WORKSHEET-59

Solutions

1.

A B

Q

Here, BQ = 5

7AB

⇒

BQ

AB =

5

7

⇒

AQ

BQ

=

7 5

5

-

=

2

5

\ AQ : BQ = 2 : 5 1

2. Points are (– 4, – 7) and (0, – 7)

\ Distance =

0 4 7 72 2

+( ) + +( )-

= 4 02

+ = 16 = 4 units 1

3. Let P divide AB in the ratio k : 1 ½

A

P

B (2, –5)k : 1

1 3,

2 2

(3/4, 5/12)

2 +1 / 2

+1

k

k =

3

4 ⇒ 8k + 2 = 3k + 3

⇒

k =

1

5 1

Required ratio = 1 : 5 ½


4. Here ∆ABC is a right angle triangle,

A (4, 7)

B ( , 3)p C (7, 3)

AB2 + BC2 = AC2 ⇒ (p – 4)2 + (3 – 7)2 + (7 – p)2 + (3 – 3)2 = (7 – 4)2 + (3 – 7)2 ½

⇒ (p – 4)2 + (– 4)2 + (7 – p)2 + 0 = (3)2 + (–4)2 1

⇒ p2 – 8p + 16 + 16 + 49 + p2 – 14p = 9 + 16 ⇒ 2p2 – 22p + 81 = 25 ⇒ 2p2 – 22p + 56 = 0 ⇒ p2 – 11p + 28 = 0 ⇒ (p – 4) (p – 7) = 0 ⇒ p = 7 or 4 p ≠ 7, p = 4 ½


5. Let point be (0, y) ½

52 + (y + 2)2 = (3)2 + (y – 2)2 ⇒ 25 + 4y + 4 = 9 – 4y + 4 1

8y = – 16 ⇒ y = – 2 ½

⇒ Point (0, – 2) 1


qqq


TOPIC-2Area of Triangle

WORKSHEET-60

Solutions

1. Area of triangle

=

1

2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]

=

1

2[0(0 – 5) + 6(5 – 0) + 0(0 – 0)]

=

1

2[6 × 5] = 15 sq. units 1


2. Area of the triangle formed by the given points A(0, 1), B(2, 3) and C(3, 4)

=

1

2|0(3 – 4) + 2(4 – 1) + 3(1 – 3)| 1

=

1

2|0 + (2) (3) + (3) ( – 2)|

=

1

2|6 – 6|

=

1

2(0)

= 0 \ The given points are collinear. 1


3. A(x1, y1), B(x2, y2), C(x3, y3)

(2, 3) (4, p) (6, – 3)

Since the points are collinear

\ Area = 0 1

1

2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

⇒

1

2[2(p + 3) + 4(– 3 – 3) + 6(3 – p)] = 0 1

⇒

1

2[2p + 6 – 24 + 18 – 6p] = 0 ½

⇒

1

2[– 4p] = 0

⇒ 4p = 0, p = 0. ½


4. Since the points are collinear \ The area of triangle = 0 ½

\Area of triangle =

1

2[x1 (y2 – y3) + x2 (y3 – y1)

+x3 (y1 – y2)] ½

⇒

1

2[5(4 – y) + (– 3) (y – 2) + x(2 – 4)] = 0

⇒

1

2[20 – 5y – 3y + 6 + (– 2x)] = 0

⇒

1

2[– 2x – 8y + 26] = 0

⇒ x + 4y – 13 = 0

Hence proved. 2


5. Area of the triangle = 1

2 |t (t + 2 – t) + (t + 2)

(t – t + 2) + (t + 3) (t – 2 – t –2)| 2

= 1

2 [2t + 2t + 4 – 4t – 12] 1

= 4 sq. units.

which is independent of t. 1


6. Using mid-point formula, co-ordinaes of P are

− − +

10 2

2

4 0

2,

= –6, 2 1

Now as P lies on line joining C and D. So, C, D, P are collinear.

Hence, Area of ∆ = 0 ...(i)

i.e.,

1

2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 1

Given points are C(– 9, – 4) P(– 6, 2) D(– 4, y) ↓ ↓ ↓ x1, y1 x2, y2 x3, y3 1 So, from (i)

Area. of ∆=

1

2[– 9(2 – y) + (– 6) (y + 4) + ( – 4)

(– 4 – 2)]

⇒1

2[– 18 + 9y – 6y – 24 + 24] = 0

⇒ 3y – 18 = 0

⇒ y = 6

So, the co-ordinates of D are now (– 4, 6)

Now, finding length of CP and PD using distance formula,


CP = 45 = 3 5

PD = 20 = 2 5

\ required ratio =

CP

PD =

3 5

2 5 = 3 : 2 1

qqq

WORKSHEET-61

Solutions

1. Since the points are collinear, then

Area of triangle = 0

1

2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0 ½

1

2[x(– 4 + 5) + (– 3) (– 5 – 2) + 7(2 + 4)] = 0

x + 21 + 42 = 0

x = – 63 ½

2. Let the points be A(2, – 2), B(–2, 1) and C(5, 2)

Applying distance formula

AB2 = (2 + 2)2 + (–2 –1)2

= 16 + 9

AB2 = 25 ⇒ AB = 5

Similarly BC2 = (–2 –5)2 + (1 – 2)2

= 49 + 1 = 50

⇒ BC2 = 50 ⇒ BC = 5 2

Also,

AC2 = (2 – 5)2 + (–2 –2)2

= 9 + 16

= 25

⇒ AC2 = 25 and AC = 5

Clearly AB2 + AC2 = BC2 1

25 + 25 = 50

Hence the triangle is right angled,

Area of ∆ABC = 1

2× Base × Height

= 1

2 × 5 × 5 =

25

2 sq unit. 1

3.

A(1, – 4)

B ( , )x y

E(2, –1)

F(0, –1)

C ( , )x y1 1

½

Let E be the mid point of AB

x + 1

2 = 2 ⇒ x = 3

y + −( ) 4

2 = – 1 ⇒ y = 2

½

⇒ B(3, 2) ½

⇒

x1 1

2

+

= 0 ⇒ x1 = – 1

y1 4

2

+ −( )

= – 1 ⇒ y1 = 2

⇒ C = (– 1, 2) ½

Now the co-ordinates are A(1, – 4), B(3, 2), C(– 1, 2) Area of triangle

= 1

2[x1(y2 – y3) + x2(y3 – y1)

+ x3(y1 – y2)] ½

=

1

2[1(2 – 2) + 3(2 + 4) – 1(– 4 – 2)]

=

1

2[0 + 18 + 6]

= 12 sq. units ½

[CBSE Marking Scheme. 2015]

4.

A (–3, 2) B (5, 4)

C (7, –6)D (–5, –4) ½

ar (∆ABD) =

1

2|–3(8) + 5(–6) + –5(2 – 4)|

= 22 sq. units 1½

ar (∆BCD) = 1

2|5(–2) + 7(–8) – 5(10)|

= 58 sq. units 1½

ar (Quad ABCD) = 80 sq. units ½


qqq


WORKSHEET-62

Solutions

1. Formed by the given points A(1, 3), B(–1, 0) and C(4, 0)

Area of triangle =

1

2[x1(y2 – y3) + x2(y3 – y1)

+ x3(y1 – y2)] ½

=

1

2[1(0 – 0) + ( –1) (0 – 3) + 4(3 – 0)]

=

1

2[3 +12] =

15

2 = 7.5 s, units ½

2. Area of ∆ABC = 0

Area of ∆ABC =

1

2

[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0 ½

=

1

2[x(7 – 5) – 5(5 – y) – 4(y – 7) = 0 1

⇒ 2x – 25 + 5y – 4y + 28 = 0 ½

⇒ 2x + y + 3 = 0


3. If three points are collinear, then area of ∆ = 0

\ Area =

1

2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] 1

Here, (x1, x2, x3) = (– 2, a, 4) and (y1, y2, y3) = (1, b, 1)

\ Area =

1

2[– 2b + 2 + 0 + 4(1 – b)]

[Q (a – b) = 1]

0 = – 3b + 3 1

⇒ 3b = 3 ⇒ b = 1

\ By given data, a – b = 1

i.e., a – 1 = 1

a = 2

\ (a, b) = (2, 1) 1 4. If the points are collinear, then Area of triangle = 0 1

\

1

2[(p + 1) [p – (2p – 6)] + (p – 1)[(2p – 6) – (2p – 2)]

+ (p – 3) (2p – 2 – p)] = 0 1

1

2[(p + 1)(– p + 6) + (p – 1) (– 4) + (p – 3) (p – 2)] = 0

⇒[– p2 + 6p – p + 6 – 4p + 4 + p2 – 2p – 3p + 6] =0 [– 4p + 16] = 0 ⇒ – 4p = – 16 \ p = 4 1

5. Area of triangle =

1

2[x1(y2 – y3) + x2 (y3 – y1) +

x3(y1 – y2)]

⇒24 =

1

2[1(2k + 5) – 4(– 5 + 1) – k(– 1 – 2k)] 2

⇒ 48 = 2k + 5 + 16 + k + 2k2

⇒ 2k2 + 3k – 27 = 0 1

⇒ (k – 3) (2k – 9) = 0

⇒ k = 3, k

−9

2 1


qqq

WORKSHEET-63

Solutions

1. The points are collinear, then area of triangle = 0

\ 1

2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0

⇒

1

2[0(2 – y) + 1(y – 0) + x(0 – 2) ] = 0

⇒

1

2[y – 2x] = 0

⇒ 2x – y = 0 1


2. Given points are (8, 1), (3, – 2k) and (k, – 5) are collinear.

⇒ Area of triangle formed = 0

⇒

1

2[8(– 2k + 5) + 3(– 5, – 1) + k(1 + 2k)] = 0 1

⇒ 2k2 – 15k + 22 = 0

⇒ (k – 2) (2k – 11) = 0

⇒ k = 2,

11

2



3. Area of quadrilateral ABCD= ar (∆ABC) + ar (ADC) 1

A (5, –2) B (–3, –1)

C (2, 1)D (6, 0)

1

ar (quad. ABCD) =1

2[(x1y2 – x2y1) + (x2y3 – x3y2)

+(x3y4 – x4y3) + (x4y1 – x1y4)]

ar (quad. ABCD) =

1

2[5(– 1) – (– 2) (– 3) + (– 3) (1)

– (– 1) (2) + (2 × 0 – 1 × 6) + 6 (– 2) – (0 × 5)]

= 1

2[– 30] = |– 15| = 15 sq. units 1


4. Finding mid-points F (5, 3), D (– 1, – 2) and E (1, – 1) 1

Finding area using the formula,

Area =

1

2[5(– 1) – 1(– 4) + 1 × (5)]

= 2 sq. units 2 [CBSE Marking Scheme, 2012]

5. Area of triangle =

1

2[x1(y2 – y3) + x2 (y3 – y1) +

x3(y1 – y2)] 1

⇒6 =

1

2[(k + 1) (– 3 + k) + 4 (– k – 1) + 7 ( 1 + 3)]

1

⇒ k2 – 6k + 9 = 0 1

⇒ (k – 3) (k – 3) = 0

⇒ k = 3 1


6. Here, Area of ∆ABC = 0

Area of triangle ABC =

1

2[(k + 1) + (2k + 3 – 5k) +

3k(5k – 2k) + (5k – 1) (2k – 2k – 3)] = 0 2

– 3k2 + 3k – 3k + 3 + 9k2 – 15k + 3 = 0

⇒ 6k2 – 15k + 6 = 0

or 2k2 – 5k – 2 = 0 1

(k – 2)(2k – 1)

⇒ k = 2 or k =

1

2 ½ + ½


qqq

WORKSHEET-64

Solutions


1. P(x, y), A(5, 1), B(– 1, 5) PA = PB ⇒ PA2 = PB2 ½ (x – 5)2 + (y – 1)2 = (x + 1)2 + (y – 5)2

⇒ x2 – 10x + 25 + y2 – 2y + 1 = x2 + 2x + 1 + y2 – 10y + 25 ½ ⇒ – 10x – 2y + 26 = 2x – 10y + 26 ½ ⇒ – 10x – 2x = – 10y + 2y ⇒ – 12x = – 8y ⇒ 3x = 2y. Hence proved. ½ [CBSE Marking Scheme, 2012]

2. Point P(6, – 6) lies on the line 3x + k (y + 1) = 0 ⇒ 18 + k (– 6 + 1) = 0 1½

⇒ k =

18

5 1½


3. Let co-ordinates of P are (x1, y1) and it divides line AB in the ratio k : 1.

\ x1 =

8 3

1

k

k

+

+,

y1 = 9 1

1

k

k

-

+ 1

Point (x1, y1) lies on line x – y – 2 = 0, so co-ordinates of P must satisfy the equation of line

8 3

1

9 1

12

k

k

k

k

+

+ +-

--

= 0 1

⇒ 8k + 3 – 9k + 1 – 2k – 2 = 0

x – y – 2 = 0

(3, –1) (8, 9)

kA B

P

l

⇒ k =

2

3 ½

So, line x – y – 2 = 0 divides AB in the ratio 2 : 3. ½ [CBSE Marking Scheme, 2012]


4. (i) Using distance formula, we have

AB = ( ) ( )6 3 4 1

2 2- -+

= ( ) ( )3 32 2

+

= 9 9+ = 18 = 3 2 units ½

BC =

( ) ( )8 6 6 4

2 2- -+

=

( ) ( )2 2

2 2+

= 4 4+ = 8 = 2 2 units ½

AC = ( ) ( )8 3 6 1

2 2- -+

=

5 52 2

( ) + ( )

= 25 25+ = 50 = 5 2 units ½

Q AB + BC = 3 2 2 2+ = 5 2 = AC

\ A, B and C are collinear.

Thus, Ashima, Bharti and Camella are seated in a line. ½

(ii) Co-ordinate Geometry. ½

(iii) Democratic values lead to equality. ½

qqq

WORKSHEET-65

Solutions


1.

P Q

( )a, b ( )b, aR

( )x, y1�

Let point R divides the line joining P and Q in the

ratio λ : 1. ½

x =

λ

λ

b a+

+ 1 and y =

λ

λ

a b+

+ 1

Adding, x + y =

λ λ

λ

b a a b+ + +

+ 1 1

=

λ

λ

a b a b+( ) + +( )+ 1

1

=

λ

λ

+( ) × +( )+

1

1

a b= a + b

⇒ x + y = a + b. Hence proved. ½


2. (i) Considering A as origin (0, 0), AB as X-axis and AD as Y-axis.

Niharika runs in the 2nd line with green flag and distance covered (parallel to AD)

= 1

4100×

= 25 m

\ Co-ordinates of green flag are (2, 25) and label it as P i.e., P(2, 25). 1

Similarly, Preet runs in the eighth line with red flag and distance covered (parallel to AD)

= 1

5100×

= 20 m

\ Co-ordinates of red flag are (8, 20) and label it as Q i.e., Q (8, 20) 1

Now, using distance formula, distance between green flag and blue flag

PQ = ( ) ( )8 2 20 25

2 2- -+

=

6 52 2

+ ( )- =

36 25+

= 61 m ½

(ii) Also, Rashmi has to post a blue flag at the mid-point of PQ, therefore by using mid-point formula, we

have 2 8

2

25 20

2

+ +

,

i.e.,

545

2,

½

Hence, the blue flag is in the fifth line, at a distance

of 45

2 i.e., 22.5 m along the direction parallel to AD.

½

(iii) Co-ordinate Geometry.

(iv) Team spirit leads to perfection. ½

qqq


WORKSHEET-66

Solutions

1. As DE || BC

\

AD

DB =

AE

EC

⇒

x

x + 1 =

x

x

+

+

3

5

⇒ x2 + 5x = x2 + 4x + 3 ⇒ x = 3 1


2. ∆ABC ∼ ∆PQR (Given)

AB

PQ =

BC

QR =

AC

PR

⇒

z

3 =

8

6 =

4 3

y

⇒

z

3 =

8

6 and

8

6

=

4 3

y

⇒ z =

8 3

6

×

and y =

4 3 6

8

×

\ z = 4 and y = 3 3

\ y + z = 4 + 3 3 2

3

. B

A

CD Q

P

RS

∆ABC ∼ ∆PQR

∠A = ∠P, ∠B = ∠Q, ∠C = ∠R 1

ar ABC

ar PQR

( )

( )

∆

∆

=

AB

PQ

2

2

...(i)

∠A = ∠P

⇒ 1

2 ∠A =

1

2∠P

⇒ ∠BAD = ∠QPS

∆BAD ∼ ∆QPS (AA similarity) 1

BA

QP =

AD

PS ...(ii)

By eq. (i) and (ii),

ar ABC

ar PQR

( )

( )

∆

∆

=

AD

PS

2

2

1


qqq

WORKSHEET-67

Solutions

1. In ∆PAO and ∆QBO

∠A = ∠B = 90° (Given)

∠POA = ∠QOB

(Vertically Opposite Angle)

∆PAO ∼ ∆QBO, (by AA)

OA

OB =

PA

QB

⇒

6

4 5. =

4

QB

⇒ QB =

4 4 5

6

× .

\ QB = 3 cm 1

2. AB2 = OA2 + OB2 = (3)2 + (4)2

(Pythagoras th.)

AB = 5 cm 1

AB2 + AC2 = (5)2 + (12)2 = (13)2 = BC2

∠CAB = 90° 1

(Converse of Pythagoras th.)

3. Let ∆ABC be an equilateral triangle of side 24 cm and AD is altitude which is also a perpendicular bisector of side BC.

Hence BD =

BC

2 =

24

2 = 12 cm

AB = 24 cm

\ AD = AB BD2 2-

= 24 12

2 2( ) ( )-

TRIANGLES

SECTION

BSECTIONCHAPTER

7


= 576 144-

= 432

\ AD = 12 3 cm

\ The length of the altitude is 12 3 cm. 2

4. (i) The given ∆'s are not similar. 1

(ii) In ∆PQR, ∠R = 180° – (45° + 78°) = 57°

In ∆LMN, ∠N = 180° – (45° + 57°) = 78°

\ ∆PQR ∼ ∆MNL (By AA similarly criterion) 2


qqq

WORKSHEET-68

Solutions

1.

A

YX

B C

AX

XB =

3

4, AY= 5, YC = 9 (Given)

AX

XB =

3

4 and

AY

YC =

5

9

AX

XB ≠

AY

YC

Hence XY is not parallel to BC. 1

2. AD2 = BD × CD

⇒

AD

CD =

BD

AD ½

\ ∆ADC ∼ ∆BDA (by SAS; Q∠D = 90°)

⇒ ∠BAD = ∠ACD;

∠DAC = ∠DBA

(Corresponding angles of similar triangles) ½

∠BAD+ ∠ACD + ∠DAC + ∠DBA = 180°

⇒ 2∠BAD + 2∠DAC = 180° ½

⇒ ∠BAD + ∠DAC = 90°

\ ∠A = 90° ½

3. Let the triangle be ABC.

A

B C

h

a/2 a/2

aa

D

1

In ∆ABD,

(a)2 =

ah

2

22

+

⇒ h2 = a

a22

4−

⇒ h2 =

3

4

2a

\ h = 3

2

a

1

4. In ∆ADE and ∆ABC, ∠A = ∠A (Common) ∠ADE = ∠ABC (Given) \ ∆ADE ∼ ∆ABC (AA similarity)

⇒

AD

AB =

DE

BC

⇒

AD

AE BE+ =

DE

BC

⇒

7 6

7 2 4 2

.

. .+

= DE

8 4.

⇒ DE =

7 6 8 4

11 4

. .

.

×

= 5.6 cm. 4

qqq

WORKSHEET-69

Solutions

1.

CD

AD =

CE

BE ⇒

x

x

+ 3

2 =

x

x2 1− ½

⇒ 5x = 3 ⇒ x = 3

5 ½




In ABC, DE||AB (Given)

CD

CA =

CE

CB

(By Thales Theorem)

⇒

CD

CD AD+

= CE

CE BE+

⇒

x

x x

+

+ +

3

3 2 =

x

x x+ 2 1- ½

⇒

x

x

+

+

3

3 3 =

x

x3 1-

⇒ (x + 3)(3x – 1) = x(3x + 3)

⇒ 3x2 – x + 9x – 3 = 3x2 + 3x

⇒ 8x – 3 = 3x

⇒ 8x – 3x = 3

⇒ 5x = 3

\

x =

3

5 ½

2. ∆ABC ∼ ∆DEF (Given)

ar ABC

ar DEF

∆( )∆( )

=

AB

DE

2

2

1

⇒

100

196

= AB2

27( )

⇒

100

196 =

AB2

49

⇒ AB2 =

49 100

196

×

⇒ AB2 =

4900

196

⇒ AB2 = 25

\ AB = 5 cm 1

3. Given : AD = 1.5 cm, \ BD = 3 cm and AB = AD + BD = 1.5 + 3.0 = 4.5 cm.

Proof : Given, In triangle ADE and ABC,

∠A is common and DE || BC

⇒ ∠ADE = ∠ABC

(Corresponding angles) 1

\ ∆ADE ∼ ∆ABC, (AA similarity)

\

ar ADE

ar ABC

( )

( )

∆

∆ =

AD

AB

2

2 =

1 5

4

2

2

.

.

( )

( )5 =

1

9

⇒

ar ADE

ar ABC ar ADE

( )

( ) ( )

∆

∆ ∆-

=

1

9 1-

1

ar ADE

ar BCED

( )

( )

∆

trapezium =

1

8

4.

A B

E

C

F

D

∆BCE ∼ ∆ACF (equilateral ∆'s are equiangular)

hence they are similar by AAA similar criterion

∆ABC is a right triangle.

\ By Pythagoras theorem,

AC2 = AB2 + BC2

⇒ AC2 = 2BC2

(\AB = BC in a square) 1

⇒ AC = 2 BC

Now ar ACB

ar DCE

∆

∆

= AC

BC

2

2

=

22

2

BC

BC

( ) = 2

\ ar (∆ACF) = 2 ar (∆BEC)

⇒ ar (∆BEC) =

1

2 ar (∆ACF) 2


qqq

WORKSHEET-70

Solutions

1. No, Angle included should be same. 1

2. According to the question, (Given) ∠QPR = 90°

\ QR2 = QP2 + PR2 1

\ PR = 26 242 2-

= 100 = 10 cm

∠PKR = 90° (Given)

\ PK = 10 82 2- = 100 64-

= 36 = 6 cm. 1


3. ∆AOB ∼ ∆COD (AA similarity)

ar

ar

∆( )∆( )

COD

AOB =

CD

AB

2

2

1

= CD

CD

2

23( ) =

CD

CD

2

29

=

1

91

ratio = 1 : 9

O

CD

BA

1


4. BC and OX bisect each other

So BXCO is a parallelogram, BE || XC and BX || CF

In ∆ABX, by B.P.T.,

AF

FB =

AO

OX ...(i)

In ∆AXC,

AE

EC =

AO

OX ...(ii)

(i) and (ii) give,

AF

FB =

AE

EC

So by converse of B.P.T.,

EF || BC

(i) give

OX

OA =

AB

AF

Adding 1 on both sides

AX

OA =

AB

AF

or OA : AX = AF : AB 4

qqq

WORKSHEET-71

Solutions

1. Let the triangles be ∆ABC and ∆DEF ½

ar

ar

( )

( )

∆

∆

ABC

DEF =

5

6

2

=

25

36

25 : 36 ½ [CBSE Marking Scheme, 2015]

2. According to the question, AO = 20 cm, BO = 12 cm, PB = 18 cm In ∆AQO and ∆BPO, ∠AOQ = ∠BOP (Vertically opposite angles) ∠A = ∠B = 90° \ ∆AQO ∼ ∆BPO (AA criterion) 1

⇒

AQ

BP =

QO

PO =

AO

BO

⇒

AQ

18 =

20

12 \ AQ = 30 cm. 1

3. A

C B

D

p

bc

a

In ∆ACB and ∆CDB ∠ACB = ∠CDB = 90° ∠B = ∠B (common) \ ∆ACB ∼ ∆CDB (by AA Similarity)

⇒

b

p

=

c

a

⇒

1

p

=

c

ab

Squaring on both sides,

12p

=

c

a b

2

2 2

⇒

12p

=

a b

a b

2 2

2 2

+

\

12p

=

1 12 2b a

+ Hence Proved.

