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Surender VermaM.Sc. (Mathematics), B.Ed.

Delhi Public School,Dwarka, New Delhi

By

1. Quadratic Equations

� Worksheets (1 to 20) ................................................................................................ 135

• Assessment Sheets (1 and 2) .................................................................................. 157

• Chapter Test ............................................................................................................ 160

2. Arithmetic Progressions

� Worksheets (24 to 37) .............................................................................................. 162


• Chapter Test ............................................................................................................ 178

3. Circles

� Worksheets (41 to 46) .............................................................................................. 180


• Chapter Test ............................................................................................................ 189

4. Constructions

� Worksheets (50 to 54) .............................................................................................. 191


• Chapter Test ............................................................................................................ 209

Second Term

– 3 –

5. Some Applications of Trigonometry

� Worksheets (57 to 62) ...............................................................................................211

• Assessment Sheets (9 and 10) ................................................................................ 219

• Chapter Test ............................................................................................................ 223

6. Probability

� Worksheets (66 to 68) .............................................................................................. 225

• Assessment Sheets (11 and 12) ............................................................................... 228

• Chapter Test ............................................................................................................ 231

7. Coordinate Geometry

� Worksheets (72 to 82) .............................................................................................. 233

• Assessment Sheets (13 and 14) .............................................................................. 247

• Chapter Test ............................................................................................................ 250

8. Areas Related to Circles

� Worksheets (85 to 89) .............................................................................................. 252


• Chapter Test ............................................................................................................ 260

9. Surface Areas and Volumes

� Worksheets (93 to 98) .............................................................................................. 262


• Chapter Test ............................................................................................................ 273

PRACTICE PAPERS (1 to 5) ................................................................................... 275

– 4 –

Solutions toSolutions toSolutions toSolutions toSolutions toPULLOUT WORKSHEETSPULLOUT WORKSHEETSPULLOUT WORKSHEETSPULLOUT WORKSHEETSPULLOUT WORKSHEETS

ANDANDANDANDANDPRACTICE PAPERSPRACTICE PAPERSPRACTICE PAPERSPRACTICE PAPERSPRACTICE PAPERS

[Summative Assessments][Summative Assessments][Summative Assessments][Summative Assessments][Summative Assessments][SECOND TERM]

135DAUQ AR CIT AUQE IT SNO

1Chapter

QUADRATIC EQUATIONS

WORKSHEET–11. (B) For real and distinct roots, D > 0

⇒ b2 – 4ac > 0 ⇒ 16 – 4p > 0⇒ 4 > p ∴ p < 4.

2. (C) p(1)2 + p(1) + 3 = 0

⇒ 2p = – 3 ⇒ p = – 32

and (1)2 + 1 + q = 0 ⇒ q = – 2

∴ pq = − 32

(– 2) = 3.

3. (A)−

+ − −−

2( 5)2( 5 )

5k = 0

⇒ –10 – 2k = 5⇒ – 2k = 15

⇒ k =152

−.

4. Given equation: 4x2 – 3x – 5 = 0

Divide throughout by coefficient of x2, i.e.,4 to get

x2 – 3 54 4

x− = 0

12

of coefficient of x =3

8−

and square of it = 364

Therefore, required constant is 3

64.

5. TrueReason: The value of t for which givenequation has real and equal roots are± 2 21 which are irrational.

6. 22 + 7 + 5 2x x = 0

⇒ 22 + 5 + 2 5 2x x x + = 0

⇒ ( ) ( )2 5 2 2 5x x x+ ++ = 0

⇒ ( ) ( )2 2 5x x ++ = 0

⇒ + 2x = 0 or +2 5x = 0

⇒ 2x = − or 5

2x

−= .

7. − +22 5 3x x = 0

⇒ 2 5 32 2

x x− + = 0

⇒ −2 52

x x = − 32

Add both side

254

⇒2

2 5 52 4

x x − + =

23 52 4

− +

⇒ −

254

x =− +24 25

16 =

116

⇒ − 54

x = ± 14

⇒ = +5 14 4

x or −5 14 4

⇒ x =32

or 1.

8. x = 2 or 1Hint: The given eqn. can be written as:

( )( )− − −+ −

7 44 7

x xx x

=1130

⇒ x2 − 3x − 28 = − 30⇒ x2 − 3x + 2 = 0.

9. Let speed of stream = x km/h∴ upstream speed = (18 − x) km/h

downstream speed = (18 + x) km/hTime taken to cover upstream distance of

24 km = 24

18 x−

136 AM T H E M A T C SI X–

Time taken to cover downstream distance of

24 km = 2418 x+

According to question,

24 2418 18x x

−− +

= 1

⇒ 24+ − +

− 218 18

324x x

x= 1

⇒ 48x = 324 − x2

⇒ x2 + 48x − 324 = 0x2 + 54x − 6x − 324 = 0

⇒ x(x + 54) − 6(x + 54) = 0⇒ (x − 6) (x + 54) = 0⇒ x = 6 or x = − 54 (Reject)∴ x = 6 km/h.

ORLet size of square be x∴ No. of students in square = x2

∴ According to question,Case I. Total students = x2 + 24Case II. Also total students = (x + 1)2 – 25∴ x2 + 24 = (x + 1)2 – 25⇒ x2 + 24 = x2 + 2x + 1 – 25⇒ 2x = 48 ⇒ x = 24Number of students = (24)2 + 24

= 576 + 24 = 600.

WORKSHEET – 2

1. (A) D = 2( 2 2)− – 4 × 2 × 1 = 8 – 8 = 0.

2. (C) Two equal and real roots⇒ D = 0⇒ k2 – 4 × 2 × 3 = 0⇒ k2 = 4 × 6

⇒ k = ± 2 6 .

3. (A) x = a must satisfy the given equation⇒ a2 – (a + b)a + k = 0 ⇒ k = ab.

4. For real and distinct roots,D > 0

⇒ b2 − 4ac > 0⇒ 16 − 4k > 0⇒ 16 > 4k ⇒ k < 4.

5. Yes.

∵ (x – 2 )(x + 3 ) = 0

⇒ x2 + ( 3 2− ) x – 6 = 0.

6. x = 3 19

5±

Hint: 5x2 − 6x − 2 = 0

⇒ x2 − 65

x = 25

Add both side

235

and use

a2 − 2ab + b2 = (a − b)2.

7. D = 0; real and equal roots, x = 13

or 13

Hint: Discriminant: D = b2 − 4ac.

8. Let y = +2 1

xx

∴ Given equation becomes

1y

y+ =

2910

⇒+2 1y

y=

2910

⇒ 10y2 − 29y + 10 = 0⇒ 10y2 − 25y − 4y + 10 = 0⇒ 5y(2y − 5) − 2(2y − 5) = 0⇒ (5y − 2) (2y − 5) = 0

⇒ y = 25

or y =52

If y = 25

, then 2 1

xx +

=25

gives x = 2

If y = 52

, then +2 1

xx

=52

gives x = − 58

.

9. Yes, length = 40 m and breadth = 20 mHint: The situation will be possible if D ≥ 0.

ORLet shorter side = x m

∴ longer side = 30 + x

∴ diagonal = + +2 2(30 )x x



+ +2 2(30 )x x = x + 60

Squaring both sidesx2 + (30 + x)2 = (x + 60)2

⇒ x2 + 900 + x2 + 60x = x2 + 3600 + 120x⇒ x2 − 60x − 2700 = 0⇒ x2 − 90x + 30x − 2700 = 0⇒ x(x − 90) + 30 (x − 90) = 0⇒ (x + 30) (x − 90) = 0⇒ x = − 30 (Rejected)

or x = 90∴ Shorter side = 90 m

Longer side = 120 m.

WORKSHEET– 31. (A) For two distinct real roots,

D > 0 ⇒ b2 – 4ac > 0 ⇒ b2 > 4ac.2. (C) Given equation is:

2 − 5x + 2x2 = 0⇒ 2x2 − 5x + 2 = 0

D = (5)2 − 4 (2)(2)D = 9.

3. (D) Since 23

and −3 are roots of equation

Sum of roots =23

− 3 =7

m−

⇒ m = 3

Product of roots =23

(− 3) = nm

⇒ n = − 6.

4. No, as given equation is:x2 + x + 8 = x2 − 4

⇒ x − 12 = 0∴ It is a linear equation.

5. False.

∵ x2 – 3x + 1 = 0is an equation with integral coefficients butits roots are not integers.

6. x = 53

, 2

Hint: See Worksheet −−−−− 1, Sol. 7 for comple-ting square method.

7. p = 7; k = 74

Hint: Put x = – 5 in first equation, whichgives p = 7And p(x2 + x) + k = 0 to have equal roots:

D = 0⇒ (p)2 − 4 (p)(k) = 0⇒ (7)2 − 28k = 0

⇒ k =74

.

8. x = 0; 2(a + b)Hint: Given equation is:

x2 – x (2a + 2b) + 4ab = 4ab⇒ x2 – 2x (a + b) = 0.

9. 10, 5Hint: Quadratic equation is:

1x

+ −1

15 x=

310

⇒− 2

1515x x

=3

10

⇒ 150 = 45x − 3x2

⇒ 3x2 − 45x + 150 = 0⇒ x2 − 15x + 50 = 0.

OR

Let Ist part = x (larger)∴ 2nd part = 16 − x (smaller)According to question,

(16 − x)2 + 164 = 2x2

256 + x2 – 32x + 164 = 2x2

⇒ x2 + 32x − 420 = 0⇒ x2 + 42x − 10x − 420 = 0⇒ x(x + 42) −10 (x + 42) = 0⇒ (x − 10) (x + 42) = 0⇒ x = 10 or x = – 42 (Reject)∴ Larger part = 10

Smaller part = 6.


WORKSHEET– 41. (D) x = 1 must satisfy both the equations

⇒ a + a + 3 = 0 and 1 + 1 + b = 0

⇒ a =3

–2

and b = – 2

⇒ ab = 3.

2. (B) For real and equal roots, D = 0⇒ (4k)2 – 4.1. (k2 – k + 2) = 0⇒ 12k2 + 4k – 8 = 0⇒ 3k2 + k – 2 = 0⇒ (3k – 2)(k + 1) = 0⇒ 3k – 2 = 0 or k + 1 = 0

⇒ k =23

or – 1.

3. (A) Putting x =23

in the given equation,

we get

322 2

3 3p +

+ 4 = 0

⇒ 4 + 2p + 12 = 0⇒ p = – 8.

4. Yes. As 16x2 − 24x − 1 = 0

⇒ x =D

2

b

a

− ±

=224 (24) + 4×16

2×16±

=24 576 + 64

32±

=24 640

32±

=3 10

4±

.

5. 1 334

− ±

Hint: Divide the equation by 2 and then

add 14

on both sides of equation.

6. a = 3, b = – 6

Hint: Use sum of roots =2

coefficient of–

coefficient of

x

x

Product of roots = 2

constant term

coefficient of x

7. k = 14Hint: For real and equal roots:

D = 0⇒ b2 − 4ac = 0Use a = k − 12, b = 2(k − 12), c = 2.

8. x = 9 57

4±

Hint: Given equation can be written as2x2 – 9x + 3 = 0∴ Quadratic formula is

x =2 4

2b b ac

a− ± −

.

9. Let unit's digit = xand ten's digit = y∴ Original number = 10y + xNow xy = 16

⇒ y =16x ...(i)

Also, according to question⇒ 10y + x − 54 = 10x + y⇒ 9y − 9x = 54⇒ y − x = 6 ...(ii)

Substituting y =16x

from (i) to (ii), we get

16x

− x = 6

⇒ 16 − x2 = 6x⇒ x2 + 6x − 16 = 0⇒ x2 + 8x − 2x − 16 = 0⇒ (x + 8) (x − 2) = 0⇒ x = 2 or x = − 8 (Reject)

∴ y =162

= 8

∴ Number is 82.


OR40 km/hHint: Let speed of train = x km/h

∴ 360 3605x x

−+

= 1

WORKSHEET– 5

1. (A) For real and equal roots,

D = 0 ⇒ 9k2 – 4 × 4 × 1 = 0 ⇒ k = ± 43

.

2. (D) 2(x2 – x) = 3 ⇒ 2x2 – 2x – 3 = 0Here, D = 4 + 4 × 2 × 3 = 28 > 0.⇒ So roots are real and distinct.

3. (A) Given equation is 2x2 – 14x – 1 = 0∴ D = b2 – 4ac

D = (– 14)2 – 4(2) (–1) = 196 + 8 = 204

∴ x = 14 204

4±

= 7 51

2±

.

4. No.We have, x2 – 2x + 8 = 0Put x = – 2, we get⇒ (– 2)2 – 2(– 2) + 8 = 0

16 = 0 which is wrong.Therefore, x = – 2 does not satisfy givenequation.

5. True.D = b2 – 4ac

Put b = 0 and a = 1D = – 4c

As c < 0 ⇒ – 4c > 0 ⇒ D > 0⇒ Roots are real

Also, sum of roots = – ba = 0

⇒ Roots are numerically equal andopposite in sign.

6. Quadratic equation written as:

(2x)2 + 2 × (2x) ×34

+2 23 3

4 4 −

+ 5 = 0

⇒23

24

x + – 9

16 + 5 = 0

⇒23

24

x + =71

16−

< 0

But23

24

x + cannot be negative for any

real value of x. So there is no real value ofx satisfying the given equation. Therefore,the given equation has no real roots.

7. Let another root be α.

Product of roots = 2α =6

2−

⇒ α =32

−

Sum of roots =2k− ⇒ 3

2− + 2 =

2k

−

⇒ k = – 1

Thus, k = – 1 and another root = 32

− .

8. Let y =−−1

2 1xx

∴ Given equation can be written as:

⇒ y +1y

=52

⇒ 2y2 + 2 = 5y⇒ 2y2 – 5y + 2 = 0⇒ 2y2 – 4y – y + 2 = 0⇒ 2y(y – 2) – 1 (y – 2) = 0⇒ (2y – 1) (y – 2) = 0

⇒ y = 12

or y = 2

⇒ −−1

2 1xx

= 12

or −−1

2 1xx

= 2

⇒ 2x − 2 = 2x − 1 Not possibleor x − 1 = 4x − 2

⇒ 1 = 3x

⇒ x = 13

.

9. 15 hrs or 25 hrsHint: Let smaller tap takes x hrs to fill thetank itself.


∴ Larger tap will take (x − 10) hrs to fillthe tank itself.

∴ Given situation can be expressed as:

−110x

+1x

= 875

⇒ 4x2 − 115x + 375 = 0

⇒ x = 154

(cannot be taken) or 25.

ORLet one number be x so another number be15 – x.According to question,

1 115 –x x

+ =3

10

⇒15 –

(15 – )x xx x

+=

310

3x(15 – x) = 150⇒ 15x – x2 = 50⇒ x2 – 15x + 50 = 0⇒ x2 – 10x – 5x + 50 = 0⇒ x(x – 10) – 5(x – 10) = 0⇒ (x – 5)(x – 10) = 0⇒ x = 5 or x = 10When x = 5, 15 – x = 10;when x = 10, 15 – x = 5Hence, the numbers are 5 and 10.