4. Given \ AB + AD = BC + CD ⇒ AD = BC + CD – AB ⇒ AD = h + d – x 1

In rt ∆ACD, AD2 = AC2 + DC2

⇒ (h + d – x)2 = (x + h)2 + d2

⇒ (h + d – x)2 – (x + h)2 = d2 1

⇒(h + d – x – x – h) (h + d – x + x + h) = d2

[\(a2 – b2) = (a – b) (a + b)] ⇒ (d – 2x) (2h + d) = d2\ ⇒ 2hd + d2 – 4hx – 2xd = d2

⇒ 2hd = 4hx + 2xd

= 2 (2h + d) x

⇒ x =

hd

h d2 + 2


qqq


WORKSHEET-72

Solutions

1. In the triangles AOB and DOC,

Since

AO

OC =

BO

OD =

1

2 and AB = 3 cm

Q

AO

OC =

BO

OD (Given)

\ ∠AOB = ∠DOC

(Vertically opposite angles)

\ ∆AOB ~ ∆DOC

(SAS similarity)

\ AO

OC =

BO

OD=AB

DC =

1

2

(\In similar triangles corresponding

sides are proportional)

\ DC = 2AB

⇒ DC = 2 × 3 = 6 cm 1

2. Since G is the mid-point of PQ,

\ PG = GQ

⇒

PG

GQ = 1 1

According to the question, GH || QR

\

PG

GQ =

PH

HR (By BPT)

⇒

1 =

PH

HR \ PH = HR. 1

Hence, H is the mid-point of PR. 3.

A

C

D

B

E

\ DE || BC

AD

DB =

AE

EC (By BPT )

⇒

x

x

+

+

2

3 16 =

x

x3 5+ 1

⇒ (x + 2)(3x + 5) = x(3x + 16) ⇒ 3x2 + 5x + 6x + 10 = 3x2 + 16x ⇒ 11x + 10 = 16x 1

⇒ 16x – 11x = 10 ⇒ 5x = 10 \ x = 2 1

qqq

WORKSHEET-73

Solutions

1. Draw AC intersecting EF at G.

D C

BA

GE F

In ∆CAB, GF || AB

⇒

AG

CG =

BF

FC (By BPT)...(i) 1

In ∆ADC, EG || DC

⇒

AE

ED =

AG

CG (By BPT)...(ii)

From equations (i) and (ii),

AE

ED =

BF

FC· 1

2. In ∆CAB, ∠A = ∠B (Given)

\ AC = CB

(By isosceles triangle property) 1

But, AD = BE (Given) ...(i) ⇒ AC – AD = CB – BE

\ CD = CE ...(ii)

Dividing equation (ii) by (i),

CD

AD =

CE

BE

By converse of BPT,

DE || AB. 1

3. Given : ABC is right angled at B and D is the mid-point of BC. ½

\ BD = DC =

1

2BC ½


A

B D C

In ∆ABD, AD2 = AB2 + BD2

(Pythagoras theorem) ...(i) ½

In ∆ABC, AC2 = AB2 + BC2

(Pythagoras theorem) ...(ii) ½

From eqn. (i), AD2 = AB2 +

BC

2

2

(D is the mid-point of BC) ½

⇒ 4AD2 = 4AB2 + BC2

⇒ BC2 = 4AD2 – 4AB2 ...(iii) ½

Using this in equation (ii),

AC2 = AB2 + 4AD2 – 4AB2

⇒ AC2 = 4AD2 – 3AB2

Hence proved. 1

qqq

WORKSHEET-74

Solutions

1. Given DE|| BC, then

AD

AB =

DE

BC(By Thales Theorem)

⇒

3

AD DB+ =

x

14

⇒ 3

3 4+ =

x

14

⇒

3

7 =

x

14

⇒ 7x = 14 × 3

\ x =

14 3

7

× = 6 cm 1

2.

D

A B

C

3 km

4 km

6 km

E 1

Given, AE =

2

3AB

=

2

36×

= 4 km

In right triangle ADE,

DE2 = (3)2 + (4)2

⇒ DE2 = 25

\ DE = 5 km. 1


3. From given figures,

PQ

ZY =

4 2

8 4

.

. =

1

2;

PR

ZX =

3 3

6 3 =

1

2;

QR

YX =

7

14 =

1

2

⇒

PQ

ZY =

PR

ZX =

QR

YX 1

⇒ ∆PQR ∼ ∆ZYX (By SSS) \ ∠X = ∠R \ ∠X = 180° – (60° + 70°) = 50° 1

4.

OW

N

E1,300

S

1,0

00

Distance covered by first aeroplane due North after two hours = 500 × 2 = 1,000 km. 1

Distance covered by second aeroplane due East after two hours = 650 × 2 = 1,300 km. 1

Distance between two aeroplane after 2 hours

NE = ON OE

2 2+

= 1000 1300

2 2( ) + ( )

= 1000000 1690000+

= 2690000

= 1640.12 km 1

qqq


WORKSHEET-75

Solutions

1. Q PQ || MN

So,

KP

PM =

KQ

QN

(By BPT)

⇒

KP

PM =

KQ

KN KQ-

⇒ 4

13 =

KQ

KQ20.4 -

⇒ 4 × 20.4 – 4 KQ = 13 KQ

⇒ 17 KQ = 4 × 20.4

\ KQ =

20.4×4

17 = 4.8

cm. 1

2. Since, XY || QR

\

PX

XQ =

PY

YR (By BPT)

⇒

1

2 =

PY

PR PY-

=

4

4PR - 1

⇒ PR – 4 = 8

⇒ PR = 12 cm

RQ

P

X Y

4 cm

\In right ∆PQR, QR2 = PR2 – PQ2

= 122 – 62

= 144 – 36 = 108

\ QR = 6 3 cm 1

3. Here, ∆APQ ∼ ∆ABC

\

ar

ar

( )

( )

∆

∆

A Q

ABC

P

=

AP

AB

2

2

⇒

1

2 =

AP

AB

2

2

⇒

1

2 =

AB BP

AB

-(QAB = AP + BP)

⇒

AB BP

AB

-

=

1

2 1

⇒ 1 –

BP

AB =

1

2 A

P Q

B C

⇒

BP

AB = 1 –

1

2 =

2 1

2

-

\BP : AB = ( 2 – 1) : 2 2

qqq

WORKSHEET-76

Solutions

1. Let CD ⊥ AB, then CD = p

Area of ∆ABC =

1

2 × base × height 1

⇒ Area of ∆ABC =

1

2 × AB × CD =

1

2 cp

Also, Area of ∆ABC =

1

2 × BC × AC =

1

2 ab

1

2cp =

1

2 ab

⇒ cp = ab. Proved 1

2. In ∆ABD, from Pythagoras theorem,A

B CD

⇒ AB2 = AD2 + BD2 1

⇒ BC2 = AD2 + BD2, (as AB = BC = CA) ⇒ (2BD)2 = AD2 + BD2, (⊥ is the median in an equilateral ∆) \ 3BD2 = AD2. 1


3. Proof : LN || CD,

⇒

AL

AC =

AN

AD ( By BPT) 1

and LM || CB,

⇒

AL

AC =AM

AB (By BPT) 1

\

AM

AB =

AN

AD 1

4. Q ∆FEC ≅ ∆GBD

⇒ EC = BD ...(i) 1 It is given that ∠1 = ∠2 ⇒ AE = AD (Isosceles triangle property)...(ii) 1

From eqns. (i) and (ii),

AE

EC =

AD

BD ⇒ DE || BC, (converse of B.P.T.) ⇒ ∠1 = ∠3 and ∠2 = ∠4 (Corresponding angles) 1 Thus in ∆ADE and ∆ABC, ∠A = ∠A ∠1 = ∠3 ∠2 = ∠4 So by AAA criterion of similarity, ∆ADE ∼ ∆ABC. Hence proved 1

qqq

WORKSHEET-77

Solutions


1. By Pythagoras theorem,

BC = 10 cm

A (∆ABC) = A (∆AOB) + A (∆AOC) + A (∆BOC) ½

⇒

1

2× ×AB AC

=

1

2( )AB r×

+

1

2( )AC r×

+1

2( )BC r×

½

⇒

1

28 6× ×

=

1

28( )r

+1

26( )r

+

1

210( )r

½

⇒ 24 = 12r

⇒ r = 2 cm. ½

2.

A

P

C QB

Since P divides AC in the ratio 2 : 1

CP =

2

3 AC

QC =

2

3 BC ½

AQ2 = QC2 + AC2

⇒

AQ2 =

4

9BC2 + AC2

⇒ 9AQ2 = 4BC2 + 9AC2 ...(i) ½

Similarly, we get 9BP2 = 9BC2 + 4AC2 ...(ii)


9(AQ2 + BP2) = 13AB2 1

3. Given : AB = BC = CD =AD = 5 cm

and AC = 6 cm

O

A

B

C

D

5 cm

AO = OC

⇒ AO = 3 cm.

∆AOB is right angled triangle as diagonals of rhombus intersect at right angles. 1

\ By Pythagoras theorem, OB = 4 cm.

Since DO = OB, BD = 8 cm.

Length of the other diagonal = 2(BO)

where BO = 4 cm

\ BD = 8 cm. 1

4. Distance from village A to B = 7 km

Distance from village B to C = 5 km

and distance from village C to A = 8 km

O

A

B C

(i) Triangle 2

(ii) Location of the well be will at the incentre of the triangle. 1

(iii) Social, Honesty, Equality. 1

qqq


WORKSHEET-78


1. Draw AL ⊥BC.

⇒ AL is median of BC (Isosceles triangle)

⇒ BL = LC = 6 cm.

In right ∆ALB, by Pythagoras theorem, AL = 8 cm.

In ∆BPQ and ∆BLA,

∠B = ∠C (Isosceles triangle)

∠BPQ = ∠BLA = 90°

\ ∆BPQ ~ ∆BLA (AA similarity)

⇒

BP

PQ =

BL

AL 1

⇒

6 - x

y =

6

8 ⇒x = 6 –

3

4

y.

Hence proved. 1

2.

A

CB

D 1

In ∆ADC, By Pythagoras theorem,

AC2 = AD2 + CD2

= AD2 + (BC + BD)2

= AD2 + BC2 + 2BC × BD + BD2

= AB2 + BC2 + 2BC × BD 1

3. (i) Let AB and CD be the two trees of height a and b metre such that the trees are p metre apart i.e., AC = p. Let the lines AD and BC meet at O such that OL = h m.

½

D

B

C A

O

x y

a

b

h

L

Let CL = x and LA = y,

then x + y = p

In ∆ABC and ∆LOC,

∠CAB = ∠CLO (each 90°)

∠C = ∠C (Common)

\ ∆CAB ~ ∆CLO (AA similarity)

⇒

CA

CL =

AB

LO

⇒ p

x =

a

h

⇒ x = ph

a ...(i) ½

In ∆ALO and ∆ACD,

∠ALO = ∠ACD (each 90° )

∠A = ∠A (Common)

\ ∠ALO ~ ∠ACD (AA similarity)

⇒

AL

AC =

OL

DC

⇒ y

p =

h

b

⇒ y = ph

d ...(ii) ½


x + y =

ph

a

ph

b+

⇒ p = pha b

1 1+

⇒ 1

h =

1 1

a b+

\ h = ab

a b+

m. ½

(ii) Triangle. 1

(iii) Trees are helpful to keep our life in this world. They should be saved at any cost. 1

qqq


WORKSHEET-79

Solutions

1. ∠OAC = 90° (as radius ⊥r tangent)

∠BOC = ∠OAC + ∠ACO

(Exterior angle property)

⇒ 130° = 90° + ∠ACO

⇒ ∠ACO = 130° – 90°

= 40° 1

2. ∠APB = 80° ½

∴ ∠AOB = 100° ½



∠PAB = 50°

∴ ∠PBA = 50° (as PA = PB)

⇒ 50 + 50 + ∠P = 180° (angle sum property) ⇒ ∠P = 80°

A

B

POC

50°

Also, ∠O + ∠P = 180°

⇒ ∠O = 180° – 80° = 100° 1

3. Given, AP = 8 and OA = 6 cm

∠OAP = 90° (as radius ⊥ tangent)

⇒ OP2 = AP2 + OA2

= 82 + 62

= 64 + 36

OP2 = 100

⇒ OP = 10 cm 1

Now in triangle OPB,

OP = 10 cm, OB = 4 cm

⇒ OP2 = BP2 + OB2

⇒ 100 = BP2 + 16

BP2 = 100 – 16 = 84

or BP = 84

= 2 21 cm. 1

4. Given that ∠SPT = 120°

⇒ ∠OPS = 120

2

° = 60°

(as OP bisects ∠SPT) ½

Also, ∠ PTO = 90°

(as radius ⊥ tangent) ½

∴ In right triangle POS.

cos ∠OPS = PS

OP

⇒ 1

2 =

PS

OP

⇒ OP = 2PS. 1

Hence proved.

5. Let PL = y, OP is ⊥ bisector of AB

⇒ AL = BL = 8 cm

OL2 = OA2 – AL2 = 102 – 82 = 36

⇒ OL = 6 cm 1

In ∆OAP, AP2 = (y + 6)2 – (10)2 ...(i)

In ∆ALP, AP2 = y2 + 64 ...(ii) 2

From (i) and (ii) y =

32

3 ½

∴ AP =

40

3 cm. ½


qqq

WORKSHEET-80

Solutions

1. On the diameter of a circle only two tangents can

be drawn. [CBSE Marking Scheme, 2012] 1

A

P

B

O50°

CIRCLES

SECTION

BSECTIONCHAPTER

8


2. Here, ∠APB = 50°

∠PAB = ∠PBA =

180 50° °-

2 = 65°

∠OAB = 90° – ∠PAB

= 90° – 65° = 25°


3.

P

O

T

A BD ½

Join PO and produce it to D. ½

OP ⊥ TP and TP || AB.

∠ADP = 90°

OD ⊥ AB

⇒ AD = BD ½

∆ADP ≅ ∆BDP, [SAS] ½

∴ PA = PB, (CPCT)


4.

A

B

PO

½

Let P be any external point from which two tangents PA and PB are drawn to a circle with centre O.

To Prove : PA = PB ½

Proof : ∠OAP = ∠OBP = 90. 1

(radius is ⊥ to tangent)

In ∆APO and ∆BPO,

∠OAP = ∠OBP (each 90°)

OA = OB (radius)

OP = OP common

∴ ∆APO ≅ ∆BPO (by RHS) 1

⇒ PA = PB (CPCT) 1

qqq

WORKSHEET-81

Solutions

1. ∠QPR = ∠50° (Given) ∴ ∠QOR = 180° – 50° = 130° ½ From ∆OQR

⇒ ∠OQR = ∠ORQ =

180 130

2

° °-

=

50

2

°

= 25° ½


2. ∠QOR + ∠QPR = 180° (Supplementary angles)

⇒ ∠QOR + 46° = 180° ⇒ ∠QOR = 180° – 46° = 134°. 1 3. Let AD = AF = x ⇒ DB = BE = 12 – x

and CF = CE = 10 – x BC = BE + EC ⇒ 8 = 12 – x + 10 – x ⇒ x = 7 1 ∴ AD = 7 cm, BE = 5 cm, CF = 3 cm 1

4. PA = PB ½

⇒ ∠PAB = ∠PBA = 60° ½

∴ ∆PAB is an equilateral triangle. ½

Hence AB = PA = 5 cm. ½


5.

P

A

N

MC B

c

O

b

a 1

Let circle touch CB at M and CA at N, AB at P

Now OM ⊥ CB and ON ⊥ AC (radius ⊥ tangant)

OMCN is square

Let OM = r = CM = CN 1

AN = AP, CN = CM, BM = BP

(tangant from external point)

AN = AP

or AC – CN = AB – BP 1

b – r = c – BM

b – r = c – (a – r)

b – r = c – a + r

∴ 2r = a + b – c

r =

a b c+ -

2.

1


qqq


WORKSHEET-82

Solutions

1.


2. It changes into a rhombus. 1 3. PQ = 6 cm, OP = OQ = 6 cm ∴ PQ = OP = OQ

∴ ∠POQ = 60° (angle of equilateral ∆.) ½

∠P = ∠Q = 90° (radius ⊥ tangent)

∴ ∠T + 90° + 90° + 60° = 360° (angle sum poperty) ½

∠T = 120° 1

O

Q

T

P

4. Since the tangent is perpendicular to the radius,

∠OAB =∠OBA (Q OA = OB)

∠OBA = 30° 1

∴ ∠AOB = 180° – (30° + 30°)

∴ ∠AOB = 120° 1

∠ABP = ∠OBP – ∠OBA

= 90° – 30° = 60° 1


5. PT = 169 25- = 12 cm

and TE = 8 cm ½ + ½

Let PA = AE = x

TA2 = TE2 + EA2 1

⇒ (12 – x)2 = 64 + x2

⇒ x = 3.3 cm. 1

Thus AB = 6.6 cm. 1


qqq

WORKSHEET-83

Solutions

1. QP = 3.8 QP = PT (Length of tangents from

external points are equal) ⇒ PT = 3.8 cm PR = PT = 3.8 cm ⇒ QR = 7.6 cm. 1 [CBSE Marking Scheme, 2012]

2. AX = AY → (1) BX = BZ → (2) CZ = CY → (3) (Tangents from an external

point to a circle are equal)½

AB = AC, (Given) ⇒ AX + XB = AY + YC ½ ⇒ XB = YC ⇒ BZ = CZ ½ ∴ Z is the mid-point of BC and Z is the point of

contact. ⇒ BC is bisected at the point of contact. ½ [CBSE Marking Scheme, 2012]

3. ∠AOQ = 58° (Given)

∠ABQ =

1

2∠AOQ

(Angle on the circumference of the circle by the same arc)


=

1

2 × 58°

= 29° 1 ∠BAT = 90° (Q OA ⊥ AT) ∴ ∠ATQ = 90° – 29° = 61° 1 [CBSE Marking Scheme, 2015]

4.

C A D

E B F

O

Let AB be a diameter of a given circle and let CD

and EF be the tangent lines drawn to the circle at A and B respectively.

AB ⊥ CD and AB ⊥ EF 1

∴ ∠CAB = 90° and ∠ABF = 90° ½

∠CAB = ∠ABF ½

∠CAB and ∠ABF are alternate interior angles. 1

∴ CD EF


qqq

WORKSHEET-84

Solutions

1. From ∆AOP

∠APO + ∠OAP + ∠AOP = 180°

⇒ ∠APO + 90° + 60° = 180°

⇒ ∠APO = 180° – 150° = 30°. 1


2. Let AF = AE = x

∴ AB = 6 + x, AC = 9 + x, BC = 15 ½A

BD

C

O

3 cm

X

F

6

X

E

9

6 9

1

2 [15 + 6 + x + 9 + x].3

= 54 1

⇒ x = 3 ∴ AB = 9 cm, AC = 12 cm ½ and BC = 15 cm. [CBSE Marking Scheme, 2015]

3. ∠POR = 90° – 60° = 30° 1

PR

OR = sin 30° =

1

2

⇒ OR = 2 PR

P

RO

Q PR = QR ∴ OR = PR + QR 1 [CBSE Marking Scheme, 2015]

4.

F

EB

D

A

C ½ AF = AD BE = BD, (tangents from external points) 1½ CE = CF

AB = AC AD + BD = AF + FC ⇒ BD = FC (Q AD = AF) BE = EC (Q BD = BE, CE = CF) ∴ E bisects BC. 1 [CBSE Marking Scheme, 2012]

qqq


WORKSHEET-85

Solutions

1. Q ∠RPQ = 50° ∴ ∠OPQ = 90° – 50° = 40° Since, OP = OQ (radii of circle) Q ∠OPQ = ∠OQR = 40° ∠POQ = 180° – (40° + 40°) = 100°. 1 [CBSE Marking Scheme, 2012]

2. CAPB will be a square CA = AP = PB = BC = 4 cm ∴ Length of tangent = 4 cm. 1

3. PA = PB, KA = KM, NB = NM, (Length of tangents from an external point are

equal) ∴ KA + NB = KM + NM 1 or AK + BN = KM + MN = KN. 1 [CBSE Marking Scheme, 2012]

4. ∠TOQ = 180° – 70° = 110° 1

T

R P70°

Q

O

⇒

∠TRQ

=

1

2 ∠TOQ =

1

2 × 110° = 55°. 1


qqq

WORKSHEET-86

Solutions

1. Length of the tangent =

d r2 2-

=

8 6

2 2( ) ( )-

= 64 36-

= 28 = 2 7 cm. 1


2. Angle at the point of intersection of tangents = 180° – 130° = 50°. 1 [CBSE Marking Scheme, 2012]

3. ∠OAP = 90° and ∠OBP = 90° (since tangent to a circle is perpendicular to the

radius through the point of contact) 1 ∴ ∠OAP + ∠OBP = 180° ½ since one pair of opposite angles of quadrilateral

AOBP are supplementary. ∴ AOBP is a cyclic quadrilateral. ½ [CBSE Marking Scheme, 2011]

4.

M

O

N

3

42

1

Q PM = PN (length of tangents are equal) ∠1 = ∠2 (angles opp. to equal sides are equal) 1 Q 180° – ∠1 = 180° – ∠2 (linear pair) ∠3 = ∠4 1 [CBSE Marking Scheme, 2015]

5. AC = 8 cm AB = 10 cm and BC = 12 cm ½

Let CF = x

∴ CF = EC = x

AF = 8 – x = AD

BE = 12 – x = BD ½

⇒ 8 – x + 12 – x = 10

20 – 2x = 10

E

C

DA B

F

– 2x = – 10 ⇒ x = 5 1

∴ AD = 3 cm

BE = 7 cm

and CF = 5 cm. 1


qqq


WORKSHEET-87

Solutions

1. Angle between the radii = 180° – 30° = 150°

(Since the sum of opposite angles = 180°). 1


2. tan 30° =

OA

AP

⇒

1

3 =

3

AP

⇒ AP = 3 × 3 = 3 cm. 1

3. ∠OPA =

1

270× °

= 35° ½

∠OAP = 90° ∴ ∠AOP = 180° – (90° + 35°) = 55° ½ ∴ ∠OAB = 35° ∠PAB = 90° – 35° = 55°.