WORKSHEET– 61. (A) x2 – 4x + p = 0

For real roots, D ≥ 0⇒ (– 4)2 – 4 × 1 × p ≥ 0⇒ 16 – 4p ≥ 0 ⇒ 4p ≤ 16 ⇒ p ≤ 4.

2. (C) For equal roots, D = 0⇒ (6k)2 – 4 × 9 × 4 = 0⇒ 36k2 = 4 × 36 ⇒ k = ± 2.

3. (B) D = ( )24 3 – 4 × 4 × 3 = 48 – 48 = 0

Since, the discriminant is zero, therefore,the given equation has real and equal roots.

4. – 5 must satisfy 2x2 + px – 15 = 0,i.e., 2 × 25 – 5p – 15 = 0 ⇒ p = 7

As p(x2 + x) + k = 0, i.e., px2 + px + k = 0 hasequal roots, D = 0 ⇒ p2 – 4pk = 0But p = 7, ∴ (7)2 – 4(7) k = 0.

⇒ 4 × 7k = 7 × 7 ⇒ k = 74

.

5. − − +23 6 6 2x x x = 0

⇒ ( ) ( )3 3 2 2 3 2x x x− − − = 0

⇒ ( ) ( )− −3 2 3 2x x = 0

⇒ −3 2x = 0 or −3 2x = 0

⇒ x =2 2

;3 3

⇒ x =6 6

;3 3

.

6. + +22 2 15 2x x = 0

⇒ x2 + 15 1

22 2x + = 0

⇒ x2 + 15

2 2x =

12

− .

Add 215

4 2

to both sides,

⇒ 2

2 15 15

2 2 4 2x x

+ +

= −

+

21 152 4 2

⇒

+

215

4 2x = − +1 15

2 32

=− +16 15

32

∴

+

215

4 2x =

132

−

which is not possible as square of any realnumber can't be negative.∴ No real roots possible.

7. x = 12

or 43

Hint: Let y =−−

2 31

xx

∴ 1y

= −−1

2 3xx

Given equation becomes: y − 4y

= 3

Now solve.


8. Let the required number has x as ten’s digitof the number.

Given: Product of the digit = 8

∴ Unit’s digit = 8x

∴ Number = 10x + 8x

If 63 is subtracted from the number thedigit interchange their places.

∴ 10x + 8x

– 63 = 10 × 8x

+ x

⇒ 10x + 8x

– 63 =80x

+ x

⇒ 9x – 72x

– 63 = 0

⇒ 9x2 – 63x – 72 = 0⇒ x2 – 7x – 8 = 0⇒ (x + 1)(x – 8) = 0⇒ x + 1 = 0, x – 8 = 0∴ x = – 1, x = 8Reject x = – 1 ∴ x = 8

∴ Required number = 10 × 8 + 88

= 81.

9. 60 km/hr

Hint: Let speed of express train = x km/h∴ Speed of another train = (x − 12) km/h ∴ According to question,

240 24012x x

−−

= 1

Now solve.

WORKSHEET–71. (D) x(px + 6) = – 1 ⇒ px2 + 6x + 1 = 0

For real and distinct roots, D > 0⇒ 62 – 4 × p × 1 > 0 ⇒ p < 9.

2. (C) D = (– 2)2 – 4 4 33 4

= 4 – 4 = 0.

3. (D) For equal roots, D = 0⇒ k2 – 4 × 4 × 9 = 0 ⇒ k = ± 12.

4. For no real roots, D < 0∴ (5m)2 – 4 × 1 × 16 < 0

⇒ 25m2 < 64 ⇒ m2 < 6425

⇒ m2 < 28

5

⇒ – 85

< m < 85

.

5. 23 9 2 6 3x x x+ + + = 0

⇒ ( ) ( )3 3 3 2 3 3x x x+ + + = 0

⇒ ( ) ( )+ +3 2 3 3x x = 0

⇒ x =−2

3 or x = −3 3 .

6. 5 x2 + 9x + 4 5 = 0

Dividing both sides by 5 to get

x2 +95

x + 4 = 0

Adding 29

2 5

to both sides to get

x2 + 95

x + 4 + 2 29 9

2 5 2 5 =

⇒ x2 +95

x +29 81

202 5 =

– 4

⇒ 2 29 1

2 5 2 5x + =

⇒ 2 29 1

2 5 2 5x + −

= 0

⇒ 9 1 9 12 5 2 5 2 5 2 5

x x + + + − = 0

⇒ x = 5− , – 45

.

7. 13, 15Hint: Two consecutive odd numbers are oftype 2x + 1, 2x + 3∴ According to question,(2x + 1)2 + (2x + 3)2 = 394.Now solve.

8. Let y = + 1x

x


∴ Given equation becomes 2y2 − 5y + 2 = 2∴ D = b2 − 4ac⇒ D = 25 − 4 (2) (2) = 25 − 16 = 9

∴ y = D

2b

a− ±

⇒ y = ±5 94

= ±5 34

= 2, 12

∴ =+

21

xx

or =+

11 2

xx

⇒ x = 2x + 2 or 2x = x + 1⇒ x = − 2 or x = 1.

9. Let the present ages (in years) of fatherand son be x and y respectively.∴ x = y2 ... (i)1 year ago, father’s age = (x – 1) years andson’s age = (y – 1) years.According to the question,

x – 1 = 8(y – 1)⇒ x – 8y + 7 = 0 ... (ii)Substituting the value of x from equation (i)in equation (ii), we have

y2 – 8y + 7 = 0(y – 1)(y – 7) = 0

⇒ y = 1 or 7Case I. If y =1 year1 year ago, son’s age = 1 – 1 = 0So, 1 year ago the father could not 8 timesas old as his son.Case II. If y = 7 years1 year ago, son’s age = 7 – 1 = 6 yearsSo, 1 year ago, the father was 8 × 6 = 48years old.∴ x = y2 = 72 = 49Hence, present ages of the father and theson are 49 years and 7 years.

ORLet number of books purchased = x

∴ Cost of 1 book =80x


80x

− +80

4x= 1

⇒ 80 + − +

4( 4)

x xx x

= 1

⇒ 320 = x2 + 4x⇒ x2 + 4x − 320 = 0⇒ x2 + 20x − 16x − 320 = 0⇒ x (x + 20) − 16 (x + 20) = 0⇒ (x − 16) (x + 20) = 0⇒ x = 16 or x = − 20 (Reject)∴ Number of books = 16.

WORKSHEET – 8

1. (C)Hint: Discriminant is: D = b2 − 4ac.

2. (B) For equal and real roots, D = 0.

∴ (– 5)2 – 4.k.k = 0 ⇒ 4k2 = 25 ⇒ k = ± 52

.

3. (D) – 4 must satisfy x2 + px – 4 = 0⇒ p = 3 ... (i)

x2 + px + q = 0 has equal roots⇒ p2 – 4q = 0 ... (ii)

From (i) and (ii), we have p = 3, q = 94

.

4. No.

LHS = − +24(3) 14(3) 16

= − +36 42 16 = 10

RHS = − +9 12 3 + −9 9 = 0

As LHS ≠ RHS ⇒ x = 3 is not a root.

5. − − +25 5 4 4 5z z z = 0

⇒ ( ) ( )− − −5 5 4 5z z z = 0

⇒ ( ) ( )− −5 4 5z z = 0

⇒ z = 45

or z = 5 .

6. 4x2 + 4 3 x + 3 = 0Divide both sides by 4.

x2 + 3x +34

= 0


Area (A1) of 1st square = x2

Area (A2) of 2nd square = y2

Perimeter of 1st square = 4x Perimeter of 2nd square = 4yAccording to question,

x2 + y2 = 468; 4x − 4y = 24 ...(i)x = 6 + y ...(ii)

∴ Use (ii) in (i)(6 + y)2 + y2 = 468

⇒ 36 + y2 + 12y + y2 = 468⇒ 2y2 + 12y − 432 = 0⇒ y2 − 6y − 216 = 0⇒ y2 + 18y − 12y − 216 = 0⇒ y(y + 18) − 12(y + 18) = 0⇒ (y – 12)(y + 18) = 0⇒ y = −18 or y = 12

Reject y = − 18 ∴ y = 12∴ x = 18 m; y = 12 m.

WORKSHEET–9

1. (D) For equal roots, D = 0

∴ b2 – 4ac = 0 ⇒ c = 2

4b

a.

2. (C) Sum of roots = 1l−

− ⇒ l + m = l

⇒ m = 0 and l can take any real value,e.g., m = 0, l = – 2.

Product of roots = 1m ⇒ lm = m

⇒ m(l – 1) = 0 ⇒ m = 0, l = 1.

3. (C) For real roots: D ≥ 0⇒ b2 − 4ac ≥ 0⇒ k2 − 4(5)(5) ≥ 0⇒ k2 − 100 ≥ 0⇒ (k − 10) (k + 10) ≥ 0⇒ k ≤ − 10 or k ≥ 10.

4. Yes, (x + 2)3 = x(x2 – 1)⇒ x3 + 23 + 3(x + 2)2x = x3 – x⇒ x3 + 8 + 6x2 + 12x = x3 – x⇒ 6x2 + 13x + 8 = 0which is a quadratic equation.

⇒ x2 + 3x +34

+2

32

–

23

2

= 0

⇒2

2 3 3 33

2 4 4x x

+ + + − = 0

⇒2

3 3 30

2 2 2x x x

+ = ⇒ + +

= 0

⇒ x = –3 3

,2 2

− .

7. x = – 52

, 32

Hint: Let y =+ 1x

x.

8. 6 km/hrHint: Let the speed of stream = x km/hrAccording to question,

24 2418 – 18x x

−+

= 1 ⇒ x = 6.

9. Let the usual speed of the plane beu km/h. Let the usual time of flight bet hours.

Distance = Time × Speed1600 = t × u ... (i)

1600 =4060

t − × (u + 400)

⇒ 1600 = tu + 400t –23

u –800

3 ... (ii)

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

400 × 1600 2 800

3 3u

u− − = 0

⇒ u2 + 400u – 960000 = 0

⇒ u2 + 1200u – 800u – 960000 = 0

⇒ (u + 1200)(u – 800) = 0

⇒ u = – 1200 or u = 800

Since, speed in negative sign is not possible

∴ Speed = 800 km/h.

OR Let side of 1st square = x and side of 2nd square = yLet x > y


5. 23 10 7 3 0x x+ + =

⇒ x2 +10

3x + 7 = 0

⇒25 25

33x + −

+ 7 = 0

⇒2 25 2

3 3x + −

= 0

⇒ 5 2 5 2 –

3 3 3 3x x + + +

= 0

⇒ x = 7

, 33

− − .

6. D = (8ab)2 − 4(3a2)(4b2)= 64a2b2 − 48a2b2

= 16a2b2

= (4ab)2 ≥ 0

∴ x =D

2b

a− ±

= 2

8 42 3ab ab

a− ±

×

= 2126

aba

− or 2

46

aba

−

⇒ x =2ba

− or x = 2

3b

a− .

7. p = 14Hint: For real and equal roots

D = 0∴ Take a = p − 12

b = 2(p − 12)c = 2

b2 − 4ac = 0.8. x = – 2 or 1

Hint: Given equation is:(4 − 3x) (2x + 3) = 5x

⇒ 8x + 12 − 6x2 − 9x = 5x⇒ 6x2 + 6x − 12 = 0⇒ x2 + x − 2 = 0.

9. Son = 2 years; Father = 22 yearsHint: Let boy's present age = x∴ Father's present age = 24 − x


14

x. (24 − x) = x + 9

Now solve.OR

Let the usual speed and time of journey beu km/h and t hours respectively

Time =Distance

Speed

t =360u

... (i)

But new speed = (u + 5) km/h andtime = (t – 1) hours

Then, t – 1 =360

5u +... (ii)

Subtracting equation (ii) from (i) to get

1 = 360 1 1

5u u − +

⇒ u(u + 5) = 5 × 360⇒ u2 + 5u – 1800 = 0

⇒ u = 5 25 + 4 × 1 × 1800

2 × 1

− ±

⇒ u = 5 85

2− ±

⇒ u = – 45 or u = 40

Neglecting u = – 45 due to negative value,hence u = 40 km/h.

WORKSHEET– 101. (A) For no real roots D < 0

⇒ b2 − 4ac < 0 ⇒ 25p2 − 64 < 0

⇒ 25p2 < 64 ⇒ p < ± 85

∴ 8 85 5

p−

< < .

2. (B) For equal roots, D = 0

∴ 22(k – 12)2 – 4(k – 12) × 2 = 0⇒ 4(k – 12)(k – 12 – 2) = 0⇒ k = 12 or k = 14But k – 12 ≠ 0 as it is the coefficient of x2.Hence, k = 14.


3. (B) For perfect square, D = 0⇒ [4(α − 3)]2 −4(α − 3)(4) = 0⇒ 16(α − 3)2 −16(α − 3) = 0⇒ 16(α − 3) (α − 3 − 1)] = 0⇒ α = 3 (Reject) or α = 4∴ Required value is α = 4.

4. Given equation is: 2x2 – 4x + 3 = 0Here a = 2, b = – 4, c = 3

∴ D = b2 – 4ac = (– 4)2 – 4.2.3= 16 – 24 = – 8 < 0.

As D < 0, so the given equation has no realroot.

5. Yes.Given equation is

(2x + 5)(5x – 3) = 16x2 – 3⇒ 10x2 – 6x + 25x – 15 = 16x2 – 3⇒ 6x2 – 19x + 12 = 0Therefore, given equation is quadraticequation.

6. Let the Shefali’s marks in Mathematics bex and marks of English be (30 – x)According to question,

(x + 2)(30 – x – 3) = 210⇒ 27x – x2 + 54 – 2x = 210⇒ x2 – 25x + 156 = 0⇒ x2 – 12x – 13x + 156 = 0⇒ x(x – 12) – 13(x – 12) = 0⇒ (x – 12)(x – 13) = 0⇒ (x – 12) = 0, (x – 13) = 0∴ x = 12, x = 13⇒ 30 – x = 18, 17.Thus, marks in Mathematics and in Englishare 12 and 18 or 13 and 17 respectively.

7. 1 13

2x x− =

− ⇒ 2

( 2)x xx x

− −−

= 3

⇒ 3x(x – 2) = – 2 ⇒ 3x2 – 6x = – 2⇒ 3x2 – 6x + 2 = 0

⇒ x = 6 36 4 3 2

2 3

± − × ××

⇒ x = 6 12

2 3±

×

⇒ x = 6 2 3

2 3±

× ⇒ x =

3 33

±.

8. D = b2 −−−−− 4ac= [− 3(a2 + b2)]2 − 4(9)(a2 b2)= 9(a2 + b2)2 − 36a2b2

= 9a4 + 9b4 + 18a2b2 − 36a2b2

= 9a4 + 9b4 − 18a2b2

= (3a2 − 3b2)2 ≥ 0

∴ x =2 2 2 2D 3( ) 3( )

2 2×9b a b a b

a− ± + ± −=

⇒ x =26

18a

or x = 26

18b

⇒ x =2

3a or x =

2

3b

.