P

A

O

B 1 [CBSE Marking Scheme, 2012]

4.


5.

O

C BA

1

Here OA = 6.5 cm and OC = 2.5 cm ½

∴ AC = OA OC2 2- ½

=

6 5 2 5

2 2. ( . )( ) -

= 42 25 6 25. .- = 6 cm 1

∴ AB = 12 cm. [CBSE Marking Scheme, 2012]

6.

O

A

B

P x y

r

r

1


∠A + ∠B + x + y = 360° 90° + 90 + x + y = 360° 1 180 + x + y = 360°

x + y = 180° 1 ⇒ Opp. angle are supplementary 1 [CBSE Marking Scheme, 2011]

qqq

WORKSHEET-88

Solutions

HOTS & Value Based Answers

1. Here, ∠QPT = 60°

∠OPQ = ∠OQP = 90° – 60° = 30°

∠POQ = 180° – (30° + 30°)

= 180° – 60° = 120°

∠PRQ =

1

2 Reflex ∠POQ 1

[Q∠POQ = 360° –120° = 240°]

=

1

2 ×

240°= 120°


2. Q DS = 5 cm

∴ DR = DS = 5 cm 1

(length of tangents are equal)

AR = 23 – 5 = 18 cm

AQ = 18 cm

QB = 29 – 18 = 11 cm = PB

or r = OP = PB = 11 cm. 1

3. OR = OP PR2 2- =

25 16- = 3 cm

P

RT

Q

O

5 cm

4 c

m

Let RT be x PT2 = PR2 + RT2 = 16 + x2 ...(i) ½ Also PT2 = OT2 – OP2 = (3 + x)2 – 25 = x2 + 6x – 16 ...(ii) From (i) and (ii) ⇒ 16 + x2 = x2 + 6x – 16 1

⇒ x =

16

3

TP = TQ =

16

256

9+

=

20

3 cm ½


4.

A 12

24

M

5

O

B

20

P

Join OB.

In rt. ∆OMB,

OB2 = 52 + 122 = 132

∴ OB = 13 1

Since OB ⊥ PB (radius ⊥ tangent)

∴ In rt. ∆OBP,

OP2 = OB2 + BP2

= 132 + 202

= 569

⇒ OP = 569 = 23.85 cm. 1


qqq


WORKSHEET-89

Solutions


1.

A

EB D

O

Join OD and AE

∠ODB = 90° (radius is

perpendicular to tangent

at point of contact)

∠AEB = 90° (anlges in a semicircle)

OD AE

(corresponding angles) ½

AE = 2 × OD

= 2 × 8 = 16 cm

In right ∆ODB, BD2 = 132 – 82 ½

= 169 – 64 = 105

BD = 105 cm ½

DE = 105 cm ½

In right ∆AED, AD2 = AE2 + DE2

= 162 +

105

2

( ) = 256 + 105 = 361 ½

AD = 19 cm. ½

[CBSE Marking Scheme 2016]

2. Let ABCD be the || gm.

∴ AB = CD, AD = BC ...(i) ½

AP + PB + DR + CR = AS + BQ + DS + CQ 1

⇒ AB + CD = AD + BC ½

From (i), 2AB = 2AD or AB = AD

A BP

CD

S Q

R 1 ⇒ ABCD is a rhombus.


3. Area of ∆ABC =

1

28 6× × = 24 cm2 1

AC = 8 62 2+ = 10 cm ½

r

O

8 cm

6 cm

A

B C

ar (∆ABC) = ar (∆OBC) + ar (∆OCA) + ar (∆OAB) = 12r 1

∴ 12r = 24

⇒ r = 2 cm. ½


qqq


TOPIC-1Division of a Line Segment in a Given Ratio.

WORKSHEET-90

Solutions

1. P divides AB internally in the ratio 4 : 4 or 1 : 1. 1


2. Minimum number of points marked will be = 2 + 5 = 7 [CBSE Marking Scheme, 2012] 1

3.

A B

P

A1

A2

A3

A4

A5

A6

A7

A8

X 1

Steps of construction :

1. Draw a line segment AB = 7 cm.

2. Draw any ray AX making an acute angle with AB.

3. Draw the point A1, A2, A3,....., A8 on AX such that AA1 = A1A2 = A2A3 = ...., A7 A8.

4. Join BA8.

5. Through the point A3, draw a line parallel to BA8.

Then AP : PB = 3 : 5 1


4. Steps of Construction :


2. Draw any ray AX making an acute angle with AB.

3. Draw the point A1, A2, A3, .... A10 on AX such that

AA1 = A1A2 = .... = A9 A10.

A BP

A1

A2

A3

A4

A5

A6

A7

A8

A9

A10X 1

4. Join BA10.

5. Through the point A3 draw a line parallel to BA10. Then AP : PB = 3 : 7. 1


5. Given : AB is a line segment of length 7 cm. ½

To construct : To divide a line segment of 7 cm length externally in the ratio 3 : 5. ½

P A B

X

B5

B4

B3

B2

B1

Steps of construction : 1


2. Draw ray BX making an acute ∠ABX.

3. Along BX, mark off B1, B2, B3, B4 & B5. Join B2 to A.

4. Through B5 draw B5P || B2A intersecting BA at P.

5. The point P so obtained is the required point which

divides AB externally in 3 : 5. 1

qqq

CONSTRUCTIONS

SECTION

BSECTIONCHAPTER

9


TOPIC-2Tangents to a Circle from a Point Outside it

WORKSHEET-91

Solutions

1. Given : A circle with centre O and a point P outside it.

Construction : We have to construct the two tangents from P to the circle.

P

Q

MO

R

1

Steps of construction :

1. Draw a line segment PO = 10 cm.

2. From the point O draw a circle of radius = 6 cm.

3. Draw a perpendicular bisector of PO. Let M be the mid-point of PO.

4. Taking M as centre and OM as radius draw a circle.

5. Let this circle intersects the given circle at the points Q and R.

6. Join PQ and PR.

Then PQ and PR are the required two tangents. 1


2. Steps of construction :

1. Draw a circle of radius OO’ = 4 cm with centre O.

2. Extend OO’ upto P (i.e.), OP = 8 cm.

3. Draw a ⊥ bisector of OP, OP intersect at S by QR. 4. Draw a circle of radius OS which meets the first

circle at Q and R. 5. Draw tangments PQ and PR

∴ PQ and PR are required tangents. 1

PQ = PR = 5 cm

Q

R

SP

O'O 8 cm

2


1. Draw a circle of radius 4 cm taking O as centre.

2. Draw two radii OA and OB inclined to each other at an angle of 120°.

60°

A

B

O 120° P

2

3. Draw AP ⊥ OA at A and BP ⊥ OB at B. They meet at P.

4. PA and PB are the required tangents inclined to each other at 60°. 2

qqq

WORKSHEET-92

Solutions


1. Draw a circle of radius 1·5 cm. Take a point P outside it.

2. Through P draw a secant PAB to meet the circle at A and B.

3. Produce AP to C such that PC = PA. Bisect CB at Q.

4. With CB as diameter and centre as Q, draw a semi-circle.

5. Draw PD ⊥ CB, cutting the semi-circle at the point D.

6. With P as centre and PD as radius an arc to cut the circle at T and T’. 1

PT and PT’ are the required tangents.


D

C P

T

QA

T�

B

1

[CBSE Marking Scheme 2011, 12]


1. Draw a circle of radius 4 cm with centre O.

2. Draw another circle of radius 6 cm with same centre O. 3. Take a point P on second circle and join OP.

A

B

PO'O

2

4. Draw perpendicular bisector of OP which intersect OP at O’.

5. Draw a circle with centre O’ which intersect inner circle at points A and B.

6. Join AP and BP.

∴ AP and BP are required tangents. 1

3.

B

P

A

120

O 4 cm

60°

1½

Steps of construction : 1. Draw a circle of radius 4 cm with O as centre. 2. Take a point A on the circumference of the circle

and join OA. Draw perpendicular to OA at point A. 3. Draw a radius OB, making an angle of 120° with

OA. 4. Draw the perpendicular to OB at point B. Let both

the perpendiculars intersect at point P. 5. Join OP. PA and PB are required tangents, which make an

angle of 60° to each other. 2½[CBSE Marking Scheme, 2015]

qqq

TOPIC-3Construction of a Triangle Similar to a Given Triangle

WORKSHEET-93

Solutions

1. When AB + BC < AC triangle cannot be drawn. 1

2. The triangle are said to be congruent. 1

3. Scale factor =

3

4 [CBSE Marking Scheme, 2012] 1

4. Following steps will be followed for constructing the similar triangle to the given triangle.

1. Draw a triangle ABC of which given sides are AB = 6·5 cm, BC = 7·5 cm and CA = 5·5 cm.

2. Draw a ray BX making an acute angle.

3. Locate 7 points B1, B2, B3, B4, B5, B6, B7 on line segment BX.

4. Join B7C. Draw a parallel line through B5 to B7C intersecting extended line segment BC at C’.

5. Through C’ draw a line parallel to AC intersecting extended line segment AB at A’.

6. ∆A’BC’ is the required triangle. 1

7·5 cm

6·5

cm

A'

B

B1

B2B

A

5·5

cm

C'C

3B4

B5B6

B 7 X 2



1. Draw a line segment BC = 8 cm.

2. Construct AM⊥ BC.

3. Taking C as centre and radius as 10 cm, draw an arc that cuts the ray BM at A’.

4. Join CA’ to obtain ∆ ABC.

5. Below BC, make an acute angle CBX.

6. Along BX mark off 5 points B1, B2, B3, B4, B5 such that BB1 = B1B2 = B2 B3 = ......... = B4 B5.

7. Join B5C.

8. From B4, Draw B4C’ B5C.

9. From aC’ draw C’A’ CA meeting BA at point A’.

Then A’ BC’ is the required triangle. 2

A

M

A�

B

B1

B2

B3

B4

B5

CC�8 cm

10 cm

2 [CBSE Marking Scheme, 2015]

qqq

WORKSHEET-94

Solutions

1. Two triangle are said to be similar when their corresponding sides are proportional and angles are equal.


2. Triangle PQR is smaller than triangle ABC. 1

(QReduced scale factor figures are smaller in size) [CBSE Marking Scheme, 2011]

3. The sum of two sides of a triangle must be greater than third side. ½

Let the sides are 2.5 cm, 4.5 cm and 6.5 cm ½



1. Draw a line segment QR = 5 cm.

2. With Q as centre and radius = PQ = 5 cm, draw an arc.

3. With R as centre and radius = PR = 5 cm, draw another arc meeting the arc drawn in step 2 at the point P.

4. Joint PQ and PR to obtain ∆PQR.

5. Below QR, construct an acute ∠RQX.

6. Along QX, mark off seven points Q1, Q2, …… Q7 such that QQ1 = Q1Q2 = Q2Q3 = …… = Q6Q7.

7. Join Q7R.

8. Draw Q6R’ || Q7R.

9. From R’ draw R’P’ || RP.

Thus, P’QR’ is the required triangle.

P

P'

R'

5cm

Q R

5cm

5 cm

Q1

Q2

Q3

Q4

Q5

Q6

Q7X

1

[CBSE Marking Scheme 2011] 1


1. Draw a line BC = 7·5 cm.

A

7·5 cm

A'

C'

3·5

cm

OB C

B1

B2

B3 B4B5

B6B 7 2


2. Draw a perpendicular bisector of BC which cut the line BC at O.

3. Cut the line OA = 3·5 cm.

4. Join A to B and C.

5. ∆ABC is the given triangle.

6. Draw a ray BX making an acute angle.

7. Locate 7 points B1, B2, .........., B7 on line segment BX.

8. Join B7C. Draw a parallel line through B4 to B7C intersecting extended line segment BC at C’.

9. Through C’ draw a line parallel to AC intersecting extended line segment AB at A’.

10. Hence, ∆A’BC’ is a required triangle. 1

qqq

WORKSHEET-95

Solutions

HOTS Answers

1. Here, AB = 5 cm, BC = 12 cm.

Let the radius of circle be x cm.

B x

x

xx

(5–x)

(12–x)Q C

P5 cm

13 cm

12 cm

A

R

O

5–x

12–x

1

∴ AC =

( ) ( )12 52 2+

= 144 25+

= 169 = 13 cm Q AC= AR + RC

∴ AC = (5 – x) + (12 – x)

⇒ 13 = 5 – x + 12 – x

⇒ 2x = 17 – 13 = 4

⇒ x =

4

2 = 2 cm 1

2. Following steps will be followed for constructing

the tangents on the given circles :

1. Draw a line segment AB of 8 cm. Taking A and B as centre draw two circles of 4 cm and 3 cm radius. ½

2. Bisect the line AB. Let mid-point of AB is C. Taking C as centre draw a circle of radius AC which will intersect the circles at point P, Q, R and S. Join BP, BQ, AS and AR. These are required tangents. 1

ABC

P

QR

S

O'

O

1


3. Steps of Construction :

1. Draw a line segment AB of 7 cm.

2. Taking A and B as centre draw two circle of 3 cm and 2 cm radius.

B

PS

3cm7cm

2cm

R

CA

Q

3. Bisect the line AB. Let mid-point of AB is C. 4. Taking C as centre draw a circle of radius AC which

will intersect the circle at point P, Q, R and S. 5. Join BP, BQ, AS and AR. These are the required tangents. 2


qqq


TOPIC-1Trigonometric Ratios and Trigonometric Ratios of Comple-mentary Angles

WORKSHEET-96

Solutions

1. sin (A + B) = 1 = sin 90° ⇒ A + B = 90° ...(i)

sin (A – B) =

1

2 = sin 30° 1

⇒ A – B = 30° ...(ii) Solving eq. (i) and (ii), A = 60° and B = 30° 1

2. C

BA

8

6

AC2 = (8)2 + (6)2 = 100 ⇒ AC = 10

∴ sin A =

8

10, cos A =

6

10 1

and sin C = 6

10, cos C =

8

10 1

∴ sin A cos C + cos A sin C

=

8

10

8

10

6

10

6

10× + ×

=

100

100 = 1. 1


3. PQ2 + QR2 = PR2

(By Pythagoras theorem) ⇒ PQ2 + 92 = PR2

⇒ PQ2 + 81 = (PQ + 1)2

⇒ PQ2 + 81 = PQ2 + 1 + 2PQ ⇒ PQ = 40 PR – PQ = 1 (Given) ⇒ PR = 1 + 40 ⇒ PR = 41

∴ sin R + cos R = 40

41 +

9

41 =

49

41 3

41

40

P

RQ 9 cm [CBSE Marking Scheme, 2015]

4. tan2 30° sin 30° + cos 60° sin2 90° tan2 60° – 2 tan 45° cos2 0° sin 90°

Expression =

1

3

1

2

1

2× + × 1 × 3 × – 2 × 1 × 1 × 1.

=

1

6

3

22+ -

=

1 9 12

6

+ -

=

- -2

6

1

3=

4


qqq

WORKSHEET-97

Solutions

1. Consider a triangle ABC with each side equal to 2a ∠A = ∠B = ∠C = 60° Draw AD is perpendicular to BC ∆BDA ≅ ∆CDA by RHS ½

BD = CD ∠BAD = ∠CAD = 30° by CPCT By Pythagoras theorem,

AD = 3a ½

In ∆BDA, cosec 30° =

AB

BD

a

a= =

22

½

INTRODUCTION TO TRIGNOMETRY AND TRIGONOMETRIC IDENTITIES

SECTION

BSECTIONCHAPTER

10


and cos 60° = BD

AB

a

a= =2

1

2 ½

A

B CD


2. =

cosec13°

sec 77°

cot 20°

tan 70°-

=

cos ( )

sec

cot( )

ta

es

90 77

77

90 70

70

° °

°

° °

°

--

-

n

=

sec 77°

sec 77°

tan 70°

tan 70°-

= 1 – 1 = 0 2

3. cos 71° – sin 57° + tan 63° = cos (90° – 19°) – sin (90° – 33°) + tan (90° – 27°) = sin 19° – cos 33° + cot 27°. 3

4. cos (40° + x) = sin 30° ⇒ cos (40° + x) = sin (90° – 60°) ⇒ cos (40° + x) = cos 60° ⇒ 40° + x = 60°

⇒ x = 60° – 40°

∴ x = 20° 3 5. cos (A + B) = 0

⇒ cos (A + B) = cos 90°

⇒ A + B = 90° ...(i)

and cot (A – B) = 3

⇒ cot (A – B) = cot 30°

⇒ A – B = 30° ...(ii)

on solving eq. (i) and eq. (ii),

A + B = 90°

A – B = 30°

2A = 120°

∴ A = 60°

Put A = 60° in eq. (i),

A + B = 90°

⇒ 60° + B = 90°

⇒ B = 90° – 60°

∴ B = 30°

Hence A = 60° and B = 30° 4

qqq

WORKSHEET-98

Solutions

1. Given sec θ.sin θ = 0

⇒

sin

cos

θ

θ = 0

⇒ tan θ = 0 = tan 0°

∴ θ = 0°

[CBSE Marking Scheme, 2016] 2. sin (36 + θ)° = cos (16 + θ)°

⇒ cos [90° – (36 + θ)°] = cos (16 + θ)° 1

⇒ 90° – 36° – θ = 16° + θ

⇒ 2θ = 90 °– 36° – 16° = 38°

∴ θ =

98°

2 = 19°. 1

3.

5 cos 60° + 4 cos 30° tan 45°

sin 30° + cos 60°

2 2 2

2 2

-

=

51

2+4

3

2(1)

1

2+

1

2

2 2

2

2 2

-

1

=

5

4+3 1

1

4+1

4

-

1

=

5

4+2

1

2

=

13

41

2

= 26

4 =

13

2 1


sin 3θ = cos (θ – 6°) 1

⇒ cos (90° – 3θ) = cos (θ – 6°) ⇒ 90° – 3θ = θ – 6° 1

⇒ 4θ = 90° + 6° = 96°

∴

θ =

96°

4 = 24° 1


5. tan (A + B) =

tan tan

tan tan

A B

A B

+

1 -


(i) tan 75° = tan (45° + 30°)

=

tan tan

tan .tan

45 30

1 45 30

° + °

° °-

=

11

3

11

3

+

-

∴ tan 75° =

3 1

3 1

+

- 2

(ii) tan 90° = tan (60° + 30°)

=

tan tan

tan tan

60 30

1 60 30

° + °

° °–

=

31

3

1 31

3

+

×-

=

3 1

3

0

+

∴ tan 90° = ∞ 2

qqq

WORKSHEET-99

Solutions

1. Given, 2 sin θ = 1

⇒ sin θ = 1

2= sin 45°

∴ θ = 45° 1

Now sec2 θ – cosec2 θ = sec2 45° – cosec2 45°

= 2

2

( ) – 22

( ) = 2 – 2 = 0. 1

2. We know that,

sec (90° – θ) = cosec θ,

tan (90° – θ) = cot θ,

cot (90° – θ) = tan θ,

cosec (90° – θ) = sec θ 1

Hence,

sin sec (90° )tan

cosec (90° )cos cot(90° )

tan (90° θ θ θ

θ θ θ

-

- --

- θθ

θ

)

cot

= sin tan

sec cos tan

cot

cot

θ θ θ

θ θ θ

θ

θ

cosec - 1

=

sin × 1

sin ×tan

1

cos ×cos tan

1

θ θ

θ

θθ θ

- = 1 – 1 = 0 1

3. LHS = 1 cos

1+cos

- θ

θ =

1

1+

-1

21

2

(Q cos 60°=1

2)

=

1

23

2

= 1

3 1

RHS = sin

1+cos

θ

θ =

sin 60°

1+cos 60°

=

3

2

11

2+

=

3

23

2

1

= 1

3 = LHS 1

Hence relation is verified for θ = 60°.

4. sin2 30° cos2 45° + 4 tan2 30° +

1

2sin2 90°

– 2 cos2 90° + 1

24

=

1

2

2

× 1

2

2

+ 41

3

2

+ 1

212

( )

– 2 (0) + 1

24 1

= 1

4

1

2

+ 4

3+

1

2 +

1

24 1

= 1

8+4

3+

1

24+1

2

= 3+32+1+12

24 1

= 48

24 = 2. 1

qqq


WORKSHEET-100

Solutions

1. Given tan 2A = cot (A + 60°)

⇒ cot (90 – 2A) = cot (A + 60)

⇒ 90 – 2A = A + 60°

⇒ 3A = 30°

∴ A = 10° 1


2. Q tan (A + B) = 3 = tan 60°

Hence, A + B = 60° ...(i) ½

Again tan (A – B) =

1

3 = tan 30°

⇒ A – B = 30° ...(ii) ½

Adding equations (i) and (ii),

2A = 90°

∴ A = 90

2

° = 45°

Putting this value of A in equation (i),

B = 60° – A = 60° – 45° = 15°

Hence, A = 45° and B = 15°. 1

3. cos (A – B) =

3

2 = cos 30°

⇒ A – B = 30° ...(i) ½

sin (A + B) = 3

2 = sin 60°

⇒ A + B = 60° ...(ii) ½ Adding equations (i) and (ii), 2A = 90° ∴ A = 45° ½

From eqn. (ii), B = 60° – A = 60° – 45° = 15° ½


4. tan A + cot A = 2

On squaring both sides,

(tan A + cot A)2 = (2)2 1

⇒ tan2 A + cot2 A + 2 tan A. cot A = 4

⇒ tan2 A + cot2 A + 2 tan AA

×1

tan = 4 1

⇒ tan2 A + cot2 A + 2 = 4

⇒ tan2 A + cot2 A = 4 – 2

∴ tan2 A + cot2 A = 2 1

5. cos θ + sin θ = 2 cos θ

⇒ sin θ = cos θ( 2 – 1)

⇒ sin θ =

cos ( 2 1) ( 2+1)

( 2+1)

θ -

1

⇒ sin θ =

cos (2 1)

2+1

θ -

⇒ ( 2 + 1)sin θ = cos θ 1

⇒ 2 sin θ + sin θ = cos θ

⇒ cos θ – sin θ = 2 sin θ . Hence proved. 1

6. 4 sin 30°+cos 60°4 4( )

– 3 cos 45° sin 90°

2 2-( )

= 4

1

2+

1

2

4 4

– 31

21

22

( )

- 1

= 4

1

16+

1

16

– 3

1

21-

1

=

1

2+3

2 1

=

4

2 = 2. 1

qqq

WORKSHEET-101

Solutions

1.

sin25°

cos65°+tan 23°

cot 67°

=

sin25°

sin( )

tan

tan( )90 65

23

90 67° °+

°

° °- -

= 1 + 1 = 2 1


2. cos 68° + tan 76° = cos (90° – 22°) + tan (90° – 14°) 1

= sin 22° + cot 14°,

[Q cos (90° – θ) = sin θ and tan (90° – θ) = cot θ ] 1

3. Given, 2 sin 2θ = 3

⇒ sin 2θ =

3

2 = sin 60° 1

⇒ 2θ = 60°

Hence, cos 2θ = cos 60° =

1

2· 1



4.

C

A

B

x x + 1

5 cm

Let AB = x

Q AC – AB = 1

⇒ AC = x + 1

Q AC2 = AB2 + BC2

⇒ (x + 1)2 = x2 + (5)2

⇒ x2 + 2x + 1 = x2 + 25

⇒ 2x = 24

⇒ x =

24

2 = 12 cm 1

Hence, AB = 12 cm, AC = 13 cm

sin C =

AB

AC=12

13

cos C =

BC

AC=5

13 1

Now

1+

1+

sin

cos

C

C =

1+12

13

1+5

13

=

25

1318

13

=25

18· 1

5. 15tan2 θ + 4sec2 θ = 23 15tan2 θ + 4(tan2 θ + 1) = 23

(∴ sec2θ = 1 + tan2

θ) 1 ⇒ 15tan2 θ + 4tan2 θ + 4 = 23 ⇒ 19tan2 θ = 19 ⇒ tan θ = 1 = tan 45° 1 ∴ θ = 45° Now, (sec θ + cosec θ)2 – sin2 θ = (sec 45° + cosec 45°)2 – sin2 45°

= 2 2

1

2

2+( )

-

2

1

= 2 2

1

2

2

( ) -

= 8

1

2=15

2- 1

qqq

TOPIC-2Trigonometric Identities

WORKSHEET-102

Solutions

1.

sec A = 1 1

12cos sinA A

=-

1

and tan A =

sin

cos

sin

sin

A

A

A

A=

12

-

1


2.

LHS =

(sin cos )

sin cos

4 4

2 21 2

θ θ

θ θ

+

-

=

(sin ) (cos )

sin cos

2 2 2 2

2 21 2

θ θ

θ θ

+

-

=

(sin cos ) sin cos

sin cos

2 2 2 2 2

2 2

2

1 2

θ θ θ θ

θ θ

+ -

-

=

1 2

1 2

2 2

2 2

-

-

sin cos

sin cos

θ θ

θ θ = 1 = RHS 2


3. LHS = cos

1 + tan

sin

1 + cot

A

A

A

A−

=

cos

sin

cos

sin

cos

sin

AA

A

AA

A1 1+

−

+

1

=

cos

cos sin

sin

sin cos

2 2A

A A

A

A A+−

+

=

cos sin

(sin cos )

2 2A A

A A

−+

1

=

(cos sin )(cos sin )

sin cos

A A A A

A A

+ −

+

= cos A – sin A

= RHS Hence proved. 1

4.

A

C

θ

a

b

B

b a


bcos θ = a

⇒ cos θ = a

b

cosec θ =

b

b a2 2-

, cot θ =

a

b a2 2-

cosec θ + cot θ = b a

b a

b a

b a

+=

+

2 2- -

3


5. (i) x

a

2

2 = sec2

θ, y

b

2

2 = tan2

θ

⇒

x

a

y

b

2

2

2

2-

= sec2

θ – tan2 θ = 1.

∴ b2x2 – a2y2 = a2b2 2

(ii) x

a

2

2 = cosec2

θ, y

b

2

2 = cot2

θ

⇒ x

a

y

b

2

2

2

2-

= cosec2

θ – cot2θ = 1

∴ b2x2 – a2y2 = a2b2 2

qqq

WORKSHEET-103

Solutions

1. = sin2 41 + sin2 49

= sin2 (90° – 49°) + sin2 49°

= cos2 49 + sin2 49°

= 1. [Qcos2 θ + sin2 θ = 1] 1

2.

cos

sin

cos

sin

θ

θ

θ

θ1 1-+

+ = 4

⇒ cos ( sin ) cos ( sin )

( sin )( sin )

θ θ θ θ

θ θ

1 1

1 1

+ +

+

-

- = 4

⇒

cos [ sin sin ]

sin

θ θ θ

θ

1 1

1 2

+ + -

- = 4

⇒ cos ( )

cos

θθ2

2 = 4

⇒ 2

cos θ = 4

⇒ 2 sec θ = 4

⇒ sec θ = 2

sec θ = sec 60°

∴ θ = 60° 2

3. sin sin

cos cos

θ θ

θ θ

−

−

2

2

3

3 =

sin ( sin )

cos ( cos )

θ θ

θ θ

1 2

2 1

2

2

-

-

=

sin (sin cos sin )

cos ( cos sin cos )

θ θ θ θ

θ θ θ θ

2 2 2

2 2 2

2

2

+ -

- -

=

tan (cos sin )

(cos sin )

θ θ θ

θ θ

2 2

2 2

-

-

= tan θ Hence proved. 3

4.

H

A

B

P

Let, tan A =

P

B, sec A =

H

B

H2 = P2 + B2

LHS = 1 + tan2 A 1

= 1

2

+

P

B =

1

2

2+P

B

=

B P

B

2 2

2

+

=

H

B

2

2

= H

B

2

= sec2 A 1

= sec2 A

= RHS Hence proved.

Equations that are true no matter what value is plugged in for the variable. On simplifying an identity equation, one always get a true statement.

3

5. On squaring, cosec2 θ + cot2

θ – 2cosec θ cot θ = 2cot2 θ

⇒ cosec2 θ – cot2 θ = 2cosec θ cot θ ⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 2cosec θ cot θ

⇒ (cosec θ + cot θ) 2 cot θ = 2cosec θ cot θ

⇒ cosec θ + cot θ = 2 cosec θ 4


qqq


WORKSHEET-104

Solutions


1. Let sec2 θ + cosec2 θ < 2. 1 ⇒ 1 + tan2

θ + 1 + cot2 θ < 2.

⇒ 2 + tan2 θ + cot2

θ < 2 ⇒ (tan2 θ + cot2 θ) < 0 which is not possible. 1 ∴ sec2

θ + cosec2 θ can never be less than 2

Hence our assumption is wrong. 2. tan A + sin A = m and tan A – sin A = n LHS = m2 – n2 = (tan A + sin A)2 – (tan A – sin A)2

= (tan2 A + sin2 A + 2 sin A. tan A) – (tan2 A + sin2 A – 2 sin A. tan A) = tan2 A + sin2 A + 2 sin tan A – tan2 A – sin2 A + 2 sin A tan A = 4 sin A tan A 1

RHS = 4 mn = 4 (tan sin )(tan sin )A A A A+ -

= 4 tan sin2 2A A-

= 4

sin sin cos

cos

2 2 2

2

A A A

A

-

= 4

sin ( cos )

cos

2 2

2

1A A

A

-

= 4

sin sin

cos

2 2

2

A A

A

×

1

= 4 sin A tan A

∴ m2 – n2 = 4 mn 1

3.

12 mRope

30°

A

BC

(i) Clearly, distance covered by the artist is equal to the length of the rope AC. Let AB be the vertical pole of height 12 m.

It is given that ∠ACB = 30°

Thus, in right-angled triangle ABC,

sin 30° = AB

AC

⇒ 1

2 =

12

AC

... AC = 24 m. 2

Hence, the distance covered by the circus artist is 24 m.

(ii) Trigonometric ratios of an acute angle of a right- angled triangle. 1

(iii) Focus helps us to gain success in life. 1

qqq

WORKSHEET-105

Solutions


1. Given,

cos

cos

α

β = m and

cos

sin

α

β = n

⇒ m2 =

cos

cos

2

2

α

β and n2 =

cos

sin

2

2

α

β

LHS = (m2 + n2) cos2 β 1

=

cos

cos

cos

sin

2

2

2

2

α

β

α

β+

cos2 β

= cos2 α

12 2

cos sinβ β

cos2 β 1

=

cos

sin

2

2

α

β= n2

∴ (m2 + n2) cos2 β = n2 1

2. 7cosec φ –3cot φ = 7

⇒ 7cosec φ – 7 = 3cot φ 1

⇒ 7(cosec φ – 1) = 3cot φ

⇒ 7(cosec φ –1)(cosec φ +1) = 3cot φ(cosec φ+1)

⇒ 7(cosec2 φ – 1) = 3cot φ(cosec φ + 1)

⇒ 7cot2 φ = 3cot φ(cosec φ + 1)

⇒ 7cotφ = 3(cosec φ + 1) 1

∴ 7cot φ – 3cosec φ = 3 1

3. (i) Let AB be the tree broken at a point C such that the broken part CB takes the position CO and strikes the ground at O. It is given that OA = 30 m and ∠AOC = 30°.

Let AC = x and CB = y, then CO = y


y

x

A

C

O30°

30 m

y

B

In ∆OAC,

tan 30° = AC

OA

⇒

1

3 =

x

30

...

x =

30

3 =

10 3

1

Again in ∆OAC,

cos 30° =

OA

OC

∴

3

2

=

30

y

⇒

y =

60

3 =

20 3

∴ Height of the tree = (x + y)

= 10 3 + 20 3

= 30 3 = 30 × 1.732 = 51.96 m 1 (ii) Trigonometric ratios of an acute angle of a right

angled triangle. 1 (iii) The problem of decreasing ratio of trees and land is

discussed here. 1

qqq


WORKSHEET-106

Solutions

1.

15 m

60°

A

B C

h Wall

Ladder

Let the height of wall be h

h

15 = cos 60°

⇒ h = 15 × cos 60°

= 15 ×

1

2

= 7.5 m 1

2. Let OA be the electric pole and B be the point on the ground.

Let BA = x

In ∆ABO,

AO

AB = sin 45° 1

A

OBx

x

45°

10 m 1

⇒

10

x =

1

2

⇒ x = 10 2

= 10 × 1.414

= 14.14 m 1


3.

45°

45°

60°

x

h

60°

tan 45° = h

x

− 50

⇒ x = h – 50 ½

tan 60° = h

x ½

⇒ x =

h

3 ½

Hence h – 50 = h

3 ½

⇒ h = 75 + 25 3 = 118.25 m. 1 [CBSE Marking Scheme, 2016]

qqq

WORKSHEET-107

Solutions

1.

In right ∆ABC,

AB

AC = sin 60°

⇒

90

x =

3

2 ½

⇒ x = 180 3

3

⇒ x = 60 3

B

x

A

90 m

60°

C

= 60 × 1.732 1½ Hence length of string = 103.92 m. [CBSE Marking Scheme, 2011]

HEIGHTS AND DISTANCES

SECTION

BSECTIONCHAPTER

11


2.

45°60°

y

x

7 m

B

½

(i)

x

y

= tan 45° = 1

⇒ x = y 1

(ii)

x

x

+ 7

= tan 60° = 3

⇒ 7 =

( )3 1− x 1+½

x =

7 3 1

2

( )+

=

7 2 73

2

( . )

= 9.6 m [CBSE Marking Scheme, 2016]

3.

45°

45°30°

30°

C

100 m

A D Byx ½

In right ∆ADC,

tan 30° =

CD

AD ½

⇒

1

3 =

100

x

⇒ x = 100 3 ...(i) ½

In right ∆BDC,

tan 45° =

CD

DB ½

⇒ 1 =

100

y

½

⇒ y = 100 m

Distance between two cars

= AB = AD + DB ½

= (100 3 + 100 m) ½

= (100 × 1.73 + 100) m

= (173 + 100) m

= 273 m ½


qqq

WORKSHEET-108

Solutions

1.

A

B

E D

C

h

Tower

60°

28.2 m1.2 m

28.2 m

From the triangle ABC

h

28 2. = tan 60°

⇒ h = 28.2 ×tan 60°

= 28.2 × 3

Height of the tower = AC + CD

= 28.2 3 + 1.2 m

= 50 m 1


2.

C

Brokenpart

B

A

30°

8 mD

Tree

½


In right ∆BCD,

cos 30° = CD

BD

⇒ 3

2 =

8

BD

⇒ BD =

16

3 ½

and tan 30° =

BC

CD

⇒ 1

3 =

BC

8

⇒ BC = 8

3

∴ Height of tree = BC + BD

= 8 16

3

+

=

24

33

= 8 3 m. 1


3.

D

A

B C

E 150 m60 m

60 m

30°

h

Let AB and CD be two towers. Let the height of the shorter tower AB = h BC = AE = 60 m, DE = DC – EC = 150 – h

In ∆ AED,

DE

AE = tan 30°

⇒

150

60

h-

= tan 30° =

1

3 1

⇒ 150 3 – h 3 = 60

⇒ 3 h = 150 3 – 60

⇒ h = 150 –

60

3 ×

3

3

⇒ h = 150 20 3-( )

m. 1


qqq

WORKSHEET-109

Solutions

1. Let the ∠ACB = θ

tan θ =

AB

BC

tan θ =

20

20 3=

1

3= tan 30° 1

⇒ θ = 30° Sun's altitude is 30°. [CBSE Marking Scheme, 2015]

2.

x

y60° 30°

30 m

DC

A

B

1

In ∆ABC, AB

BC= °tan30

⇒ x

30 = tan 30° =

1

3

⇒ x = 10 3

In ∆ABD, AB

BD= °tan60

10 3

y

= tan 60°= 3

⇒

y = 10 m

1

Hence the length of shadow = 10 m.



3.


4.

30°60°P

y

5 m

x

Tower

Flagstaff

1

(i)

x

y

= tan 30° =

1

3

⇒ y = 3x 1

(ii)

x

y

+ 5

= tan 60° =

3

or

x

x

+ 5

3 =

3

1+½

Height of Tower = 2.5 m

Distance of P from tower = (2.5 × 1.732) or 4.33 m. [CBSE Marking Scheme, 2016] ½

qqq

WORKSHEET-110

Solutions

1. Let height of flagstaff = x m

∴

tan 30° =

AB

AP

⇒

1

3

10=AP

AP = 10 3 1

i.e., distance of the building

= 10 × 1.732

= 17.32 m ½

AP

B

D

x

10 m

30°

45°


tan 45° = AD

AP or 1 =

10

10 3

+ x

or length of flagstaff x = 10 3 1-( )

= 7.32 m. ½


2. Let AD be the height (h) of the light house and BC is the distance between the ships

A

B CD x

30° 45°

45°30°

h

100 m

100 – x

Given, BC =100 m

tan 45° = h

DC

⇒ DC = h ...(i) 1

In ∆ABD, tan 30° = h

DC100-

⇒

1

3 =

h

x100 - ∴ 100 – x = h 3 1

100 – h = h 3 [By (i)]

⇒ 100 = h + h 3

⇒ 100 = h 1 3+( ) h =

100

1 3+

⇒ h = 100 3 1

3 1

( )-

-

=

50 3 1( )-

1

= 50(1.732 – 1) = 50 × 0.732 ∴ Height of tower = 36.60 m.

3.

1

Let PX = x and PQ = h m ∴ QT = (h – 40) m In ∆PQX,

tan 60° = h

x

⇒ 3 = h

x

⇒ h = 3 x ... (i) 1

In ∆QTY, tan 45° = h

x

- 40

⇒ 1 =

h

x

- 40

⇒ x = h – 40 ... (ii) 1 Solving (i) and (ii),

x = 3 x – 40

or 3 1-( ) x = 40

or x = 40

3 1-

= 20 3 1+( ) m

∴ h = 3 × 20 3 1+( ) = 20 3 3+( ) m 1

qqq

WORKSHEET-111

Solutions

1.

45° 60°

A

C B O

100 m

45°60°

1

Let A be a point and B, C be two objects.

In ∆ACO,

AO

CO = tan 45°

⇒

100

CO = 1

⇒ CO = 100 m

Also in ∆ABO,

AO

OB = tan 60°


⇒

100

OB =

3 1

⇒ OB =

100

3

∴ BC = CO – OB

= 100 –

100

3

= 100

1

1

3-

m. 1


2.

A

90 mh

OB

� �

1

Let A be the position of kite and AB be the string. Since it is given that

tan θ =

15

8

∴ sin θ =

15

17

U g Pythagoras

Theorem

sin

.

1

In ∆ABO,

AO

AB =

sin θ

⇒

h

90

=

15

17

⇒ h = 15 90

17

×

= 79.41 m 1

Hence, height of kite is 79.41 m.


3.

A

Q P R

B

x x

80 – y y

30°

30°

60°

80 m

1

Let the distance of Pole PR = y and PQ = 80 – y

From ∆BPR,

tan 60° =

x

y

⇒ x = y 3 ...(i) 1 From ∆APQ,

tan 30° =

x

y80 -

⇒ 3 x = 80 – y ...(ii) 1 Solving (i) and (ii)

y = 20, x = 20 3 m.

Height of pole = 20 3 m.

Distance of Pole, PR = 20 m. QP = 80 – 20 = 60 m. 1


qqq

WORKSHEET-112

Solutions 1. It will be less than 45°.

(Since base is greater than perpendicular of the triangle) 1

2.

A

20 m

60°

BC60°

x

Let C be the point where the ball is ∠C = 60° (alternate angles)

In ∆ ABC, tan 60° = AB

BC

⇒

3 =

20

x

1

⇒ x = 20

3

= 203

3

½

Hence, required distance is 1 = 11.53 m. ½ [CBSE Marking Scheme, 2011]


3. Let two men are standing at A and C. BT is the tower.

B CA

T

20 m

30° 60°

In rt. ∆ABT,

tan 30° =

BT

AB

⇒

1

3 =

20

AB

⇒ AB = 20 3 In rt. ∆BTC,

tan 60° = BT

BC

⇒ 3 =

20

BC 1

⇒ BC = 20

3 1

∴ Distance between two men = AB + BC

= 20 3 +

20

3

= 60 20

3

+

=

80 3

3 m.

1

Hence, distance between the two men is 80 3

3 m.

4. In ∆ABE, ∠AEB = 30° ½

tan 30° = h

x

h = x

3

In ∆CDE, ∠CED = 60°

60°30°

E 80 – xx DB

h

A C

h

1

tan 60° =

h

x80 -

⇒ 3 = h

x80 -

⇒ h = 3 (80 – x)

⇒

x

3 =

3 (80 – x)

⇒ x = 60 m 1

h = 60

3

Hence, height of pole = 20 3 = 34.64 m. ½


qqq

WORKSHEET-113

Solutions


1. Here, AB = 50 m, ∠ADB = 45°, ∠ACM = 30°

tan 45° =

AB

BD = 1

⇒ AB = BD = 50 m. 1

(i) Distance of pole from bottom of tower = 50 m

Now, tan 30° =

AM

MC

AM

BD=

A

M

B D

C30°

45°30°

45°

50 m

½

⇒ AM =

50

3or 28.87 m. 1

(ii) Height of pole = CD = BM

= 50 – 28.87 = 21.13 m. ½


2. (i) Let P be the position of the balloon when its angle of elevation from the eyes of the girl is 60° and Q be the position when angle of elevation is 30°.

P Q

30°

1·2 m

60°

L'

L

M'

M

88·2 m

O

1

In ∆OLP, tan 60° =

PL

OL


⇒ 3

=

PL' LL'

OL

-

=

88.2 1.2-

OL ½

⇒

3

=

87

OL

⇒ OL = 87

3

In ∆OMQ, tan 30° = QM

OM

= QM' MM'

OM

-

⇒ 1

3 =

88 2 1 2. .-

OM

⇒ OM = 87 3 ½

∴ Distance travelled by the balloon, PQ = LM = OM – OL

= 87 387

3-

m

= 87

31

3−

m

= 87 2

3

×

=

174

3 m

=

174 3

3 =

58 3

m

1

(ii) Height and distance. ½ (iii) Concentration. ½

qqq

WORKSHEET-114

Solutions


1.

B C

L M60°

30°

A x

1500 3

½

Let, AL = x,

∴

BL

x = tan 60°

Q BL = CM = 1500 3

⇒

1500 3

x

= 3 ⇒ x = 1500 m. 1

CM

AL LM+ =

tan30

1

3° =

⇒ 1500 + LM = 1500 (3) = 4500 1

⇒ LM = 3000 m.

∴ Speed =

Distance

Time=

3000

15

= 200 m/s

= 720 km/hr. ½


2.B (Bird)

30°45°

D

CAx

x

80 m80 m

y

Correct Figure 1

tan 45° = 80

y

⇒ y = 80 ½

tan 30° = 80

x y+

⇒ x + y = 80 3 ½

∴ x = 80 ( 3 – 1) = 58.4 m.

Hence, speed of bird = 58 4

2

. = 29.2 m/s. 1


3.

30°A

30° 60°

P Q Ox

60°


(i) Let OA be the cliff and P be the initial position of the boat when the angle of depression is 30°. After 6 minutes the boat reaches to Q such that the angle of depression at Q is 60°.

Let PQ = x m. In ∠POA and ∠QOA, we have

tan 30° = OA

OP

and tan 60° = OA

OQ

⇒ 1

3 =

OA

OP and 3

=

OA

OQ

½

⇒ OA = OP

3 and OA = 3OQ

⇒

OP

3 =

3OQ

⇒ OP = 3OQ ⇒ PQ = OP – OQ ½

=OP – OP

3

= 2

3 OP

[... OQ =

1

3OP] ½

Let the speed of the boat be v m/minute, then PQ = Distance travelled by the boat in 6 minutes ⇒ PQ = 6v ½

⇒

2

3(OP)

= 6v [... PQ =

2

3OP]

⇒ OP = 9v ∴ Time taken by the boat to reach at the shore is

given by,

T = OP

v

⇒ T = 9v

v minutes

= 9 minutes 1 (ii) Height and distance. ½ (iii) Uniformity in action. ½

qqq

WORKSHEET-115

Solutions


1.

AB

C

D

x30°

60°

h

20m 20m

From, ∆ABC,

h

x = tan30

1

3° =

⇒ x h= 3 . ...(i) ½

From ∆ABD,

40 + h

x = tan 60° =

3

⇒ x =

40

3

+ h

...(ii) ½

From (i) and (ii),

∴ 3 h =

40

3

+ h

⇒ h = 20 m. 1

∴ x = 20 3 m

∴ AC = ( ) ( )BC AB

2 2+

= ( ) ( )20 20 3

2 2+

= 400 1200+

= 40 m. 1


2. Let the height be ‘h’ m and breadth be ‘b’ m.

60° 30°

A

h

B C Db 20 m 1

In ∆ABC ,

h

b = tan 60° = 3

⇒ h = 3 b 1

In ∆ABD,

h

b + 20 = tan 30° =

1

3

⇒ h =

b + 20

3

∴

b 3

=

b + 20

3

3b = b + 20 ⇒ 2b = 20 ⇒ b = 10 m


h = b 3 = 10 × 1.73 = 17.3 m 1

∴ Height of tree is 17.3 m and breadth of river is 10 m. [CBSE Marking Scheme, 2012]

3.

A

B C

D

60°

30°

Tower 40 m

Chimney

Suppose AB = 40 m be the height of the tower and CD be the height of smoking chimney.