9. Let the breadth of the rectangular park beb metres.Then its length = (b + 3) metresArea of the rectangular park = b(b + 3) sq. mArea of the triangular park

=12

× base × altitude

= 12

× b × 12

= 6bNow,area of rectangular park – area of triangularpark = 4

b(b + 3) – 6b = 4⇒ b2 + 3b – 6b – 4 = 0⇒ b2 – 3b – 4 = 0⇒ (b – 4)(b + 1) = 0 ⇒ b = –1, 4Reject b = –1 as breadth is not possible innegative.∴ b = 4 m and b + 3 = 7 mHence, length = 7 m and breadth = 4 m.

OR

Let first number be x and second numberbe x + 4.According to question,

1 14x x

−+

=421

⇒4

( 4)x xx x

+ −+

=421


⇒ 24

4x x+ =

421

⇒ 21

4x x+ =

121

⇒ x2 + 4x – 21 = 0⇒ x2 + 7x – 3x – 21 = 0⇒ x(x + 7) – 3(x + 7) = 0⇒ x = 3 or x = – 7.Reject x = – 7∴ First number = 3

Second number = 7.

WORKSHEET – 11

1. (A) We have, kx2 − 2kx + 6 = 0For real, equal roots D = 0⇒ 4k2 − 24k = 0⇒ 4k(k − 6) = 0⇒ k = 0 or k = 6∴ k = 6.

2. (D)Hint: D = b2 − 4ac.

3. (B) If x = – 2 is a root of the equation, thenk(– 2)2 + 5(– 2) – 3k = 0 ⇒ k – 10 = 0⇒ k = 10.

4. Yes.

At x = 23

, 9x2 – 3x – 2 = 922

3

– 3 23

– 2

= 4 – 4 = 0

At x = – 13

, 9x2 – 3x – 2 = 921

3 − – 3

13

− – 2

= 2 – 2 = 0

Clearly, both the values of x =2 1

,3 3

−

satisfy the equation 9x2 – 3x – 2 = 0, so,

x = 2 1,

3 3 − are the roots of it.

5. Consider, α + β = 4

⇒ 2 − 3 + β = 4

⇒ β = 2 + 3

∴ Equation is: (x − 2 + 3 ) (x − 2 − 3 ) = 0

⇒ (x − 2)2 − ( )23 = 0

⇒ x2 + 4 − 4x − 3 = 0⇒ x2 − 4x + 1 = 0.

6. − + −− −1 2 4

( 2)( 1)x xx x

=6x

⇒ (3x − 5)x = 6 (x2 − 3x + 2)⇒ 3x2 − 5x = 6x2 − 18x + 12⇒ 3x2 − 13x + 12 = 0⇒ 3x2 − 9x − 4x + 12 = 0⇒ 3x(x − 3) − 4(x − 3) = 0⇒ (3x − 4) (x − 3) = 0

⇒ x = 43

or x = 3.

7. Discriminant for 3 x2 + 10x – 8 3 = 0 isgiven by

D = 102 – 4 × 3 × (– 8 3 )= 100 + 96 = 196

⇒ D > 0As D > 0, the given equation has real roots.

Now, x =10 196 10 14

2 3 2 3− ± − ±=

⇒ x = 4 3− , 23

Thus, the given equation has real roots

which are 4 3− and 23

.

8. Let P's obtained marks in Mathematicsbe x then obtained marks in Science be(28 – x).Consider the question,P’s new marks in Mathematics = (x + 3)and P’s new marks in Science = (28 – x – 4)

= (24 – x)According to question,

Product of new marks = 180⇒ (x + 3)(24 – x) = 180⇒ 24x – x2 + 72 – 3x = 180⇒ x2 – 21x + 108 = 0⇒ x2 – 12x – 9x + 108 = 0⇒ x(x – 12) – 9 (x – 12) = 0


⇒ (x – 9)(x – 12) = 0⇒ x – 9 = 0, x – 12 = 0∴ x = 9, x = 12If x = 9; 28 – x = 19, so, marks obtained inMathematics and Science are 9 and 19respectively.If x = 12; 28 – x = 16, so, marks obtained inMathematics and Science are 12 and 16respectively.

ORLet the sides of the two squares be a and b.Now, sum of areas = 640⇒ a2 + b2 = 640 ... (i)Difference of perimeters = 64⇒ 4a – 4b = 64⇒ a – b = 16⇒ b = a – 16 ... (ii)From equations (i) and (ii), we have

a2 + (a – 16)2 = 640⇒ 2a2 – 32a – 384 = 0⇒ a2 – 16a – 192 = 0⇒ a2 – 24a + 8a – 192 = 0⇒ (a – 24)(a + 8) = 0⇒ a = 24 or a = – 8Rejecting a = – 8 (negative length),... a = 24 mUsing equation (ii), b = 8 m.Hence, sides of the two squares are 24 mand 8 m.

9. 3 hr 30 min.Hint: Let average speed = x km/h∴ Distance = 2800 km

∴ Original time (duration) = 2800

x

∴ New time = −

2800100x

∴ −

2800100x

− 2800

x =

12

Now solve.

WORKSHEET–12

1. (A) Let x = 12 12 12 ....+ + + ...(i)

⇒ x = 12 x+[Using equation (i)]

⇒ x2 = 12 + x (Squaring)⇒ x2 – x – 12 = 0⇒ x = 4, – 3 (Reject) ∴ x = 4.

2. (B) For equal roots, D = 0∴ { – 2b(a + c)}2 – 4(a2 + b2)(b2 + c2) = 0⇒ 4a2b2 + 4b2c2 + 8ab2c – 4a2b2 – 4c2a2

– 4b4 – 4b2c2 = 0⇒ b4 – 2ab2c + c2a2 = 0 ⇒ (b2 – ca)2 = 0⇒ b2 = ac.

3. (Α) α – β = ( )2 4α + β − αβ

= ( ) ( )22 4 1− = 0.

4. 2 6p ≤ − or 2 6p ≥Hint: Use D ≥ 0i.e., b2 − 4ac ≥ 0.

5. D = p2 – 4 × 1 × 2q = p2 – 8q∵ q < 0 ∴ 8q < 0 ⇒ – 8q > 0 ...(i)Further p2 ≥ 0 ...(ii)Adding equation (i) and (ii) to getp2 – 8q > 0 ⇒ D > 0 ⇒ Real and distinctroots.

6. 4x2 + 4bx – (a2 – b2) = 0

⇒ x2 + bx14

− (a2 – b2) = 0

⇒ x2 + bx +2 2 1

4 4 4b b

− − (a2 – b2) = 0

⇒2

2b

x + –

2

2a

= 0

⇒2 2 2 2b a b a

x x + + + − = 0

[∵ A2 – B2 = (A + B)(A – B)]

⇒2 2

a b a bx x

+ − + − = 0

⇒ x = – , .2 2

a b a b+ −

7. Hint: D = 0... b2 – 4ac = 0

(c – a)2 – 4(b – c) (a – b) = 0⇒ (c + a – 2b)2 = 0⇒ c + a = 2b


∴ 200 2005l l

−+

= 2 (Accordingly question)

⇒ 200 × 5

( 5)l ll l+ −

+= 2 ⇒ l2 + 5l – 500 = 0

⇒ l2 + 25l – 20l – 500 = 0⇒ (l – 20)(l + 25) = 0⇒ l = 20 (Rejecting l = – 25)

Rate =20020

= ` 10.

Hence, the original length of the piece is20 m and its original rate is ` 10 per metre.

WORKSHEET –13

1. (D) For real roots, D ≥ 0∴ (– 3p)2 – 4 × 4 × 9 ≥ 0 ⇒ 9p2 ≥ 4 × 4 × 9⇒ p ≥ 4 or p ≤ – 4.

2. (A) x = 52

must satisfy 2x2 – 8x – m = 0

∴ 2 25 5

82 2

− – m = 0 ⇒ m =

152

− .

3. (A) Let y = 6 6 6 6 .... .+ + + +

∴ y = 6 y+ ⇒ y2 − y − 6 = 0

⇒ (y − 3) (y + 2) = 0⇒ y = 3 or − 2 (Reject).

4. 5x2 + (3p + 2)x + 15 = 0Putting x = 5, we have5 × 25 + (3p + 2) × 5 + 15 = 0⇒ 125 + 15p + 10 + 15 = 0⇒ 15p = – 150⇒ p = – 10.

5. 4x2 + 4 3x + 3 = 0 ⇒ x2 + 3x +34

= 0

⇒ x2 + 3x +2 2

3 3 3– +

2 2 4

= 0

⇒ 2

32

x

+ = 0 ⇒ x +

32

= 0

⇒ x = – 3

2.

⇒ b =2

a c+.

8.+ + ++ +2 2 2

( 1)( 2)x xx x

= +4

4x

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

∴ x =± +4 16 32

2

x =±4 48

2

= 2 ± 2 3 .

9. Let 1st pipe fill cistern in x minand 2nd pipe fill cistern in (x + 3) min.The part of cistern filled by 1st pipe in

1 min = 1x

.

The part of cistern filled by 2nd pipe in

1 min = +1

3x.


++

1 13x x

= 1340

23

3x xx x+ +

+=

1340

⇒ 80x + 120 = 13x2 + 39x⇒ 13x2 − 41x − 120 = 0⇒ 13x2 − 65x + 24x − 120 = 0⇒ 13x (x − 5) + 24(x − 5) = 0⇒ (13x + 24) (x − 5) = 0

⇒ x =24

13−

(Rejected), x = 5

∴ Required time = 5 min, 8 min.OR

Let the original length of the piece bel metres.

∵ CostLength

= Rate


6. D = b2 − 4ac

= ( 3 + 1)2 − 4(1)( 3 )

= 3 + 1 + 2 3 4 3−

= 4 − 2 3

= ( 3 1− )2 > 0

∴ x = D

2b

a− ±

⇒ x = ( )3 1 3 1

2

+ ± −

x = 3 , 1 .

7. 2x2 + 5x − 4 = 0

⇒ α + β = 5

2−

; α.β = − 2

(a) +α ββ α

=2 2α + βαβ

=( )2 .+ 2α β − α β

αβ

=

254

22

− + −

= − 418

.

(b) α3 + β3 = (α + β)3 − 3 αβ (α + β)

= − − +

125 56

8 2

= − −125 120

8 =

− 2458

.

8. Given equation is

2 2

2 22 2

62 2

x xx x

+ −+− +

= 5 ... (i)

Putting y = 2

2

2

2xx

+−

so that2

21 2

2x

y x−=+

,

equation (i) reduces to

y + 6y

= 5 ⇒ y2 – 5y + 6 = 0

⇒ (y – 3)(y – 2) = 0 ⇒ y = 2 or 3

Case I. If y = 2

2

222

xx

+−

= 2 ⇒ x2 + 2 = 4x2 – 8

⇒ 3x2 = 10 ⇒ x = ± 103

Case II. If y = 3

2

2

2

2xx

+−

= 3 ⇒ x2 + 2 = 9x2 – 18

⇒ 8x2 = 20 ⇒ x = ± 52

Hence, x = ± 10 5,

3 2± .

9. 25 min and 20 min.

Hint: Use: ++

1 15x x

= 9

100.

OR750 km/hr

Hint: Use: = 1500 1500250x x

−+

= 12

where x = usual speed.

WORKSHEET – 141. (B) For equal roots, D = 0

∴ 4(a2c2 + b2d2 + 2abcd) – 4(a2c2 + a2d2 + b2c2

+ b2d2) = 0⇒ 8abcd = 4(a2d2 + b2c2)⇒ a2d2 + b2c2 – 2abcd = 0⇒ (ad – bc)2 = 0 ⇒ ad = bc.

2. (D) For no real roots, D < 0∴ k2 – 4 × 1 × 1 < 0 ⇒ k2 – 22 < 0⇒ (k – 2)(k + 2) < 0 ⇒ – 2 < k < 2.

3. (A) Let us find the discriminant of equation

x2 – 4x + 3 2 = 0.

D = (– 4)2 – 4 × 1 × 3 2 = 16 – 12 2= 16 – 16.97

⇒ D < 0.

Therefore, x2 – 4x + 3 2 has no real roots.4. Given equation is:

x2 + ax – 4 = 0D = b2 – 4ac

= a2 – 4(– 4)= a2 + 16 > 0

As D > 0, two real and distinct roots exist.


5. Let the required whole number be x.

∴ x – 20 = 69 ×1x

⇒ x2 – 20x = 69⇒ x2 – 20x – 69 = 0⇒ x2 – (23 – 3)x – 23 × 3 = 0⇒ (x – 23)(x + 3) = 0⇒ x = 23 or x = – 3But – 3 is not a whole number∴ x = 23.

6. x = 23

a b+ , 2

3a b+

Hint: See solved example 5(ii).

7. (x − 5) (x − 6) = 2

25(24)

⇒ x2 − 11x = 225

(24) − 30

Add 211

2

to both sides.

⇒ x2 − 11x +

2112

= 225

(24) − 30 +

2112

⇒ −

2112

x = 225

(24) − 30 +

1214

= 225

(24) +

14

= 25 144576+ =

21324

⇒ x112

− = 1324

± .

⇒ x = 112

± 1324

⇒ x = 132 13

24±

⇒ x = 14524

;11924

⇒ x = 1

624

; 234

24.

8. Let the tap of larger diameter takes x hoursto fill the tank. Therefore, the other tap willtake (x + 10) hours to fill the same tank.The tap of larger diameter will fill the tank1x part in one hour and the other one will

fill 110x +

part in the same time.

According to the question,

1 110x x

++

=13

98

⇒2 10( 10)x

x x++

= 8

75

⇒ 4x2 – 35x – 375 = 0⇒ 4x2 – 60x + 25x – 375 = 0⇒ 4x(x – 15) + 25(x – 15) = 0

⇒ x = 15, 254

−

Rejecting x = 254

− hours due to negative

time, we havex = 15 hours and x + 10 = 25 hours.Hence the tap of larger diameter and ofsmaller diameter can separately fill the tankin 15 hrs and 25 hrs respectively.

OR

7 m, 4 mHint: See Worksheet −−−−− 10, Sol. 9.

9.1 1

a b x x−

+ +=

1 1a b

+

⇒( )

x a b xx a b x

− − −+ +

= b aab+

⇒ − ab = x (a + b) + x2

⇒ x2 + x (a + b) + ab = 0⇒ (x + a) (x + b) = 0

⇒ x = − a or − b.

OR

Hint:D = 4a(a3 + b3 + c3 – 3abc)∴ D = 0

a = 0 or a3 + b3 + c3 = 3abc.


WORKSHEET – 15

1. (C) For real roots, D ≥ 0∴ (– k)2 – 4 × 5 × 1 ≥ 0 ⇒ k2 ≥ 20

⇒ k ≤ – 20 or k ≥ 20 .