Considering right angled ∆ABC, we have

tan 30° = BC

AB

⇒ 1

3 =

BC

40

⇒ BC = 40

3 1

Again, considering right triangle DBC, we have

tan 60° = DC

BC 1

⇒

3 =

DC

BC 1

DC =

3 ×

40

3

⇒

DC = 40 m

∴ The height of chimney is 40 m. 1

qqq


WORKSHEET-116

Solutions

1. Q Circumference of the outer circle 2πr1 = 88 cm

∴ r1 = 88 7

2 22

×

× = 14 cm.

Q Circumference of the inner circle 2πr2 = 66 cm

∴ r2 = 66 7

2 22

21

2

××

=

cm = 10.5 cm

∴ Width of the ring = r1 – r2

= 14 – 10.5 cm = 3.5 cm. 1

2. Diameter of the circle = diagonal of square = 16 cm Let x be the side of square

∴ x 2 = 16 ⇒ x = 8 2 1

Area of square = x2 = 8 2

2

( ) = 128 cm2 1 [CBSE Marking Scheme, 2015]

3.

O

P

AB

As OA = 17 cm, AP = 15 cm and ∆OPA is right triangle

∴ Using Pythagoras theorem,

OP = 8 cm 1

Area of the shaded region

= Area of the sector AOBA

– Area of ∆OPA 1

=

60

360

1

2

2× × ×πr b h-

=

60

360

22

717 17

1

28 15

°

°× × × × ×-

= 151.38 – 60 = 91.38 cm2. 1

4.

D C

A B

1

Let r cm be the radius of each circle.

Area of square – Area of 4 sectors =

24

7 cm2

½

⇒ (2r)2 – 490

360

2°

°×

πr

=

24

7

1

⇒

4

22

7

2 2r r-

=

24

7 ½

⇒

28 22

7

2 2r r-

=

24

7

⇒ 6r2 = 24

⇒ r2 = 4 1

⇒ r =± 2

⇒ Radius of each circle is 2 cm (r cannot be negative)


qqq

WORKSHEET-117

Solutions

1. Area of circle = πr2, Let the radius of circle with centre C = R

According to question, π(8)2 + π(15)2 = πR2

⇒ 64π + 225π = πR2

⇒ R2 = 289 ⇒ R = 17 cm Circumference of circle = 2πr = 2π×17 = 34π cm 1 [CBSE Marking Scheme, 2012]

AREAS RELATED TO CIRCLES

SECTION

BSECTIONCHAPTER

12


2. Area of remaining part = Area of rectangle – Area of semi-circle ½

A B

CD

20 cm

14 cm

= 20×1422×7×7

7×2-

1

= 280154

2-

= 280 – 77 = 203 cm2. ½ [CBSE Marking Scheme, 2012]

3.

O

A B

∠AOB = 60°

Area of shaded region = Area of major sector + (Area of ∆AOB – Area of minor sector)

=

300

360

22

7×

× (6)2 1

+3

4(12)

60

360×22

7×62 2

-

1

= 660

736 3

132

7+ -

= 36 3+

528

7 cm2 1


4.

A

B

PO

5

θ

5 cm

CD

10 cm

cos θ = 1

2 ⇒ θ = 60° ½

Reflex ∠AOB = 240° ½

∴ ADB = 2 3 14 5 240

360

× × ×. = 20.93

cm 1

Hence length of elastic in contact = 20.93 cm

Now, AP = 5 3 cm

Area (∆OAP + ∆OBP) = 2 5 3 = 43.25 cm2 ½

Area of sector OACB = 25 3 14 120

360

× ×. = 26.16 cm2

½

Shaded Area = 43.25 – 26.16 = 17.09 cm2 1 [CBSE Marking Scheme, 2016]

qqq

WORKSHEET-118

Solutions

1. Distance covered in 1 revolution = circumference of wheel

= πd

= π × 1.26 m.

Distance covered in 500 revolutions

= 500 × π × 1.26

= 500 ×

22

7 × 1.26

= 1980 m. 1


2. Area of the sector = 1

2 × (length of the

corresponding arc) × radius ½

=

1

2 × l × r

=

1

2 ×12 × 6

= 36 cm. 1½


3. Area of sector OAPB = 90

360π (10)2 = 25π cm2

Area of ∆AOB =

1

2 × 10 × 10 = 50 cm2 ½

∴ Area of minor segment AQBP = (25π – 50) cm2


= 25 × 3.14 – 50 = 78.5 – 50 = 28.5 cm2 ½ Also area of circle = π(10)2

= 3.14 × 100 = 314 cm2 1 Area of major segment ALBQA = 314 – 28.5 = 285.5 cm2 1

4.

106 cm

100 m

A

B C

HE

F

D

G

O O’

We have OB = O’C = 30 m ½

AB = CD = 10 m

OA = O’D = (30 + 10) m

= 40 m. ½

ar (track) = ar (rect. ABCD) + ar (rect. EFGH) + 2 (ar semi-circle of radius 40) – 2 ar (semi-circle of radius 30)

= [(10 × 106) + (10 × 106)]

+ × × × ×

2

1

2

22

740

1

2

22

730

2 2( ) ( )-

1

= (2120 + 2200) m2

= 4320 m2. 2


qqq

WORKSHEET-119

Solutions

1. A line intersecting the circle at two distinct points is called a secant. 1


2. Perimeter of sector = r + r + 2

360

π θr

°

= 7 + 7 + 2 ×

22

77

90

360× ×

= 14 + 11 = 25 cm. 1


3. Area of the remaining cardboard = Area of rectangular cardboard

– 2 × Area of circle 1

7 cm 7 cm7 cm

14 cm

= 14×7 2×22

7×

7

2-

2

= 9844

7×49

4-

= 98 – 77

= 21 cm2. 1

4.


½

½

½

½

½

½


5. Area of be shaded region

=

1

2

22

77

7

227

2

22 2

2× +

−

cm

2

=

1

2

22

749

49

4

49

8× + −

=

1

2

22

749

9

8× ×

1½

=

693

8 sq. cm or 86.625 cm2 ½


qqq

WORKSHEET-120

Solutions

1. AB = ( ) ( )16 12

2 2+

(From Paythagoras theorem)

= 256 144+

= 400 20= cm

∴ Radius of circle = 10 cm. Perimeter of shaded region = πr + AP + PB = 3 × 10 + 12 + 16 = 30 + 12 + 16 = 58 cm. 1 [CBSE Marking Scheme, 2012]

2. Name of the line that touches a circle at one point is called tangent. 1


3. Perimeter = πr + 2r ½

= (π + 2)r = 36

⇒

36

7r = 36 ⇒ r = 7 1

∴ Diameter = 14 cm. ½


4.


½

½

½

½

½

½



Given, AB = 13 and AC = 12

Since AB is a diameter, ∠ACB = 90°

(angle in a semicircle)

Applying Pythagoras theorem in ∆ABC

AB2 = AC2 + BC2

⇒ (13)2 = (12)2 + BC2

or BC2 = 25

⇒ BC = 5 1

Area of triangle ∆ACB = 1

2 × BC × AC

Area of ∆ABC = 1

2 × 5 × 12 = 30 cm2 ...(i) ½

Area of Semicircle = 1

2 πr2 where, r

AB= =

2

13

2

Area of Semicircle = 1

2 × 3.14 ×

13

2

2

= 66.33 cm2 ...(ii) ½

∴ Area of Shaded region = (66.33 – 30) cm2

= 36.33 cm2. 1

5. Radius of circle with centre O is OR

Let OR = x

∴ x2 + x2 = (42)2 ⇒ x =

21 2m

1

Area of the flower bed = Area of segment of circle with centre angle = 90°

=

22

721 2 21 2

90

360× × ×

–

1

221 2 21 2× ×

1

= 693 – 441

= 252 m2 ½ + ½ + ½

Area of flower bed = 2 × 252

= 504 m2. ½


qqq

WORKSHEET-121

Solutions

1. Radius of the circle =

10

2 = 5 cm

Area of the circle = π × r2

= π × (5)2

= 25π cm2 1


2. Perimeter of the circle = Perimeter of square Let side of square be x cm. 2π r = 4x

⇒ 2 ×

22

7 × 77 = 4x

∴ x = 2 22 11

4

× ×

= 121

Side of the square = 121 cm. 1 [CBSE Marking Scheme, 2012]

3. Perimeter of (semi-circular arc) + diameter = 72 cm 1 πr + 2r = 72 cm

⇒

r22

7+2

= 72 cm

⇒

r22+14

7

= 72

⇒

36

7r = 72

r = 14 cm ½

∴ Area of protractor = 1

2πr 2

=

1

2×22

7×14×14

= 308 cm2. ½ [CBSE Marking Scheme, 2012]


4. Here, r = 14 cm, θ = 60°

Area of minor segment =

π

θθr r2 2

360

1

2 - sin

½

=

22

714 14

60

360

1

214 14

3

2× × × × × × -

½

=

( )308

349 3- cm2 = 17.89 cm2 = 17.9 approx. 1

Area of major segment = πr2 –

( )308

349 3-

=

1540

249 3+ =598.10 cm2

= 598 cm2 approx. 1


5. Let r be the radius of incircle BC = 10 = 8 – r + 6 – r 1 (By using the tangent properties) ⇒ 2r = 8 + 6 – 10 ⇒ 2r = 4 ⇒ r = 2 cm

∴ Area of circle = πr2 =

22

7 × 2 × 2

=

88

7 = 12.57 cm2 1

Now, area of ∆ABC =

1

2 × 8 × 6

= 24 cm2 1 Area of shaded region = Area of ∆ABC

– Area of circle = 24 – 12.57 = 11.43 cm2 1 [CBSE Marking Scheme, 2015]

qqq

WORKSHEET-122

Solutions

1. Area of the circle = sum of areas of two circles

πR2 = π × (40)2 + π(9)2

⇒ R2 = 1600 + 81

⇒ R = 1681 = 41 cm.

∴ Diameter of required circle = 41 × 2 = 82 cm. 1


2.

A

B C

D

r

Side of square = 8 cm

∴ Radius of circle, r =

8

2 = 4 cm

Area of circle = πr2

= π × 4 × 4

= 16π cm2 1


3. Radius of circle (r) = 10 cm, central angle = 90° ½

∴ Area of minor segment

=

1

2× 10 2 ×

3 14 90

18090

.sin

×°

-

½

=

1

2 × 100 × [1.57 – 1] = 28.5 cm2. 1


4.

90°

A

B C

P R

7 cm 7 cm

3 cm 3 cm

From the given figure

Area of right-angled ∆ABC =

1

2 × 10 × 10

= 50 cm2 1

Area of quadrant APR of the circle of radius 7 cm

1

4 × π × (7)2 Area of quadrant =

1

4 Area of circle

=

1

4

22

749× ×

= 38.5 cm2 1


∴ Area of base PBCR = Area of ∆ABC– Area of quadrant APR

= 50 – 38.5 = 11.5 cm2. 1 [CBSE Marking Scheme, 2015]

5. Let the radii of the circles are r1 cm and r2 cm

∴ r1 + r2 = 14 ...(i)

And, sum of their areas = πr12 + πr2

2

130 = π(r12 + r2

2)

⇒ 130 π = π(r12 + r2

2)

∴ r12 + r2

2 = 130 ...(ii) 1

(r1 + r2)2 = r12 + r2

2 + 2r1r2

⇒ (14)2 = 130 + 2r1r2

⇒ 2r1r2 = 196 – 130

= 66 1

(r1 – r2)2 = r12 + r2

2 – 2r1r2

= 130 – 66 = 64

⇒ r1 – r2 = 8 ...(iii) 1

From (i) and (iii), 2r1 = 22

⇒ r1 = 11 cm

r2 = 14 – 11

= 3 cm. 1


qqq

WORKSHEET-123

Solutions

1. Perimeter of the circle = area of the circle. Q 2πr = πr2

∴ r = 2 units 1


2. Since ∆ABC is an equilateral triangle.

C

E

D

F10 cm

A

B

∴ ∠A = ∠B = ∠C = 60°

Area of sector, AFEA = θ

π360

2× r cm2

= 60

3605

2°

°×

π( )

cm2 1

= 25

6π

cm2

Q Areas of all three sectors are equal.

∴ Total area of shaded region = 325

6π

cm2

1

= 25 3 14

2

× .

= 39.25 cm2.


3. Let radius of the circle is r cm

Diameter = 2r cm

Circumference = 2πr ½

Circumference = Diameter + 16.8 ½

⇒ 2πr = 2r + 16.8

⇒ 2

22

7

r = 2r + 16.8

⇒ 44r = 14r + 16.8 × 7

⇒ 30r = 117.6

⇒ r =

117 6

30

.

1

∴ r = 3.92 cm 1


4. Area of shaded region

= Area of circle – (Area of semi-circle + Area of ∆ABC) 1


=

22

721 21 1386× × = cm2 1

Area of semi-circle =

πr2

2

1386

2693= = cm2

Area of triangle =

1

2×42 × 21

= 441

cm2 1

Area of shaded region = 1386 – (693 + 441)

= 1386 – 1134

= 252 sq. cm. 1


qqq


WORKSHEET-124

Solutions


1. Area of shaded region = Area of sector OCBAO – Area of ∆ODC

= 90°

360°× ×(7)

2 1

2×7×4π -

= 49

414

π-

= 24.5 cm2. 1

A

B

C

D

O7 cm

4 cm

1

2. OB = 20 2 cm ⇒ radius = 20 2

Area of shaded region

= Area of sector OQBPO – Area of square OABC 1

= 90°

360°×3.14×20×20×2 (20)

2-

= 2(314) – 400 = 628 – 400

Required Area = 228 cm2. 1

3. Side of square = 50 cm ∴ Area of square = 50 × 50 = 2500 cm2 ½

A B

CD

21 cm

15 cm

50 cm

Radius of quadrant = 15 cm.

Area of 4 quadrants = 4 ×

1

4 πr2 = πr2

= π × 15 × 15

=

22

7 × 225

= 707.14 cm2 ½


=

22

7

21

2

2

×

=

22

7

21

2

21

2× ×

= 346.5 cm2 ½

Area of remaining sheet = Area of square

– 4(area of quadrant) – Area of circle ½

= 2500 – 707.14 – 346.5

= 1446.36 sq. cm 1

Value : Importance of Water. 1


qqq

WORKSHEET-125

Solutions


1. Area of shaded region = Area of square – 4× area of quadrant ½

= lr22

44

- ×π

½

= 14 1422

77 7× × ×-

½

= 196 – 154 = 42 cm2. ½ [CBSE Marking Scheme, 2012]

2. ∠POQ = 60° ½

Area of segment PAQM

=

100

6

100 3

4

2π−

cm .

1

Area of semicircle = 25

2

2π cm ½

Area of shaded region =

25

2

50

325 3

π π− −

.

=

25 36

2−

π cm .

1



3.

20 m

14 m

Radius of the well =

7

2 m = 3.5 m

Volume of the earth taken out

= π7

2

2

× 10

= 22

7 ×

7

2 ×

7

2 × 10

= 385 m3 1

Area of the rectangular field = 20 × 14

= 280 m2 1

Area of the top of the well = π

7

2

2

=

77

2 m2

Area of the remaining field = 280 –

77

2

=

483

2m2 1

Let ‘h’ is the rise in the level of the field

∴ h =

385

483

2

= 1.6 m

(approx)

Value : Kind heartedness, Generous, Considerate, Understanding and Compassionate are the values depicted by farmer. He also believes in giving back to society. [CBSE Marking Scheme, 2015] 1

qqq

WORKSHEET-126

Solutions


1.

A B

D C

X YO

½ Area of square = 196 cm2

Area of semicircles AOB + DOC

=

22

7 × 49

= 154 cm2 ½

Hence area of two shaded parts

(X + Y) = 196 – 154 = 42 cm2 1

Therefore area of four shaded parts = 84 cm2. 1


2. Long hand makes 24 rounds in 24 hours

Short hand makes 2 round in 24 hours 1 Distance travelled by long hand in 24 rounds = 24 × 12π

= 288π ½ Distance travelled by short hand in 2 round

= 2 ×8π

= 16π ½

Sum of the distances = 288π + 16π = 304π

= 304 × 3.14 = 954.56 cm. 1


3. (i) Radius of the Gold scoring area

= 21

2 = 10.5 cm

∴ Area of the Gold scoring region

= 22

7 × 10.5 × 10.5 (A = πr2)

= 346.5 cm2

Radius of combined Gold and Red region

= 10.5 + 10.5

= 21 cm

∴ Area of Red scoring region = Area of combined Gold and Red regions – Area of the Gold region

=

22

7 × 21 × 21 – 346.5

= 1386 – 346.5 = 1039.5 cm2 ½

Radius of combined Gold, Red and Blue regions

= 21 + 10.5 = 31.5 cm

∴ Area of Blue scoring region = Area of combined Gold, Red and Blue region – Area of combined Gold and Red region


=

22

7 × 31.5 × 31.5 – 1386

=

3118.5 – 1386 = 1732.5 cm2 ½

Radius of combined Gold, Red, Blue and Black region

= 31.5 + 10.5 = 42 cm

∴ Area of black scoring regions = Area of combined Gold, Red, Blue and Black regions – Area of combined Gold, Red and Blue regions

=

22

7 × 42 × 42 – 3118.5

= 5544 – 3118.5

= 2425.5 cm2

Radius of combined Gold, Red, Blue, Black and White regions

= 42 + 10.5 = 52.5 cm

∴ Area of white scoring region = Area of combined Gold, Red, Blue, Black and White regions – Area of combined Gold, Red, Blue and Black regions.

=

22

7 × 52.5 × 52.5 – 5544

= 8662.5

– 5544 = 3118.5 cm2 1

(ii) Areas related to circle. ½

(iii) Logical proportion is required everywhere. ½

qqq


TOPIC-1Surface Areas and Volumes

WORKSHEET-127

Solutions

1. Curved Surface of cylinder = 2πrh

Volume of cylinder = πr2h

π

π

r h

rh

2

2 =

924

264 2

7

2⇒ =

r

∴ r = 7 m

2πrh = 264

⇒

2 ×

22

7 × 7 × h = 264

⇒ h = 6 m

∴

h

r2

6

14=

=

3

7 1

2. Volume of cylinder = π(5)2 × 4 cm3

= 100π cm3. 1

Volume of cone =

1

3×3 ×8

2π ½

= 24π

∴ Required ratio = 100π : 24π

= 25 : 6. ½


3.

h l

3.5

15.5 cm

h = 15.5 – 3.5 = 12 cm ½

l = 144 12 25+ . = 12.5 cm ½

TSA = πrl + 2πr2

= 22

7 × 3.5 × 12.5 + 2 ×

22

7 × 3.5 × 3.5 1

= 137.5 + 77

= 214.5 cm2. 1


4. Here r = 3, πrl = 47.1

∴ l = 47 1

3 3 14

.

.× = 5 cm 1

h = 5 32 2- = 4 cm ½

Volume of cone = 1

3 × 3.14 × 3 × 3 × 4 ½

= 37.68 cm3. 1


5. Given, Depth of well = 14 m, radius = 2 m.

Volume of earth taken out = πr2h

=

22

72 2 14× × ×

= 176 m3 1

Let r be the width of embankment

The radius of outer circle of embankment

= 2 +r

Area of upper surface of embankment

= π[(2 + r)2 – (2)2]

Volume of embankment = Volume of earth taken out 1½

⇒ π[(2 + r)2 – (2)2] × 0.4 = 176

⇒ π[4 + r2 + 4r – 4] × 0.4 = 176

⇒

r2 + 4r =

176 7

0 4 22

×

×.

⇒ r2 + 4r = 140

⇒ r2 + 4r – 140 = 0

⇒ (r + 14) (r – 10) = 0

⇒ r = 10 m 1½


qqq

SURFACE AREAS AND VOLUMES

SECTION

BSECTIONCHAPTER

13


WORKSHEET-128

Solutions

1. Here, h = 40 cm, circumference = 22 cm

2πr = 22

⇒

r =

22 7

2 22

××

⇒ r = 7

2

= 3.5 cm 1

2. Diameter of sphere = Radius of hemisphere

= 6 cm

⇒ Radius of sphere = 3 cm ½

V =

4

3πr3

½

=

4

3×22

7×3

3

cm3. ½

= 113.14 cm3. ½


3. Here r + h = 37 and 2πr(r + h) = 1628 ½+½

⇒ 2πr = 1628

37

⇒ r = 7 cm ½

and h = 30 cm. ½

Hence volume of cylinder = 22

7 × 7 × 7 × 30

= 4620 cm3. 1


4. Height of cylinder = 2.1 m

Radius of cylinder and cone = 3

2 m 1

Slant height of cone = 2.8 m Surface area of tent = C.S.A of cone + C.S.A. of cylinder. ∴ Area of canvas required = πrl + 2πrh ½

A = 22

7

3

2×

× 2.8 + 2 ×

22

7

3

2× × 2.1

= (13.2 + 19.8) m2

= 33 m2 1 Total Cost = 33 × 500 ` = 16,500 ` ½

5. Let the radius of hemisphere = r

∴

Therefore, r =

l

2 1

Now, the required surface area

= Surface area of cubical block – Area of base of hemisphere + Curved surface area of hemisphere. 1

= 6(side)2 – πr2 + 2πr2

=

6

2+2

2

2

ll l2

- π π

2

1

=

6

4+2

2l

ll

22

-π π

=

6 +

4l

l22π

Surface area =

1

4(24+ )π l2

units. 1


qqq

WORKSHEET-129

Solutions

1. Diameter of hemisphere = Side of cubical block 2R = 7

⇒

R =

7

2

Surface area of solid = Surface area of the cube – Area of base

of hemisphere + curved surface area of hemisphere

= 6l2 – πR2 + 2πR2 1

= 6 × 49 – 11 ×

7

2 + 77 ½

= 332.5 cm2 ½ [CBSE Marking Scheme, 2012]

2. Largest possible diameter = 10 cm.

Diameter of hemisphere = 10 cm 1

∴ Radius = 5 cm Total surface area = S.A. of Cube – Area of base of

hemisphere + C.S. area of hemisphere

= 6l2 – πr2 + 2πr2

= 6l2 + πr2


= 6 × (10)2 + 3.14 × (5)2

= 678.8 cm2 1

Cost of Painting =

678 5 5

100

3392 5

100

. .×=

= ` 33.925 = ` 33.93 1


3. Volume of bowl = 2

3

3πr

Volume of liquid in bowl

=

2

318

3 3π× ( ) cm

½

Volume of liquid after wastage

=

2

318

3 3π× ( ) ×

90

100 cm

½

Volume of bottle = πr2h

Volume of liquid in 72 bottles = π × (3)2 × h

× 72 cm2 ½

Volume of liquid in bottles = volume after wastage

π × (3)2 × h × 72 =

2

318

3π× ( ) × 90

100

⇒ h =

2

318

3 72

3

2

π

π

×

× ×

( )

( )

× 90

100

= 5.4 cm. ½ + 1


4.