2. (A) 3(2)2 – 2p(2) + 2q = 0

and 3(3)2 – 2p(3) + 2q = 0⇒ 4p – 2q = 12 and 6p – 2q = 27

⇒ p = 152

, q = 9.

3. (B) D = ( )24 3 – 4 × 3 × 4.

= 48 – 48 = 0⇒ Two roots are real and equal.

4. False.There can be quadratic equation which haveno real roots e.g. x2 + 2x + 7 = 0; This equationhas no real roots because D = – 24 < 0.

5. No.Let their ages be x years and y years.Then x + y = 20 ... (i)And (x – 4)(y – 4) = 48 ... (ii)Consider equation (ii).

xy = 112 ... (iii )From equations (i) and (iii), we have

x2 – 20x + 112 = 0Here, D < 0Hence, the given situation is not possible.

6. D = b2 −−−−− 4ac= 16a4 − 4(4) (a4 − b4)= 16a4 − 16a4 + 16b4

= (4b2)2 ≥ 0

∴ x =D

2b

a− ±

= ±×

2 24 42 4

a b

⇒ x =2 2

2a b+

; 2 2–

2a b

.

7. (α – 3) x2 + 4 (α – 3) = 4⇒ (α – 3) x2 + 4(α – 3) – 4 = 0 ... (i)Since equation (i) has real and equal roots,∴ Discriminant (D) = 0⇒ b2 – 4ac = 0

∴ a = α – 3, b = 0, c = 4 (α – 3) – 4 = 4 (α – 4)

∴ D = 0 – 4(α – 3) × 4 (α – 4) = 0⇒ α = 3 or α = 4But α ≠ 3, i.e., α – 3 ≠ 0, as (α – 3) is theconstant of the leading term.Hence, α = 4.

8. Yes; 5 m and 12 mHint: Let distance of pole P from gate B bex m and from A, (x + 7) m.Therefore, x2 + (x + 7)2 = 132

Now solve.

OR

Let the snake is caught at a distance of x mfrom the pillar base∴ From figure, AC2 = 92 + x2

(Using Pythagoras Theorem)and CD = 27 − x.Since their speed are same so,

AC = CD (∵ Distance covered will beequal in equal time)

⇒ AC2 = CD2

⇒ 81 + x2 = (27 − x)2

81 + x2 = 729 + x2 – 54x54x = 648

∴ x = 12m.


4. Yes.(x – 1)3 = x3 – 2x + 1

⇒ x3 – 1 + 3x(–1)(x – 1) = x3 – 2x + 1⇒ x3 – 1 – 3x2 + 3x = x3 – 2x + 1⇒ 3x2 – 5x + 2 = 0That is a quadratic equation.

5. (2x – 1) (x + 7) = 9⇒ 2x2 + 14x – x – 7 = 9⇒ 2x2 + 13x – 16 = 0⇒ D = 169 – 4 (2) (–16)

= 169 + 128= 297

∴ x = 13 297

4− ±

.

6. 4x2 – 2(a2 + b2) + a2b2 = 0Here, A = 4, B = – 2(a2 + b2) and C = a2b2

Now,

x = 2B B 4AC

2A− ± −

=2 2 2 2 2 2 22( ) 4( ) 4 × 4 ×

2 × 4

a b a b a b+ ± + −

=2 2 4 4 2 2 2 22( ) 2 2 4

2 × 4

a b a b a b a b+ ± + + −

=2 2 2 2 2( )

4a b a b+ ± −

=2 2 2 2 2 2 2 2

,4 4

a b a b a b a b+ + − + − +

∴ x = 2 2

,2 2a b

.

7. k = − 109

or k = 2

Hint: Put D = b2 − 4ac = 0Take a = 1, b = −2(1 + 3k)

c = 7(3 + 2k).8. 81

Hint: See Worksheet – 6, Sol. 8.

9. x2 + +a

xa b

+ 1a b

xa+ + = 0

⇒ x ++a

xa b

+ +a ba

a

xa b

+ += 0

⇒ a bx

a+ +

ax

a b

+ + = 0

∴ x = a

a b−+

or −a b

a+

OR4x2 + 4bx = a2 − b2

⇒ x2 + bx = −2 2

4a b

⇒ x2 + bx +

2

2b =

−2 2

4a b

+ 2

4b

⇒ +

2

2b

x = 2

4a

⇒ +2b

x = ±2a

x = 2

b a− ±.

WORKSHEET – 161. (C) For real and equal roots, D = 0.

∴ (4k)2 – 4 × 12 × 3 = 0⇒ 16(k2 – 32) = 0⇒ (k – 3)(k + 3) = 0⇒ k = ± 3.

2. (B) α + β = 21

and αβ = – 3

⇒ (α + 2) + (β + 2) = 2 + 4and (α + 2)(β + 2) = αβ + 2(α + β) + 4

= – 3 + 2 (2) + 4⇒ S = 6 and P = 5Required equation: x2 – Sx + P = 0,

i.e., x2 – 6x + 5 = 0.

3. (B) (b)2 – (a + b) b + p = 0⇒ b2 – ab – b2 + p = 0⇒ p = ab.


OR25 studentsHint: Let the number of students attendedpicnic = x

∴ Per head contribution = 500x


500 500– 5x x

− = 5

∴5

500( 5)

x xx x

− + −

= 5

⇒ 500 = x2 – 5x⇒ x2 – 5x – 500 = 0.

9.2 1 3 9

3 2 3 ( 3)(2 3)x x

x x x x++ +

− + − += 0

⇒2 (2 3) 3 (3 9)

( 3)(2 3)x x x x

x x+ + − + +

− += 0

⇒24 10 6

( 3)(2 3)x x

x x+ +

− += 0

⇒ 2x2 + 5x + 3 = 0⇒ 2x2 + 3x + 2x + 3 = 0⇒ (2x + 3)(x + 1) = 0

⇒ x = – 1, 32

−

But x ≠ 32

− . Therefore, x = – 1.

ORThe given quadratic equation is

x2 – ( )2 1 2x+ + = 0

Here, a = 1, b = – ( )2 1+ , c = 2

Now, x =2 4

2b b ac

a− ± −

=2 1 2 1 2 2 4 2

2+ ± + + −

=2 1 2 2 2 1

2+ ± − +

=( )2

2 1 2 12

+ ± −

=2 1 2 1

2+ + −

, 2 1 2 1

2+ − +

=2 2 2

,2 2

∴ x = 2 , 1

Hence, the required roots are 2 and 1.

WORKSHEET – 17

1. (A) Let roots be α and β such that

α = 3 2

3+

and β = 3 2

3−

.

Required equation would bex2 – (α + β)x + αβ = 0

⇒ x2 – 2x + 79

= 0

⇒ 9x2 – 18x + 7 = 0.

2. (B) α2 + β2 – 4αβ = (α + β)2 – 6αβ

= 25

2−

– 6 (– 2) =

254

+ 12 = 734

.

3. (D) Given equation is: + + −

− +2 1 1

( 1) (2 1)x x

x x = 3

⇒ x = 2x2 − x −1⇒ 2x2 − 2x − 1 = 0∴ D = (– 2)2 – 4 × 2 × (– 1) = 12.

4. – 12 < k < 12Hint: D < 0⇒ k2 – 144 < 0⇒ (k – 12) (k + 12) < 0⇒ − 12 < k < 12.

5. D = 0⇒ (11 + m)2 − 4(3m + 1)(9) = 0⇒ 121 + m2 + 22m − 108m − 36 = 0⇒ m2 − 86m + 85 = 0⇒ m2 − 85m − m + 85 = 0⇒ m(m − 85) − 1(m − 85) = 0⇒ m = 1 or m = 85.