6 cm. depth = 1.5 m

canal

8 cm

Field ½

Water flows in 1 hr = 10 km

Water flows in

1

2hr =

10

2 = 5 km = 5000 m

Now volume of water flows in

1

2 hr

= lbh 1

= 5000 × 6 × 1.5 m3 = 45000 m3. ½ According to the question,

Volume of water in

1

2 hr

= area of field ×

8

100m 1

⇒ 45000 = Area×

8

100

∴ Area =

45000×100

8= 562500

= 56.25 hectare. 1 [CBSE Marking Scheme, 2012]

qqq

WORKSHEET-130

Solutions

1.

Total surface area of hemisphere

Square of its radius

=

3π πr2

r23

1=

∴ Total surface area of hemisphere : Square of radius = 3π : 1 [CBSE Marking Scheme, 2012] 1

2. Let the height of water raised measured = h cm ½ ∴ Volume of water displaced in cylinder = π(10)2h ½ Volume of cube = 8 × 8 × 8 cm3 ½

∴ π(10)2h = 8 × 8 × 8

∴

h =

8×8×8×7

22×10×10 ½

= 1·629 cm.


3. Volume of cylinder = πr2h = π(3)2 × 5

= 45π cm3 ½

Volume of conical hole =

1

3

1

3

3

2

8

9

2

2

π πr h =

×

=

2

3

3p cm

1


Metal left in cylinder = 45π

− 2

3π

=

133

3

3p cm

1

Required ratio =

Volume of metal left

Volume of metal taken out

=

133

3π

π2

3

= 133 : 2. ½


4. Volume of water in cylinder = πr2h

= π × (60)2 × 180

= 648000π cm3 1

Volume of solid cone =

1

3

2πr h

=

1

330 60

2π× ×( )

= 18000π cm3 1

Volume of water left in cylinder

= 648000π – 18000π

= 630000π cm3

=

630000 22

1000000 7

3×

×m

= 1.98 m3. 1


5. Radius of pipe =

10

100=.10 m

½

∴ Radius of cylinder = 5 m 1 h = 2 m Volume of cylinder = πr2h = π × 5 × 5 × 2 = 50π m3

Volume of water in pi pe = πr2h Volume of water in cylinder π × (10)2 × h = 50π

⇒

h =

50

0.10×0.10 = 5000 m 1½ Now, 3000 m water flows in 60 minutes

1 ,, ,, ,, =

60

3000

5000 ,, ,, ,, =

60

3000×5000

1

= 100 minutes. [CBSE Marking Scheme, 2012]

qqq

WORKSHEET-131

Solutions

1. Side of the cube, a = 83 = 2 cm Now the length l of cuboid = 4 cm breadth, b = 2 cm height, h = 2 cm Surface area of cuboid = 2(l × b + b × h + h × l) ½

= 2(4 × 2 + 2 × 2 + 2 × 4) = 2 × 20 = 40 cm2 ½


2. Length of the cuboid so formed = l cm

∴ l = 5 + 5 = 10 cm, b = 5 cm ; h = 5 cm.

Surface area = 2(l × b + b × h + h × l) 1

= 2(10 × 5 + 5 × 5 + 5 × 10)

= 2(50 + 25 + 50)

= 2 × 125

= 250 cm2. 1


3. Volume of water collected in cylindrical vessel

=

4

51

7

2

2× × ×

π ( ) m3

1

=

44

5

3 m

1

Let the rainfall is h m. Rain water from roof = 22 × 20 × h m3

⇒

22 × 20 × h =

44

5

⇒

h =

44

5

1

22 20

1

50×

×

= m

=

1

50100 2× = cm

1


4. Height of cylindrical pipe “h” = 21 dm

= 210 cm

External Radius “R“ =

10

2 = 5 cm


Internal Radius “r” = 6

2 = 3 cm 1

Volume of copper used in making the pipe = (Volume of External Cylinder) – (Volume of

Internal Cylinder) = πR2h – πr2h 1 = πh (R2 – h2)

=

22

7210 5 3

2 2× −( )

=

22

7210 8 2× × ×

= 10560 cm3. 1 [CBSE Marking Scheme, 2015]

5. R = 8 cm, r = 6 cm

½ Surface area = 2πR2 + 2πr2 + π(R2 – r2) 1 = π[128 + 72 + 28] = 228 × 3.14 2

= 715.92 cm2 ∴ Total cost = 715.92 × 5 = ` 3579·60. ½ [CBSE Marking Scheme, 2012]

qqq

WORKSHEET-132

Solutions

1. Q Whole surface of each part

= 2πr2 + πr2 = 3πr2 ½

∴ Total surface of two parts

= 3πr2 + 3πr2 = 6πr2 ½


2. Given, TSA of hemisphere = 462 cm2.

(i.e.), 3πr2 = 462 ½

⇒

22

7×r

2

=

462

3

⇒ r2 =

462×7

22×3 = 49

⇒ r = 7 cm. ½

∴ Volume of hemisphere

=

2

3πr3

½

=

2

3×22

7×7×7×7

= 2156

3

= 718.67 cm3. ½

3. Side of cube a = 7 cm

The diameter of the largest possible

sphere is 7 cm same as side of cube

Radius = 7/2 cm. 1

Volume of the wood left = volume of cube – volume of sphere

= a3 – 4

3 πr3 1

= 7 × 7 × 7 – 4

3

22

7

7

2

7

2

7

2× × × ×

= 7 × 7 (7 – 11/3)

= 7 × 7

21 11

3

−

= 7 × 7 ×10/3 1

=

490

3

= 163.3 cm3.

4. Since a cone, a hemisphere and a cylinder are on equal bases.

Let the radius of each of them be r and height also be equal to r.

Then volume of cone, V1 =

1

3

2πr h

=

1

3

2πr r×

1

=

1

3

3πr

Volume of hemisphere, V2 =

2

3

3πr

1

Volume of cylinder, V3 =πr2h = πr2. r = πr3 1

Ratio in their volumes

V1 : V2 : V3 =

1

3

2

3

3

3

3 3 3π π πr r r: :

= 1 : 2 : 3 1


qqq


WORKSHEET-133

Solutions

1. Here r = 7 cm, h = 10 cm, Volume of cylinder = πr2h

=

22

77 102

× ×( )

= 1540 cm3 1


2. Given, radius and height = 7 m & 24 m.

Slant height (l) = r +h2 2 = 7 +24

2 2

= 625 = 25 m.

C.S.A. = πrl

=

22

7×7×25 = 550 m2. 1

Let x m of cloth is required C.S.A.

= area of cloth.

⇒

5x = 550 ⇒ x =

550

5 = 110 m.

∴ 110 m of cloth is required.

Cost of cloth = 25 × 110 = ` 2750. 1

3.


4. Volume of water flowing through pipe in 1 sec = πR2H = π × (1)2 × 0.4 × 100 cm3 1

Volume of water flowing in 30 min (30 × 60 sec)

= π × (1)2 × 0.4 × 100 × 30 × 60 1

Volume of water in cylindrical tank in 30 min

= πr2h = π × (40)2 × h ½

π × (40)2 × h = π × (1)2 × 0.4 × 100 × 30 × 60 ½ Rise in water level

h =

π

π

×(1) ×0.4×100×30×60

×40×40

2

= 45 cm. ∴ Level of water in the tank is 4·5 cm. 1

qqq

WORKSHEET-134

Solutions

1.

Volume of reduced cylinder

Volume of original cylinder=

×πr

2

2

2

h

r hπ

=

1

4 = 1 : 4

1


2. Each one of the plate is also a cylinder. Its volume is, V = πr2h = π × (·75)2(·2)

=

9

80

π

cm3. 1

The right circular cylinder has volume

V = π(2·25)2(10) =

405

8

π

cm3.

½

½

1

1


∴ Number of plates =

405

8

9

80

π

π

= 450 plates. 1

3.

6 cm

13 cm

14 cm

Radius of hemisphere =

14

2 = 7 cm

Height of cylinder = 13 – 7 = 6 cm T.S.A. of vessel = S.A. of hemisphere + S.A. of cylinder = 2πr2 + 2πrh 1 = 2πr(r + h)

=

2×

22

7×7(7+6)

1

= 44 × 13 = 572 cm2. 1

4.

18 cm 18 cm

44 cm

44 cm

18 cm

The paper is rolled along length, therefore, 44 cm forms the circumference of base of cylinder 1

∴ 2πr = 44 ⇒ r = 7 cm ∴ Volume of the cylinder

= πr2h =

22

77 18

2× ×( )

1

= 2772 cm3. 2 [CBSE Marking Scheme, 2012]

WORKSHEET-135

Solutions

1. Let the length, breadth and height of the cuboid is l, b and h respectively.

h

l

b

X = l × b

Y = b × h

Z = l × h

XYZ = l2 b2 h2

Volume of cuboid = l × b × h

∴ l2 b2 h2 = XYZ

⇒ lbh = XYZ 1


2. Volume of remaining solid

=Volume of cylinder

– Volume of cone ½

=

π πr h r h

2 2-1

3

=

2

3πr h2

1

=

2

3×22

7×0.7×0.7×2.4

1.4 cm

2·4 cm

½ = 44 × 0.1 × 0.7 × 0.8 = 4.4 × .56 = 2.464 cm3. ½ [CBSE Marking Scheme, 2012]

3. Let the radii of two cylinders be 2x and 3x and their heights be 5y and 4y respectively. ½

Ratio of their curved surface areas

=

2 ×2 ×5

2 ×3 ×4=5

6

π

π

x y

x y

1

Q Their curved surface areas are in the ratio of 5 : 6.

∴ Ratio of their volumes =

π

π

×(2 ) ×5

×(3 ) ×4

2

2

x y

x y

1

=

5×4

4×9


=

5

9 ½

i.e., their volumes are in the ratio of 5 : 9. [CBSE Marking Scheme, 2012]

4.

12 cm

15.5 cm

3.5 cm

Slant height of cone, l = h r2 2+

= 12 +3.5 =12.52 2

1

Total surface area of the toy = Surface area of hemisphere + Curved surface area of cone

= 2πr2 + πrl 1

=

2×

22

7×

7

2

2

+22

7×7

2×12.5

= 214.5 cm2. 1[CBSE Marking Scheme, 2012]

qqq

TOPIC-2Problems Involving Converting One Type of Metallic Solid into Another

WORKSHEET-136

Solutions

1. Let the radius of spherical ball = R. Volume of spherical ball = Volume of three balls

4

3π R3 =

4

3π [(3)3 + (4)3 + (5)3]

⇒ R3 = 27 + 64 + 125

⇒ R3 = 216 ⇒ R = 6 cm 1

2. Volume of cylinder = Volume of Balls

R2 × H =

4

3× r3 × N 1

3.5 × 3.5 × 14 = N × × × ×

4

3

7

12

7

12

7

12 1

N = 648[CBSE Marking Scheme, 2016]

3. Let the water level raised in cylindrical vessel be h cm

Volume of Sphere = Volume of water displaced in cylinder ½

4

3π(3)3 = π(6)2h 1

4

3× 27 = 36h

36 = 36h h = 1 cm ½ [CBSE Marking Scheme, 2016]

5.


½

½

1

1


5. Given, depth of well = 21 m

Radius of well =

6

2 = 3 m

Volume of earth dugout from the well = πr2h

=

22

73 3 21× × ×

= 594 m3 1

Let the height of the platform = h m.

Volume of platform = l × b × h

= 27 × 11 × h 1

Volume of plateform = Volume of earth dugout

27 × 11 × h = 594 1

⇒ h =

594

27 11

594

297×=

∴ Height of the platform, h = 2 m. 1


qqq

WORKSHEET-137

Solutions

1. No. of spheres = 12

Radius of cone, r = 1 cm

Height of the cone = 48 cm

∴ Volume of 12 spheres = Volume of cone

Let the radius of sphere be R cm

12 ×

4

3π R3 =

1

3π r2h

⇒

12 ×

4

3π R3 =

1

3π

× (1)2 × 48

⇒ R3 = 1

⇒ R = 1 cm 1


2. Volume of coin = πr2h

=

22

7 × (0.75)2 × 0.2 cm3 ½

Volume of cylinder =

22

7× (2.25)2 × 10 cm3 ½

No. of coins =

Volume of cylinder

Volume of coin ½

=

22

72 25 10

22

70 75 0 2

2

2

× ×

× ×

( . )

( . ) .

= 450 ½


3. Volume of sphere = Volume of cone

∴

4

313πr

=

1

322πr h

½

4

313

× r =

( )6

24

3

2×

⇒ 4r13 = 36 × 24 ½

⇒ r13 = 63

⇒ r1 = 6 Hence, radius of sphere

= 6 cm. 1

4. Volume of cylinder = πr2h

=

22

7

42

10

42

1010

3× × × cm

½

= 554.40 cm3 ½

Volume of metal scooped out

=

4

3

3πr

=

4

3

22

7

42

10

3

× ×

½

= 310.46 cm3 ½ Volume of rest of cylinder = 554.40 – 310.46

22

7

7

10

7

10× × × l

= 243.94 cm3 ½

10 cm.

4.2 cm.

If l is the length of wire, then

πr2l = 243.94

= 243.94 1

⇒ l =

243 94 10 10

22 7

. × ×

×

= 158. 4 cm ½


qqq


WORKSHEET-138

Solutions

1. Volume of sphere = Volume of cone

Let the radius of cone be R cm.

∴

4

3π r3 =

1

3π R2 × r

⇒ 4r3 = R2r

⇒ R2 = 4r2

⇒ R = 2r 1


2. Volume of cylinder = Volume of sphere, ½

πr2h = π

where r and h are radius of base and height of cylinder ½

(0.5)2 h = 1

1

2

2

h = 1

h = 4 cm. 1 [CBSE Marking Scheme, 2012]

3. Volume of sphere = Volume of cylinder ½

4

3

3πR = πr2h ½

4

34 2 3× ( . )

= 62 × h

h = 4 4 2 4 2 4 2

3 6 6

× × ×

× ×

. . .

Hence, height of cylinder h = 2.744 cm. 1


4. Volume of earth dug out = π × 2 ×2 × 21 = 264 m2 1 Volume of embankment = π(25 – 4) × h

= 66h m3 1 ∴ 66h = 264 ½ ⇒ h = 4 ½


5. Diameter of spherical marble = 14 cm

Radius r1 = 1.4/2 = 0.7 = 7/10 cm

Diameter of cylindrical vessel = 7 cm

Radius R = 7/2 cm

Let h be the rise in water level then

Volume of 150 spherical marbles =volume of water rises

⇒

100

4

3

7

10

7

10

7

10× × × × ×π

=

π× × × ×

7

2

7

2

7

2h

⇒ h =

4 7

5

×

⇒

28

5 = h

⇒ h = 5.6 cm The rise is the level of water, h = – 5.6 cm

qqq

TOPIC-3Frustum

WORKSHEET-139

Solutions

1.

16

cm

h

8 cm

12 cm

20 cm

l

d

Slant height of the frustum

= h d2 2+

= ( ) ( )16 12

2 2+

= 256 144+

= 400

= 20 cm. 1



2. Here, l = 26 cm, upper radius = 18 cm,

lower radius = 8 cm

d = difference in radius = 18 – 8 = 10 cm.

Let h be the height of bucket

∴ h = l d2 2-

= ( ) ( )26 102 2

-

= 676 100-

= 576 = 24 cm. 1


3. Let r be the radius of the top., h = 12 – 4 = 8 cm

4

r =

12

6

∴ r = 2 cm ½

l = h R r2 2+ ( )-

=

( ) ( )8 6 22 2+ -

= 64 16+ = 80

= 4 5 = 4 × 2.236

= 8.944 cm ½

Total surface area of frustum

= π[R2 + r2+ l (R + r)]

=

22

7 [(6)2 + (2)2 + 8.944 (6 + 2)]

=

22

7 [36 + 4 + 71.552]

=

22

7 × 111.552

= 350.59 cm2. 1


4. Here R = 20, r = 12, V = 12308.8

12308.8 = 1

3 × 3.14 (400 + 240 + 144) h 1

⇒ h = 15 cm ½

l = ( )20 12 152 2

- + = 17 cm ½

Total area of metal sheet used = CSA + base area

= π [(20 + 12) × 17 + 12 × 12] 1

= 2160.32 cm2 1


qqq

WORKSHEET-140

Solutions


1. Ratio of volumes

Volume of I sphere

Volume of II sphere

st

nd

=

4

34

3

3

3

π

π

r

R

=

8

72

⇒ r

R =

2

3

∴ R =

3

2r

∴ (R – r) : r =

3

2r r r−

:

=

rr

21 2: :=

1


2. Surface area of block

=

216

22

7

3 5

2

3 5

22

22

7

3 5

2

3 5

2− × × + × × ×

. . . .

1+½+½

= 225.625 cm2. [CBSE Marking Scheme, 2016] 1


Area of six faces of cube = 6a2

= 6(6)2

= 216 cm2 1

Area of circular part covered by hemisphere

= πr2

=

22

7

3 5

2

2

×

.

= 9.625 cm2 1

Area of curved surface of hemisphere

= 2πr2

=

2

22

7

3 5

2

2

×

.

= 19.25

∴ Required area = 216 – 9.625 + 19.25 cm2

= 225.625 cm2. 1


3.


qqq

WORKSHEET-141

Solutions


1.

A

B C

D

15

20

12 D 12

15

C

A

BE

15

20 20

(i) AC2 = 202 + 152 = 625

⇒ AC = 25 cm

(ii) ar (∆ABC) = ar (∆ABC)

1

2 × BC × AB =

1

2 × AC × BD

⇒ 15 × 20 = 25 × BD

⇒ BD = 12 cm 1

Volume of double cone

= Volume of upper cone + Volume of lower cone

=

1

3( ) × +

1

3( ) ×

2 2π πBD AD BD CD

=

1

3( ) { }=

1

3( ) ( )

2 2π πBD AD+CD BD AC

½

½

½

½

½

½


=

1

3×3.14×144×25=3768 cm

2

1

Surface area = C.S.A. of upper cone

+ C.S.A. of lower cone

= π(12)(20) + π(12)(15)

= 12π{20 + 15}

= 12 × 3.14 × 35

= 1318.8 cm2. 1


2. Radius of lower cylinder = R = 12 cm

Radius of upper cylinder = r = 8 cm

Height of upper cylinder = h = 60 cm

Height of lower cylinder = H = 220 cm

Volume of solid iron pole = πR2H +πr2h 1

= 3.14 × (12)2 × 220 + 3.14 × (8)2 × 60

= 111532.8 cm3 1

Mass of the pole = 111532.8 × 8 g

= 892.2624 kg. 1


3. Volume of wheat in the form of cone

=

1

3×22

7×3×3×3·5

= 11 × 3 = 33 m2 1

l = 3 3 5 4 6092 2

+ =. . m 1

Canvas required to cover the heap

= πrl

=

22

73 4 609× × .

= 43.45 m2. 1

qqq

WORKSHEET-142

Solutions


1. Volume of the milk container =Volume of frustum

=

1

3

2 2πh R r Rr[ ]+ +

=

1

330 40 20 40 202 2π× + + ×( )

= 10π (1600 + 400 + 800) 1

=

10

22

72800× ×

= 88000 cm3

= 88 litre 1

Number of containers needed =

880

8810=

Cost of milk = ` 88 × 10 × 35

= ` 30800

Volume : Helping the flood victims. 1


2. Volume of water in cone

=

1

3

2πr h

=

1

35 8

2π× ×( )

=

200

3

3π cm

½

Volume of water flows out

=

1

4

200

3

50

3

3× =π π cm

1

Let the radius of one spherical ball be r cm 1½

∴

4

3100

3πr × =

50

3π

r3 =

50

4 100

1

8×=

⇒ r =

1

2= 0.5 cm 1


3.

30 cm

r

r1

h

½

r

r1

=

h

30 ⇒ h =

30 1× r

r 1

1

31

330

12

2

π

π

r h

r

×

×

=

1

27

or

r r

r12

1

3

30

30

× ×

×

=

1

3 1½


∴ r

r1 =

1

3

⇒ h = 10 cm 1

∴ The section is made 20 cm above base.


qqq

WORKSHEET-143

Solutions


1.

2 cm

2 cm

1.5 cm

1.5 cm

2 cm

8 cm

2 cm

(i) Here, radius of two cones and cylinder=

3

2 cm

= 1.5 cm

Height of each cone = 2 cm

∴ Height of cylindrical portion = 12 – 2 – 2 = 8 cm

∴ Volume of the air = Volumes of cylindrical part

+2 × Volume of cone

= π(1.5)2 × 8 + 2

1

3(1.5) ×2

2π

1

=

22

7×(1.5) 8+

4

3

2

=

22

7×2.25×

28

3 1

= 66 cm3.

(ii) Volume (Mensuration). ½

(iii) Sincerity. ½

2.

P

H G

B C

E FA

2 cm

½ Let BPC is a hemisphere and ABC is a cone.

Radius of hemisphere = Radius of cone

= 4

2 = 2 cm

h = Height of cone = 2 cm

Volume of toy = 2

3

1

3

3 2π πr r h+

=

2

33 14 2

1

33 14 2 23 2× × + × × ×. .

= 25.12 cm3 ...(i) Let right circular cylinder EFGH circumscribe the

given solid toy. 1½ Radius of cylinder = 2 cm, Height of cylinder = 4 cm

Volumeofrightcircularcylinder=πr2h = 3.14 × (2)2 × 4 cm3 ...(ii)

= 50.24 cm3 1

∴ Required volume = Volume of cylinder

– Volume of toy = 50.24 – 25.12

= 25.12 cm3. 1 [CBSE Marking Scheme, 2012]

qqq


TOPIC-1Mean, Median and Mode

WORKSHEET-144

Solutions

1. Given, Median = Mean + 3 Also, we know that, Mode = 3 Median – 2 Mean = 3 (Mean + 3) – 2 Mean ⇒ Mode = Mean + 9 Hence Mode exceeds Mean by 9.

2. Given, Mode = 50.5

Median = 45.5

3 Median = Mode + 2 Mean

⇒ 3 × 45.5 = 50.5 + 2 Mean

⇒ Mean =

136 5 50 5

2

. .−

= 43


3.

C.I. fi xi ui = x a

h

i-

fiui

20 – 30 8 25 – 2 – 16

30 – 40 6 35 – 1 – 6

40 – 50 x 45 = a 0 0

50 – 60 11 55 1 11

60 – 70 y 65 2 2y

Total Sfi = 25 + x + y Sfiui = 2y – 11

Mean = SSf u

fi i

i

× h + a

⇒ 48 =

2 11

50

y − × 10 + 45 1

⇒ 15 = 2y – 11

⇒ y = 13

Also Sfi = 25 + x + y = 50 1

⇒ x + y = 25

⇒ x = 25 – 13 = 12

\ x = 12 and y = 13 [CBSE Marking Scheme 2016] 1

STATISTICS

SECTION

BSECTIONCHAPTER

14


4.

C.I. xi ui fi fiui

35 – 40 37.5 – 5 1 – 5

40 – 45 42.5 – 4 2 – 8

45 – 50 47.5 – 3 3 – 9

50 – 55 52.5 – 2 x – 2x

55 – 60 57.5 – 1 y – y

60 – 65 62.5 =A 0 6 0

65 – 70 67.5 1 8 8

70 – 75 72.5 2 4 8

75 – 80 77.5 3 2 6

80 – 85 82.5 4 3 12

85 – 90 87.5 5 2 10

Total Sfi =31 + x + y Sfiui = 22 – 2x – y

Here, Sfi = 31 + x + y = 40

2

⇒ x + y = 9 ...(i) Sfiui = 22 – 2x – y

\ Mean = A + ×Σ

Σ

f u

fhi i

i

1

⇒ 63.5 = 62 5

22 2

405.

( )+

− −×

x y

⇒ 2x + y = 14 ...(ii) Solving eqns (i) and (ii), x = 5 and y = 4. [CBSE Marking Scheme, 2016] 1

qqq

WORKSHEET-145

Solutions

1.

C.I. 1400 – 1550 1550 – 1700 1700 – 1850 1850 – 2000

f 8 15 21 8

c.f. 8 23 44 52

Σ f

2 = 26 ⇒ Median class = 1700 – 1850 [CBSE Marking Scheme, 2015] 1

2. Modal class is 30 – 35, l = 30, f1 = 25, f0 = 10, f2 = 7, h = 5

Mode = l + f f

f f fh1 0

1 0 22

-

- -

×

⇒ Mode = 30 +

25 10

50 10 7

-

- - × 5

= 30 + 2.27 or 32.27 approx.


3. By short-cut Method (any method) For making correct table A = 187.5

∑ fiui = – 12; N = 50, h = 25

Mean = A + ∑ f y

Ni i × h

⇒ Mean = 187.5 + −1250

× 25

= 187.5 – 6 = 181.5. [CBSE Marking Scheme, 2015] 3



By short-cut Method (any method) A = 187.5

Class-Interval

Height (in cm)

Frequency

fixi

ui = x a

h

i-

fiui

50 – 75 5 62.5 – 5 – 25

75 – 100 6 87.5 – 4 – 24

100 – 125 3 112.5 – 3 – 9

125 – 150 4 137.5 – 2 – 8

150 – 175 3 162.5 – 1 – 3

175 – 200 7 187.5 0 0

200 – 225 5 212.5 1 5

225 – 250 4 237.5 2 8

250 – 275 8 262.5 3 24

275 – 300 5 287.5 4 20

∑ fi = 50 ∑ fiyi = –12

Here, ∑ fiui = – 12; N = 50, h = 25

Mean = A + ∑ f y

Ni i × h

Mean = 187.5 + −1250

× 25 = 187.5 – 6 = 181.5 1

4.

C. I. fi c.f. xi ui =x a

n

i− fiui

05–07

07–09

09–11

11–13

13–15

15–17

17–19

70

120

32

100

45

28

5

70

190

222

322

367

395

400

6

8

10

12

14

16

18

–3

–2

–1

0

1

2

3

–210

–240

–32

0

45

56

15

Sf= 400 Sfiui = – 366

a = Assumed mean = 12

2

Mean, x = a +

Σ

Σ

f u

fhi i

i

×

Mean = 12 +

- 366

400 × 2 = 10.17

∑ f

2 = 200 ⇒ Median class = 09 – 11

Median = l +

nc f

fh2

- . .

×

⇒ Median = 9 +

200 190

32

-

× 2 = 9.63 [CBSE Marking Scheme, 2015] 2

qqq


WORKSHEET-146

Solutions

1.

C.I. 5 – 10 10–15 15–20 20–25

f 7 15 6 2


2. Writing the data as discrete frequency distribution, we get

xi fi 13 1

15 3 17 1

18 3

20 3

λ 1

25 3

For 20 to be mode of the frequency distribution, λ = 20. [CBSE Marking Scheme, 2015] 2

3. Here, maximum frequency = 9, hence modal class is 60 – 90 1

Mode = L hf f

f f f+

−

− −

1 0

1 0 22

Here, L = 60, f1 = 9, f0 = 6, f2 = 6 and h = 30.

\ Mode = 60 309 6

2 9 6 6+

−

× − −

1

= 60

30 3

6+

×= 60 + 15

= 75


4.

C. I. xi fi ui =x a

n

i− fiui

40–44

44–48

48–52

52–56

56–60

60–64

64–68

68–72

42

46

50

54

58

62

66

70

4

6

10

14

10

8

6

2

–3

–2

–1

0

1

2

3

4

–12

–12

–10

0

10

16

18

8

Sf= 60 Sfiui = 18

Let a = Assumed mean = 54

2

Mean, x = a +

Σ

Σ

f u

fhi i

i

×

Mean = 54 +

18

60 × 4 = 55.2

Maximum frequency = 14 ⇒ Modal class = 52 – 56, l = 52, f1 = 14, f0 = 10, f2 = 10, h = 4

Mode = 52 +14 10

28 10 10

-

- - × 4 = 54 [CBSE Marking Scheme, 2015] 2

qqq


WORKSHEET-147

Solutions

1.

Height Frequency c.f.

140 – 145 5 5

145 – 150 15 20

150 – 155 25 45

155 – 160 30 75

160 – 165 15 90

165 – 170 10 100

Sf = 100

½

N = 100

⇒

N

2 =

100

2 = 50 ½

Hence, Median class is 155 – 160.

2. Mean = Σ

Σ

fx

f

⇒ 50 =

Σfx

100

⇒ Sfx = 5000

Correct, Sfx' = 5000 – 100 + 110 ½

= 5010

\ Correct Mean = 5010

100

= 50.1 ½

Median will remain same i.e. median = 52


3.

Expenditure f (families) c.f.

0–1000 150 150

1000–2000 200 350

2000–3000 75 425

3000–4000 60 485

4000–5000 15 500

Sf = 500

N = 500,

N

2 = 250

Median class = 1000 – 2000,

Median = l +

Nc f

f2- . .

× h

= 1000 +

250 150

200

-

× 1000

= 1000 + 500 = 1,500 3

\ Median Expenditure = ` 1500/week.

4. Mode = l +

f f

f f f1 0

1 0 22

-

- - × h 1

= 45 +

33 31

66 31 17

-

- - × 10 1

= 45 +

2

18 × 10 1

= 46.1


qqq

WORKSHEET-148

Solutions

1.

xi fi fixi

3 3 9

4 4 16

5 8 40

7 5 35

10 10 100

Total ∑fi = 30 ∑fixi = 200

Mean = ∑

∑

f x

fi i

i

=200

30 = 6.67 2


2. Height Frequency c.f.

100 – 120

120 – 140

140 – 160

160 – 180

180 – 200

12

14

8

6

10

12

26

34

40

50

Total 50

Here, N = 50 ⇒

N

2

50

2=

= 25

So, Median Class = 120 – 140

Median = l +

Nc f

f2- . .

× h

= 120 + 25 12

14

-

× 20

= 120 +

260

14

= 120 + 18.57

\ Median = 138.57


3. Here, Modal class = 11 – 13

l = 11, f1 = 95, f0 = 41, f2 = 36, h = 2

Mode = l + f f

f f f1 0

1 0 22

−− −

× h 1

= 11 + 95 41

190 41 36

-

- - × 2

= 11 + 54

113 × 2

\ Mode = 11 + 0.95 = 11.95

Now, let us calculate Mean :

Age xi fi fixi

5–7

7–9

9–11

11–13

13–15

15–17

17–19

6

8

10

12

14

16

18

67

33

41

95

36

13

15

402

264

410

1140

504

208

270

Sfi= 300 Sfixi = 3,198

Mean = x =

∑∑f x

fi i

i 1

= 3 198

300

,

= 10.66 1

qqq

WORKSHEET-149

Solutions

1.

Class Interval Frequency

0 – 20 2

20 – 40 2

40 – 60 3

60 – 80 12

80 – 100 18

100 – 120 5

120 – 140 2

Total 44

Here, Modal Class = 80 – 100

l = 80, f1 = 18, f2 = 5, f0 = 12, h = 20

\ Mode = l + f f

f f f1 0

1 0 22

-

- -

× h

= 80 + 18 12

36 12 5

-

- -

× 20

= 80 + 6

19 × 20

= 80 + 6.31

= 86.31 2


2.

C.I. f ui fiui

0 – 10 5 – 3 – 15

10 – 20 x – 2 – 2x

20 – 30 10 – 1 – 10

30 – 40 12 0 0

40 – 50 7 1 7

50 – 60 8 2 16

Total 42+ x – 2x – 2

A = Assumed mean = 35

Mean = Af u

fi i

i

+ ×Σ

Σ10

⇒ 31.4 = 352 2

4210+

− −

+×

x

x

⇒ (2x + 2)10 = (42 + x)(3.6)

⇒ 20x + 20 = 151.2 + 3.6x

16.4x = 131.2

\ x = 8

3.

Class Interval Frequency Cumulative frequency

0 – 100

100 – 200

200 – 300

300 – 400

400 – 500

500 – 600

600 – 700

700 – 800

800 – 900

900 – 1000

2

5

x

12

17

20

y

9

7

4

2

7

7 + x

19 + x

36 + x

56 + x

56 + x + y

65 + x + y

72 + x + y

76 + x + y

N = 100

Hence, 76 + x + y = 100 2

⇒ x + y = 100 – 76 = 24 ....(i)

Given, Median = 525,

⇒ Median class = 500 – 600

Now,

Median = l +

nc f

f2- . .

× h

⇒

525 = 500 +

100

236

20

- ( )+

x

× 100

⇒ 25 = (50 – 36 – x) 5

⇒

(14 – x) =

25

5 = 5

⇒ x = 14 – 5 = 9 1

Substituting the value of x in equation (i),

y = 24 – 9 = 15 1

qqq


WORKSHEET-150

Solutions

1.

xi fi xifi

3 5 15

9 4 36

15 1 15

21 6 126

27 4 108

Total ∑fi = 20 ∑ xifi = 300

Mean =

∑∑f x

fi i

i

= 300

20 = 15 2

2.

2

Class-Interval Frequency

0 – 50

50 – 100

100 – 150

150 – 200

200 – 250

250 – 300

8

15

32

26

12

7

Total 100

3.

Life time (in hours) Frequency c.f.

0–250 6 6

250–500 10 16

500–750 11 27

750–1000 15 42

1000–1250 10 52

1250–1500 5 57

Here, N = 57

⇒ N

2 =

57

2 = 28.5 3

Hence, median life time is 750 – 1000. 4.

1

C. I. f c.f.

0 – 175 10 10

175 – 350 14 24

350 – 525 15 39

525 – 700 21 60

700 – 875 28 88

875 – 1050 7 95

1050 – 1225 5 100

Sf

2 =

100

2 = 50

\ Median class = 525 – 700

\ Median = 525 + 175

21 × [50 – 39]

= 525 + 91.6 = 616.6

Mode = Lf f

f f f+

−

− −

1 0

1 0 22

1

= 70028 21

2 28 21 7175+

−

× − −

×

1

= 7007

28175+ ×

= 700 + 43.75

= 743.75 1

qqq

WORKSHEET-151

Solutions

1.

Marks No. of students

c.f.

0 – 10

10 – 20

20 – 30

30 – 40

40 – 50

5

15

30

8

2

5

20

50

58

60

N = 60

Here,

N

2 =

60

2 =30

So, Median class = 20 – 30

l = 20, f = 30, c.f. = 20, h = 10

Median = l +

Nc f

f2- . .

× h 1

= 20 +

30 20

30

−

× 10


= 20 +

100

30 = 20 +

10

3

= 20 + 3.33

\ Median = 23.33 1

2.

C.I. fi xi ui = x a

h

i-

fiui

10 – 30 15 20 – 2 – 30

30 – 50 18 40 – 1 – 18

50 – 70 25 60 = a 0 0

70 – 90 10 80 1 10

90 – 110 2 100 2 4

Total Sf = 70 Sfiui = – 34

a = 60

Mean = a +SSf u

fi i

i

× h

= 60

34

7020+

−×

= 60 – 9 .71

= 50.28

3.

Class-Interval c.f. No. of students

0 – 10 7 7

10 – 20 21 14

20 – 30 34 13

30 – 40 46 12

40 – 50 66 20

50 – 60 77 11

60 – 70 92 15

70 – 80 100 8

2

From table, Maximum frequency = 20. So, Modal class = 40 – 50 Here, l = 40, f1 = 20, f2 = 11, f0 = 12, h = 10

Mode = l +

f f

f f f1 0

1 0 22

-

- -

× h 1

= 40 +

20 -

- -

12

40 12 11

× 10

= 40 +

8

17 × 10

= 40 +

80

17 = 40 + 4.7 = 44.7 1

qqq

WORKSHEET-152

Solutions

1.

Class 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70

Frequency 1 3 5 9 7 3

Cumulative Frequency 1 4 9 18 25 28

Median class : 40 – 50 ⇒ Lower limit = 40

Modal class : 40 – 50 ⇒ Upper limit = 50 1

Their sum = 40 + 50 = 90 1




Mode = 24.5

and Mean = 29.75

The relationship connecting measures of central tendencies is :

3 Median = Mode + 2 Mean ½

⇒ 3 Median = 24.5 + 2 × 29.75

= 24.5 + 59.50 ½

⇒ 3 Median = 84.0

\ Median =

84

3 = 28 1

3.

Class- Interval Frequency

150–155 15

155–160 8

160–165 20

165–170 12

170–175 5

Total 60

1 Here, Modal Class = 160 –165

l = 160, f1 = 20, f0 = 8, f2 = 12, h = 5

Mode = l +

f f

f f f1 0

1 0 22

-

- -

× h

= 160 +

20 8

40 8 12

-

- -

× 5

= 160 +

12

20

× 5

= 163

\ Modal height = 163 cm. 2

4.

2

Class xi (class mark) fi fixi

0 – 100 50 12 600

100 – 200 150 16 2400

200 – 300 250 6 1500

300 – 400 350 7 2450

400 – 500 450 9 4050

Total Sfi = 50 Sfixi = 11,000

Mean = ΣΣx f

fi i

i

=11000

50

1

= 220

\ Average daily income = ` 220 1

qqq

WORKSHEET-153

Solutions

1. Modal class = 20 – 30

Here, l = 20, f1 = 40, f0 = 24, f2 = 36, h = 10 ½

Mode = lf f

f f fh+ ×

( )1 0

1 0 22

-

- - ½

= 20 +

( )40 24

80 24 36

-

- - × 10 ½

= 20 +

16 10

20

×

= 28

[CBSE Marking Scheme, 2012] ½

2. From the table, 12 + a = 25 ⇒ a = 25 – 12 = 13 ½ 25 + 10 = b ⇒ b = 35, ½ b + c = 43

⇒ c = 43 – b = 43 – 35 = 8 ½

and 48 + 2 = d ⇒ d = 50 [CBSE Marking Scheme, 2012] ½


3. Mode = l + f f

f f f1 0

1 0 22

−− −

× h 1

Here, Modal Class = 50 – 60

l = 50, f1 = 25, f0 = 15, f2 = 15, h = 10

= 50 +

25 15

50 15 15

-

- - × 10

= 50 +

10

20 × 10

= 50 + 5

= 55. 2

4. Let assumed mean, A = 649.5 and h = 100

Life time

(in hrs)xi ui =

x a

h

i -

fi fiui

399.5 – 499.5 449.5 – 2 24 – 48499.5 – 599.5 549.5 – 1 47 – 47599.5 – 699.5 649.5 0 39 0699.5 – 799.5 749.5 1 42 42799.5 – 899.5 849.5 2 34 68899.5 – 999.5 949.5 3 14 42

Total Sfi = 200 Sfiui =57

\ Mean, x = A +

ΣΣf u

fhi i

i

×

1

= 649.5 +

57

200× 100

= 649.5 + 28.5 = 678 1 \ Mean life time of a bulb is 678 hours.

qqq

WORKSHEET-154Solutions

1. Modal class = 35 – 40 Here, l = 35, f1 = 50, f2 = 42, f0 = 34, h = 5 ½

Mode = l

f f

f f fh+ ×

( )1 0

1 0 22

-

- - ½

=

35

50 34

100 34 425+ ×

-

- - ½

= 35 +

16 5

24

×

= 38.33 [CBSE Marking Scheme, 2012] ½

2. From the cumulative frequency distribution,

15 + x = 28

⇒ x = 28 – 15 = 13 1

and 43 + 18 = y

⇒ y = 61 1

Hence, x = 13 and y = 61 [CBSE Marking Scheme, 2012]

3. Here,

Modal class is 6000 – 8000

f0 = 8, f1 = 10 f2, = 2, h = 2000


\ Mode =

L +−

− −

×

f f

f f fh1 0

1 0 22

1

=

6000

10 8

20 8 22000+

−

− −

×

= 6000 +

2

10 × 2000

= 6000 + 400

= 6400 2

4.

xi

(Class marks)fi fixi

15 12 180

45 21 945

75 x 75x

105 52 5460

135 y 135y

165 11 1815

Total Sfi = 150 Sfixi = 8400 + 75x + 135y

x + y = 54 1

\

x

= Σ

Σ

f x

fi i

i

⇒

91 = 8400 75 135

150

+ +x y

⇒ 13650 = 8,400 + 75x + 135y

⇒ 75x + 135 y = 5250

⇒ 5x + 9y = 350 ...(i) 1

From table, 96 + x + y = 150 ...(ii) 1

⇒ x + y = 54

Solving eqns. (i) and (ii), x = 34 and y = 20 1

qqq

WORKSHEET-155

Solutions

1.

Classes Frequency Less than c.f.

0 – 10 4 4

10 – 20 4 8

20 – 30 8 16

30 – 40 10 26


40 – 50 12 38

50 – 60 8 46

60 – 70 4 50

Total N = 50

Here,

N

2 =

50

2 = 25

1

Hence, median class is 30 – 40. 1

2.

Mode Mode = l +

f f

f f f1 0

1 0 22

−− −

× h 1

Here Modal Class = 6 – 9

l1 = 6, f1 = 5, f0 = 4, f2 = 1, h = 3

Mode = 6 +

5 4

10 4 1

-

- - × 3

= 6 + 1

5 × 3

= 6 + 0.6

= 6.6 2

3.

2

Classes f c.f.

5 – 10 2 2

10 – 15 12 14

15 – 20 2 16

20 – 25 4 20

25 – 30 3 23

30 – 35 4 27

35 – 40 3 30

Total ∑f = 30 = N

Since,

N

2 = 15,


Median = l +

Nc f

f2- . .

× h 1

From table, l = 15, N = 30, c.f. = 14, f = 2, h = 5

Median = 15 +

15 14

2

-

× 5

= 15 + 2.5

= 17.5 1

qqq


WORKSHEET-156

Solutions

1.

Age Number of Patients

Less than 20 60

Less than 30 102

Less than 40 157

Less than 50 227

Less than 60 280

Less than 70 300

2.

C. I. f c.f.

0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60

5x

2015y5

5x + 5x + 25x + 40

x + y + 40x + y + 45

Sf = 60

From table N = 60 = x + y + 45

⇒ x + y = 60 – 45 = 15

Since, Median = 28.5 1

Median class = 20 – 30

Median = l +

Nc f

f

2−

. .

× h

⇒

28.5 = 20 +

30 5− +[ ]( )x

20 × 10

⇒

8.5 =

25

2

− x

1

⇒ 25 – x = 17 ⇒ x = 25 – 17 = 8

From (1), y = 15 – 8 = 7 1

3.

1

C.I. f c.f.

0 – 10 5 5

10 – 20 x 5 + x

20 – 30 6 11 + x

30 – 40 y 11 + x + y

40 – 50 6 17 + x + y

50 – 60 5 22 + x + y

Here from table, N = 22 + x + y = 40

⇒ x + y = 18 ...(1)

Since, Median = 31,



Median = l +

Nc f

f2- . .

× h

⇒

31 = 30 +

20 11- ( )+

x

y × 10 1

⇒ 1 =

( )9 10- x

y

×

⇒ y = 90 – 10x

From (i), 10x + y = 90 ...(ii) 1

x + y = 18

– – –

(On subtraction) 9x = 72

⇒ x = 72

9 = 8

From (i), y = 18 – 8 = 10 1

qqq

WORKSHEET-157

Solutions

1.

1

xi fi xifi

1 1 1

3 2 6

5 1 5

7 5 35

9 6 54

11 2 22

13 3 39

Total 20 162

Mean =

Σ

Σ

x f

fi i

i ½

⇒

x� =

162

20= 8.1

\ Mean number of plants per house is 8.1. ½

2.

2

Classes Frequency

0 – 20 17

20 – 40 5

40 – 60 7


60 – 80 8

80 – 100 13

Total 50

2

3.

1

C.I. fi c.f. xi ui =

x - a

h

i fiui

100 – 120 12 12 110 – 2 – 24

120 – 140 14 26 130 – 1 – 14

140 – 160 8 34 150 0 0

160 – 180 16 40 170 1 6

180 – 200 10 50 190 2 20

Sf = 50 Sfiui = – 12

a = assumed mean = 150

Mean x = a+

ΣΣf u

fhi i

i

×

= 150 +

−×

12

5020

= 150 – 4.8 = 145.2

N

2 =

50

2 = 25 1

⇒ Median class = 120 – 140

l = 120, f = 14, c.f. = 12

Median =

lc f

fh

N

+

−

×2. .

=

12025 12

1420+

−

×

= 120 + 18.57 138.57 1

Mode = 3 Medain – 2 Mean

= 3 × 138.57 – 2 × 145.2

= 415.71 – 290.4 = 125.31 1

Hence, mean = 145.2, median = 138.57, mode = 125.31. [CBSE Marking Scheme, 2009]

qqq


TOPIC-2Cumulative Frequency Graph

WORKSHEET-158

Solutions

1. Median. 1

2. Cumulative frequency distribution table (more than type) 1

1

Daily income of workers (in `) Number of workers

1

More than or equal to 200 100







3.

1

Age of student C.I. c.f. f

Less than 6 4 – 6 2 2

Less than 8 6 – 8 6 4

Less than 10 8 – 10 12 6

Less than 12 10 – 12 22 10

Less than 14 12 – 14 42 20

Less than 16 14 – 16 67 25

Less than 18 16 – 18 76 9

N = 78 2

Σf

2 =

76

2

= 38 1


1

80

70

60

50

40

30

20

10

02 4 6 8 10 12 14 16 18 20

3.8

13.6

qqq


WORKSHEET-159

Solutions

1.

1

Students c.f.

Less than 7 20

Less than 9 38

Less than 11 60

Less than 13 85

Less than 15 105

Less than 17 120

Less than 19 130

130

120

110

100

90

80

70

60

50

40

30

20

10

02 4 6 8 10 12 14 16 18 20

(19, 130)

(17, 120)

(15, 105)

(13, 85)

(11, 60)

(9, 38)

(7, 20)

1

This curve is the required cumulative frequency curve or an ogive of the less than type.

Here, N = 130,

So,

N

2 =

130

2 = 65 1

Now, we locate the point on the ogive whose ordinate is 65. The x-co-ordinate corresponding to this ordinate is 11.4.

Therefore, the required median on the graph is 11.4. 1


2.

Apples c.f.

More than 50 160

More than 60 55

More than 70 39

More than 80 29

More than 90 10

More than 100 6

More than 110 2

80

70

60

50

40

30

20

10

0 10 20 30 40 50 60 70 9080 100 110

(50, 60)

(60, 55)

(70, 39)

(80, 29)

(100, 6)(90, 10)

(110, 2)

Cum

ula

tive

Fre

quency

120

Y

X

Number of Apples

This curve is the required cumulative frequency curve on an ogive of the ‘more than type’.

1

Here N = 60,

So

N

2 =

60

2 = 30 1

Now, we locate the point on the ogive whose ordinate is 30. The x-co-ordinate corresponding to this ordinate is 79.

Therefore, the required median on the graph is 79. 1

qqq

WORKSHEET-160

Solutions

1.

2

x y

More than 10 100

More than 20 90

More than 30 75

More than 40 45

More than 50 13

More than 60 5

More than 70 0


‘More than’ ogive is shown below :

100

90

80

70

60

50

40

30

20

10

0x

y

10

Scale :

x-axis 1 cm = 10 units

y-axis 1 cm = 10 units

20 30 40 50 60 70

Marks

(10, 100)

(20, 90)

(30, 75)

(40, 45)

(50, 13)

(60, 5)

(70, 0)

Fre

quency

2.

1

More than c.f.

0 100

10 90

20 72

30 32

40 12

100

90

80

70

60

50

40

30

20

10

0 10 20 30 40x

y

Class

2


From graph,

N

2 =

100

2 = 50

Hence, Median = 25 1

qqq

WORKSHEET-161

Solutions

1.

1

Daily income (Classes) No. of workers (c.f.)

Less than 250 10

Less than 300 15

Less than 350 26

Less than 400 34

Less than 450 40Less than 500 50

50

45

40

35

30

25

20

15

10

5

250 300 350 400 450 500

Class limits

Cu

mu

lati

ve

freq

uen

cy Scale :

-axis 1 cm = 50 units


x

y

y

x

2

From graph,

N

2 =

50

2= 25

Hence, Median daily income = ` 345. 1

2.

2

Less than c.f. More than c.f.

30 10 20 100

40 18 30 90

50 30 40 82

60 54 50 70

70 60 60 46

80 85 70 40

90 100 80 15


100

90

80

70

60

50

40

30

20

10

0 10 20 30 40 50x

y

Cu

mu

lati

ve

freq

uen

cy

From the Graph, Median = 58

60 70 80 90

(20, 100)

(30, 90)

(40, 82)

(50, 70)

(60, 46)

(70, 40)

(80, 15)

(90, 10)

(80, 85)

(70, 60)

(60, 54)

(50, 30)

(40, 18)

(30, 10)

More than ogive

Less than ogive

2

qqq

WORKSHEET-162

Solutions

1.

2

More than or equal to c.f.

0 60

10 55

20 46

30 36

40 24

50 16

60 9

70 4

More than ogive is as shown below :

60

50

40

30

20

10

010 20 30 40 50 60 70

(0, 60)

(10, 55)

(20, 46)

(30, 36)

(40, 24)

(50, 16)

(60, 9)

(70, 4)

Scale :

-axis

-axis

x

y1 cm = 10 unit

Lower Limits

80

y

x

c.f.

2


2.

2

Wages c. f.

More than 80 200

More than 100 180

More than 120 150

More than 140 130

More than 160 90

x

y

Scale :



x

y

Lower Lts

200

190

180

170

160

150

140

130

120

110

100

90

80

70

60

50

40

30

20

10

080 100 120 140 160

More than ogive

(80, 200)

(100, 180)

(120, 150)

(140, 130)

(160, 90)

2

qqq

WORKSHEET-163

Solutions


1. We have,

( 50)xi -i

n

= 1

∑ = – 10 and

( 46)xi -

i

n

= 1

∑

= 70 1

x ni - 50

i

n

= 1

∑

= – 10 ...(i)

and x ni

i

n

- 46

=1

∑

= 70 ...(ii) 1

Subtracting (ii) from (i), – 4 n = – 80 ⇒ n = 20

xii

n

= 1

∑ – 50 × 20 = – 10


⇒

xii

n

= 1

∑

= 990

\

Mean =

1

nxi

i

n

= 1

∑

=

990

20= 49.5

Hence, n = 20 and mean = 49.5 1

2. To prove

i

n

= 1

∑ (xi – x )= 0 / algebraic sum of deviation

from mean is zero

We have, x =

1

nxi

i

n

= 1

∑

n x = xi

i

n

= 1

∑

1

Now, (

=

xii

n

x- )1

∑

= (x1 – x ) + (x2 – x ) + ……… +

(xn – x )

⇒ ( =

xii

n

x- )1

∑ = (x1 + x2 + ……… + xn) – n x

1

⇒

( =

xii

n

x- )1

∑

= xii

n

= 1

∑ – n x

⇒

( =

xii

n

x- )1

∑

= n x

– n x = 0

Hence,

( =

xii

n

x- )1

∑

=

0 1

3. (i)

No. of children

(xi)

No. of

famlies (fi) fi xi

0 5 0

1 11 11

2 25 50

3 12 36

4 5 20

5 2 10

Total ∑fi = 60 ∑fi xi = 127

Mean =

∑

∑

f x

fi i

i

1

=

127

60 = 2.12 approx. 1

(ii) Mean of ungrouped data. 1

(iii) For progress, we should decrease population growth. 1

qqq


WORKSHEET-164

Solutions

1. S = {1, 2, 3, 4, 5, 6}

n(S) = 6

A = {1, 2}

n(A) = 2

\ P(A) =

n A

n S

( )

( ) =

2

6 =

1

3 1

2. Total No. of cases = 200

Favourable cases = 200 – 12 = 188

\ Required probability = 188

200

= 47

50 1

3. S = {1, 2 ............... 80}

n(S) = 80 ½

A = {1, 4, 9, 16, 25, 36, 49, 64}

n(A) = 8 ½

P(A)

=

n A

n S

( )

( ) =

8

80

=

1

10 1

4. No. of possible outcomes = 6 + 5

= 11 balls 1

p(not red) = 11 – 6 = 5

\ = 5

11 1

5.


6. Total number of cards = 48

Probability of an event

=

total number of favourable outcomes

Total number of outcomess 1

Number of cards divisible by 7 = 7

P(cards divisible by 7) =

7

48 1

Number of cards having a perfect square = 6

P(cards having a perfect square) =

6

48

1

8=

1

Number of multiples of 6 from 3 to 50 = 8

P(multiple of 6 from 3 to 50) =

8

48

1

6=

1


qqq

PROBABILITY

SECTION

BSECTIONCHAPTER

15

1

1

1


WORKSHEET-165

Solutions


Number of red cards = 26

Number of queens which are not red = 2

\ Cards which are neither red nor queen

= 52 – [26 + 2] = 24 ½

\ Required Probability = 24

52 =

6

13 ½

2. In the English language there are 26 alphabets. Consonant are 21. The probability of chosen a

consonant =

21

26. [CBSE Marking Scheme, 2015]

1

3. Total cards = 30 Number divisible by 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27,

30 = Total number 1

4. (i) Even numbers occur is (2, 2) (2, 4) (2, 6) (4, 2) (4, 4) (4, 6) (6, 2) (6, 4) (6, 6)

P (number of each die is even) = 9

36 =

1

4

(ii) Sum of numbers is 5 in (1, 4) (2, 3) (3, 2) (4, 1)

P (sum of numbers appearing on two dice is 5)

=

4

36 =

1

9.

1

5. Total cards = 49

(i) P(odd number) = 25

49 1

(ii) P(multiple of 5) = 9

49 1

(iii) P(even prime) = 1

49 1

6. (i) Favourable outcomes are 1, 3, 5, 7 i.e. outcomes. 1

\ P(an odd number) = 4

8

1

2or ½

(ii) Favourable outcomes are 4, 5, 6, 7, 8 i.e. 5 outcomes 1

P(a number greater than 3) = 4

8

1

2or ½

(iii) Favourable outcomes are 1, 2, 3...8

P(a number less than 9) = 8

8 = 1 1


qqq

WORKSHEET-166

Solutions

1. Possible outcomes are 4, 9, 16, 25, 36, 49, i.e. 6. ½

⇒ P(perfect square number) =

6

48

1

8 or

½

2. Total number of cases = 20 ⇒ n(s) = 20 A = favourable cases = {3, 6, 7, 9, 12, 14, 15, 18} \ n(A) = 8 ½

\ Required probability = P(A) = n A

n S

( )

( )= =8

20

2

5

½

3. (i)

5

26

1

(ii)

21


4. Possible outcomes as HH, TT, TH, HT

(i) P(E1) = 3

4 1

(ii) P(E2) = 2

4

1

2=

1


5. (i) 3

4 S = {HH, HT, TH, TT} (optional) 1

(ii)

3

4 1

(iii)

1


6. Given, x = {1, 2, 3, 4}

⇒ n(x) = 4 ½

y = {1, 4, 9, 16}

n(y) = 4 ½

Total number of possible products = 4 × 4 = 16. ½

Products x.y which are less than 16 are {1 × 1, 1 × 4, 1 × 9, 2 × 1, 2 × 4, 3 × 1, 3 × 4, 4 × 1} 1

n (x.y) = 8 ½

\ Required probability = 8

16

1

2= . 1

qqq


WORKSHEET-167

Solutions

1. There are 365 days in a non-leap year. Q 365days = 52 weeks + 1 day \One day can be M, T, W, Th, F, S, S =7 ½

\P(53 Mondays in non-leap year) =

1

7 ½


2. Product of 6 are (1, 6); (2, 3); (6, 1); (3, 2) No. of possible out comes = 4

Total number of chances = 6 × 6 = 36

P (Product of 6) =

4

36

1

9=


3. Total balls = 5 + 8 + 7 = 20

(i) P(white ball) =

7

20

P(not white) = 1

7

20

13

20- =

1

(ii) P(green or red) =

8 5

20

13

20

+=

P(neither green nor red) = 1

13

20-

=

7

20. 1


4. Sample space for three coins tossed is {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

⇒ n(s) = 8

(i) Exactly two heads = {HHT, HTH, THH}

n(P1) = 3

\ P1 = n P

n s

( )

( )1 3

8= 1

(ii) At least two heads {HHT, HTH, THH, HHH}

n(P2) = 4

P2 = n P

n s

( )

( )2 4

8

1

2= = 1

(iii) At least two tails {TTH, THT, HTT, TTT}

n(P3) = 4

P3 = n P

n s

( )

( )3 4

8

1

2= = 1

5. (i) No. of cards remaining = 52 – 3 × 2

= 52 – 6 = 46

No. of red cards = 26 – 6 = 20 1

P (a red colour) = 20

46 =

10

23

(ii) No. of queen = 4 – 2 = 2

P (a queen) = 2

46 =

1

23 1

(iii) No. of ace = 4

P (as ace) = 4

46 =

2

23 1

(iv) No. of face cards = 12 – 6 = 6

P (a face card) = 6

46 =

3

23. 1

qqq

WORKSHEET-168

Solutions

1. Total number of points = 8

Total number of possible outcomes

= (1 × 8), (2 × 4), (8 × 1), (4 × 2)

\P(Factor of 8) =

No. of favourable outcomes

Total no. of possible outcomes ½

= 4

8=1

2 ½


2. QProbability of winning the game

P(E) =

5

11 ½ \Probability of losing the game

P( E ) = 1 – P(E)

= 1–

5

11 =

6

11 ½

3. When two dice are thrown

Possible outcomes = 36 1

If sum of both faces should be 10, they are,

{(4, 6), (6, 4), (5, 5)} = 3

\ P(E) = 3

36

1

12=

1

4. Total number of candles = 3 + 4 +5 = 12 ½

P(candle is red) =

3

12

1

4=

½


P(candle is not red) = 1 – P (candle is red)

= 11

4

4 1

4

3

4− =

−

=

1


5. S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n (s) = 8 Same result on all the tosses (A) = {HHH, TTT}, 1 n (A) = 2 1

P (Ramesh will lose the game) = 8 2

8

- =

6

8

3

4=

1


6.


qqq

WORKSHEET-169

Solutions

1. Q P(E) =

3

7

\ P(not E) = 1 – P(E)

= 1 –

3

7

=

4

7 1

2. P(impossible event) =

1

∞ = 0 1


3. Sample space S = GG, GB, BG, BB (optional) 1

P(atleast one girl) =

3

4 1


1

1

1

1


4. 366 days = 52 weeks + 2 days ½

2 days can be MT, TW, WTh, ThF, FS, SS, SM = 7 ½

⇒ P = 2

7 1


5. (i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) i.e. 9 outcomes. 1

P (a prime number on each die) = 9

36

1

4or

½

(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6)

(6, 5) i.e. 6 outcomes 1

P (a total of 9 or 11) = 6

36

1

6or

½


6. Since all the black face cards are removed, the total number of remaining cards = 46

(i) P(face card) = 6

46

3

23=

1

(ii) P(red card) = 26

46

13

23=

1

(iii) P(black card) = 20

46

10

23=

1

(iv) P(king card) = 2

46

1

23=

1

qqq

WORKSHEET-170

Solutions

1. The numbers divisible by 2 and 3 both = 6, 12, 18, 24, = 4

\ P(number divisible by 2 and 3) =

4

25 1

2. Prime numbers are = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, = 10 ½

No. of possible outcomes = 30

P(prime no.) =

10

30

1

3=

½

3. Total number of outcomes = 36 ½

Favourable outcomes are (2, 6), (3, 5),

(4, 4), (5, 3), (6, 2) = 5 1

\ Required probability =

5

36 ½


4. Let blue balls = x and red balls = 5

\ Total balls = 5 + x

P(red ball) = 5

5 + x

P(blue ball) = x

x5 + 1

\

x

x5 + =

3

5

5 + x ⇒ x = 15 1


5. (i) Number of good shirts = 88 1

P (Ramesh buys the shirt) = 88

100

22

25or

½

(ii) Number of shirts without Major defect = 96 1

P (Kewal buys a shirt) = 96

100

24

25or ½


6. No. of possible outcomes = 20 1 (i) Total no. divisible by 2 or 3 = 6, 12, 18 =3

P(divisible by 2 or 3) =

3

20 1½

(ii) Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19 = 8

P(a prime no.) =

8

20

2

5=

1½


qqq

WORKSHEET-171

Solutions

1. No. of possible outcomes = 90

Prime numbers less than 23 = 2, 3, 5, 7, 11, 13, 17, 19

= 8 ½

P(prime no. less than 23) =

8

90

4

45=

½

2. P(E) = 0.20

\ P(not E) = 1 – P(E)


= 1 – 0.20

= 0.80 1


3. Possibilities are HH, HT, TH, TT 1

P(HH or TT) = 2

4

1

2=

1


4. Face cards = 12

(i) P(non-faces) = 40

52

10

13=

1

(ii) P(black king) = 2

52

1

2=

6 1


5. (i) P (single digit number) =

9

100 1

(ii) P (perfect square) =

1

10 1

(iii) P (a number which is divisible by 7) =

7

50 1


(i) P (one digit number) =

4

65 1

(ii) No. divisible by 5 = 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70

= 13

P (a number divisible by 5) =

13

65

1

5=

1

(iii) Odd no. less than 30 = 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

= 12

P (an odd number less than 30) =

12

65 1

(iv) P (a composite number between 50 and 70)

=

15

65

3

13=

1


qqq

WORKSHEET-172

Solutions

1. Total outcomes = 6 Prime numbers = 2, 3, 5, = 3

P(prime no.) =

3

6

1

2=

1


2. Total number of possible outcomes = 62 = 36 ½

E : (doublets are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Outcomes favourable to E = 6 ½

\P(a doublet)

= Number of outcomes favourable to E

Total number of outcomes

=

6

36

1

6=

1


3. Total no. of pens = 144

Defective one = 20

Good ones = 144 – 20 = 124

Probability of purchasing pen

= 124

144

31

36=

1

Probability of not purchasing pen

= 20

144

5

36=

1


4. Sample space = {1, 2, 3 ... 99, 100} n(s) = 100 ½

(i) Number divisible by and perfect square are A = {9, 36, 81} n(A) = 3 ½

\ Required probability P(A) = n B

n s

( )

( )=

3

100

1

(ii) Prime numbers greater than 80 and less than 100 are B = {83, 89, 97}

\ n(B) = 3

\ Required probability P(B) = n B

n s

( )

( )=

3

100 1

5. (i) Cards of spade or an ace = 13 + 3 = 16

Total no. of cards = 52

P (spade or an ace) =

16

52

4

13=

1

(ii) Black kings = 2

P (a black king) =

2

52

1

26=

1


(iii) Jack or king = 4 + 4 = 8

P (neither jack nor a king) =

52 8

52

44

52

11

13

-= =

1

(iv) King or queen = 4 + 4 = 8 1

P (either a king or a queen) =

8

52

2

13=


qqq

WORKSHEET-173

Solutions

1. P(winning the game) = 0.08

P(losing the game) = 1 – 0.08 = 0.92 1


2. Here, P(bad eggs) = 0.035

P(bad eggs) = No. of bad eggs

Total no. of eggs

0.035 =

No. of bad eggs

400

\ Noof bad eggs = 400 × 0.035

=14 1


3. Let the number of blue balls = x

Total number of balls = x + 5

Number of red balls = 5 1

\

x

x + 5 = 2

5

5x +

⇒ x = 10. 1


4. In a leap year number of days

= 366 days

= 52 weeks + 2 days

\ Two days can be SM, MT, TW, WTh, ThF, FS, SS = 7 1

\ Out of these 7 calenders, two calenders will have 53 Wednesdays

\ P(53 Wednesdays in a leap year) =

2

7 1


5. Perfect squares are 4, 9, 16, 25, 36, 49, 64, 81, 100.

(i) P(Perfect square) = 9

100 1½

(ii) P(odd number not less than 70) = 16

100

4

25=

1½



20

20

− x

= 2

15

15

−

x

½

⇒

120

-x

=

2

2

15-

x

⇒

2

15 20

x x−

= 2 –1

⇒

8 3

60

x x-

= 1

⇒ 5x = 60

\ x = 12 ½

\ Blue ball = 12 and red ball = 3

(i) P(red ball) =

3

15

1

5=

1

(ii) P(blue ball) =

12

15

4

5=

1

(iii) P(blue ball if 5 red balls are added) =

12

20

3

5=

1


qqq

WORKSHEET-174

Solutions

1. Q Bag contains only lemon flavoured candies. \ P(orange flavoured candies) = 0 1


2. Odd numbers = 1, 3, 5, Numbers less than 4 = 1, 2, 3,

\ P(an odd no. or a no. <4) =

4

6

2

3=

1



3. Since the red face cards are removed,

\ No. of card = 52 – 6 = 46

(i) P(a red card) = 20

46

10

23=

1

(ii) P(a face card) =

6

46

3

23=

1

(iii) P(a card of clubs) = 13

46 1


4. Total possible outcomes : 36

(i) The possible outcomes are (2, 3); (3, 2); (1, 4); (4, 1)

= 4 1

\ Required probability P(E) =

4

36 =

1

9 ½

(ii) The possible outcomes are :

(2, 2); (2, 4); (2, 6); (4, 2); (4, 4); (4, 6); (6, 2); (6, 4); (6, 6) 9 1

\ Required probability P(E1) =

9

36

1

4=

½


5. Total number of three digit numbers are : 459, 495, 549, 594, 945, 954 = 6 1

(i) P(multiple of 5) = 2

6

1

3=

1

(ii) P(multiple of 9) = 6

61=

1

(iii) P(ending with 9) = 2

6

1

3=

1

6.


qqq

WORKSHEET-175

Solutions

HOTS & Value Based Answes

1. Favourable outcomes = 5

[|–2|, |–1|, |0|, |1|, |2|] < 3 1

Total outcomes = 9 1

P( |x| < 3) =

5

9 1


2. (i) P(ball not red) =

1

20

20

20-

-x x or

1

(ii) Total number of balls = 24, red balls = x + 4

P (red ball) =

x + 4

24 ½


1

2

1


=

x + 4

24 =

5

4 20×

x

1

x = 8 ½


3. For

a

b >

1, when a = 1, b can not take any value, a

= 2, b can take 1 value, a = 3, b can take 2 values, a = 4, b can take 3 values 2½

When a = 5, b can take 4 values, a = 6, b can take 5 values.

Total Possible outcomes = 36 ½

\

P

a

b>

1 =

1 2 3 4 5

36

+ + + +

=

15

36 or

5

12


4. (i) Total cards = 25

Number divisible by 3 = 3, 6, 9, 12, 15, 16, 21, 24,

Number divisible by 5 = 5, 10, 15, 20, 25, 1

Number divisible by 3 or 5 = 12 1

\P(no. divisible by 3 or 5) =

12

25

(ii) Perfect square = 1, 4, 9, 16, 25, 1

P(a perfect square no.) =

5

25=1

5 1


qqq

WORKSHEET-176

Solutions

HOTS & Value Based Answes 1. When a dice is rolled twice, the total outcomes = 36 1

\ (i) P(5 will not come up either time) =

25

36 (ii) [(1, 5), (2, 5) (3, 5) (4, 5) (5, 1) (5, 2) (5, 3) (5, 4) (5, 6)

(6, 5)]

P(5 will come up exactly one time) =

10

36

5

18=

3

2. Number of red balls = 6 Let number of blue balls = x

Total number of balls = 6 + x

P(red ball) =

6

6 + x ½

P(blue ball) =

x

x6 + 1


x

x6 + =

2 6

6

×+ x

½

⇒ x = 12 \ Number of blue balls = 12. 1 [CBSE Marking Scheme, 2012]

3. Remaining cards = 52 – 3 = 49 ½

(i) P(heart) = 13

49 ½

(ii) P(king) = 3

49 ½

(iii) P(club) = 10

49 ½

(iv) P(diamond) = 10

49 ½

(v) P(Jack) = 3

49 ½


4. (i) Total number of coins = 100 + 50 + 20 + 10

= 180 ½

\ Total number of possible outcomes of a coin will fall out = 180 ½

Number of 50 p coins = 100

\ Number of favourable outcomes relating to fall out of a 50 p coin = 100

Now, P(of getting a 50 p coin)

=

Number of favourable outcomes

Total number of possible outcoomes=

100

180

5

9=

1

(ii) P(not a ` 5 coin) = 1 – P(` 5 coin) =

1

10

180

17

18- =

½

(iii) Probability ½

(iv) Saving money for future. 1

5. (i) P (extremely patient) =

3

12

1

4=

1

(ii) P (extremely kind or honest) =

6 3

12

9

12

3

4

+= =

1

(iii) Extremely Honest. 1

qqq

b1 real numbers topic-1 euclid s division lemma … s division lemma and fundamental theorem of...

Documents