6.22

2 2

34,

aba b

−

Hint: Use x = 2B B 4AC

2A− ± −


7. Let the triangle be ABC.According to the question,

AC = 50 cm ...(i)AB + BC + AC

= 112 cm ...(ii)

From equations (i) and (ii), we haveAB + BC = 62 cm ... (iii)

Using Pythagoras theorem in ∆ABC,we have

AC2 = AB2 + BC2 ... (iv)From equations (i), (iii) and (iv), we have

(50)2 = AB2 + (62 – AB)2

⇒ 2AB2 – 124AB + 3844 – 2500 = 0⇒ AB2 – 62AB + 672 = 0⇒ AB2 – 48AB – 14AB + 672 = 0⇒ (AB – 48)(AB – 14) = 0⇒ AB = 48 cm or AB = 14 cm⇒ BC = 14 cm or BC = 48 cm

[Using (iii)]

Now, ar(∆ABC) =12

× 14 × 48

= 336 cm2.

8. −4 3x = − +6 2 3x

Squaring both sides,

4x − 3 = 36 + 2x + 3 − +12 2 3x

⇒ 2x − 42 = − +12 2 3x

⇒ x − 21 = − +6 2 3x

Again, squaring both sides, we get,x2 + 441 − 42x = 36(2x + 3)

⇒ x2 + 441 – 42x = 72x + 108⇒ x2 − 114x + 333 = 0⇒ x2 − 111x − 3x + 333 = 0⇒ x(x − 111) − 3(x − 111) = 0⇒ (x – 3)(x – 111) = 0∴ x = 3 or x = 111 (Reject).

9. Let required number of soldiers in eachrow be x.When the number of soldiers in each rowis 2 more than the total number of rows,

Number of soldiers in each row = xand number of rows = x – 2

Therefore, total number of soldiers= x(x – 2) ... (i)

When total number of rows are doubled,number of soldiers in each row

= x – 7and number of rows = 2(x – 2)Therefore, total number of soldiers

= 2(x – 2)(x – 7) ... (ii)According to the question, we arriveequation (ii) – equation (i) = 160⇒ 2(x – 2)(x – 7) – x(x – 2) = 160⇒ x2 – 16x – 132 = 0⇒ x2 – 22x + 6x – 132 = 0⇒ (x + 6)(x – 22) = 0⇒ x = – 6, 22Negative value of x cannot be taken, sorequired number of soliders is 22 in eachrow.

OR25

.

Hint: ++ =

+2 1 29

2 1 10x x

x x

⇒ 8x2 – 11x – 10 = 0

⇒ x = 2 or 5

8−

.

WORKSHEET– 18

1. (Β) − 5 is a root of 2x2 + px − 15 = 0∴ 2(−5)2 +p (−5) −15 = 0⇒ p = 7∴ p(x2 + x) + k = 0⇒ 7x2 + 7x + k = 0For equal roots, D = 0⇒ 49 − 28k = 0

⇒ k =74

.

2. (A) D = b2 – 4ac

= (– 6)2 – 4 × 7 × (– 13 7 )

= 36 + 364 = 400.


3. (A) For no real roots, D < 0.

∴ 25k2 – 64 < 0 ⇒ k2 – 28

5

< 0

⇒ 85

− < k < 85

.

4. The given equation is a perfect square, ifD = 0.⇒ (2p + 4)2 – 4(4 – p)(8p + 1) = 0⇒ 4p2 + 16p + 16 + 32p2 – 124p – 16 = 0⇒ 36p2 – 108p = 0⇒ p(p – 3) = 0 ⇒ p = 0 or 3.

5. No.

If x = 3− is a solution of x2 + 2 2x + 3 = 0

x = 3− satisfies it.

So, LHS = ( )23− + 2 2 ( )3− + 3

= 3 – 2 6 + 3 = 6 – 2 6 ≠ 0

Hence, x = 3− is not a solution of thegiven equation.

6. x = 2

2

qp

, – 1

Hint: D = b2 − 4ac= (p2 – q2)2 – 4 × p2 × –(– q)2

= (p2 + q2)2

x = D2

ba

− ±.

7. If the quadratic equationx2 + β(4x + β – 1) + 2 = 0, that is x2 + 4βx +β2 – β + 2 = 0 has real and equal roots, thenD = 0. i.e., b2 – 4ac = 0Here, a = 1, b = 4β, c = (β2 – β + 2)∴ (4β)2 – 4 × 1 × (β2 – β + 2) = 0⇒ 16β2 – 4β2 + 4β – 8 = 0⇒ 12β2 + 4β – 8 = 0⇒ 3β2 + β – 2 = 0⇒ 3β2 + 3β – 2β – 2 = 0⇒ (β + 1)(3β – 2) = 0⇒ β + 1 = 0, 3β – 2 = 0

⇒ β = – 1, 23

.

8. x = 3, – 1, 1 2±Hint:Let y = x2 − 2x∴ Given equation becomes y2 − 4y + 3 = 0.

9. 58Hint: Let numerator = x∴ Denominator = x + 3

∴ Fraction =+ 3x

xAccording to question,

+ −+ +

14 3

x xx x

=1

24.

OR

84Hint: Let unit's place = x

ten's place = y∴ Original number = 10y + xAccording to question,

10y + x = 7(y + x) ... (i)and 10y + x = 3(xy) − 12 ... (ii)Substitute the value of y from (i) to (ii) andsolve.

WORKSHEET – 191. (C) S = 8 + 2 ⇒ 10 = − a ⇒ a = − 10

(For 1st eqn.)P = 3 × 3 ⇒ 9 = b ⇒ b = 9 (For 2nd eqn.)∴ x2 − 10x + 9 = 0 ⇒ x = 9, 1.

2. (A) For equal roots, D = 0.∴ 64k2 – 4 × 9 × 16 = 0 ⇒ k = ± 3.

3. No.At x = 1, x2 + x + 1 = 12 + 1 + 1 = 3 ≠ 0At x = – 1, x2 + x + 1 = (– 1)2 – 1 + 1 = 1 ≠ 0Hence, neither x = 1 nor x = – 1 is a solutionof the equation x2 + x + 1 = 0.

4.a b

x a x b+

− − = 2c

x c−

⇒ ( )( )

ax ab bx abx a x b− + −

− − =

2cx c−

⇒ (ax + bx – 2ab)(x – c)= 2c(x2 – ax – bx + ab)


⇒ (a + b – 2c) x2 – 2abx + bcx + cax = 0⇒ x[(a + b – 2c) x – (2ab – ca – bc)] = 0

⇒ x = 0 or x = 2

2ab ac bca b c

− −+ −

.

5. x = ± 2 , ± 2

Hint: Let y = (5 + 2 6 )x2 – 3

∴1y

= (5 – 2 6 )x2 – 3

∴ y + 1y

= 10

∴ y = 10 96

2 × 1±

(Using: D = b2 – 4ac)

⇒ y = 5 ± 2 6 .Now compare the exponent.

6. x = 1 5

2±

Hint: Use a2 + b2 = (a – b)2 + 2ab2

–1

xx

x +

+ 2x 1

xx

+= 3

⇒22

1x x x

x

+ − +

+ 2 2

1x

x + = 3

⇒22

1x

x

+

+ 2 2

1x

x

+

= 3

Let y =2

1x

x +⇒ y2 + 2y – 3 = 0⇒ y = 1 or y = – 3

⇒2

= 11

xx +

or 2

1x

x + = – 3

x2 – x – 1 = 0 or x2 + 3x + 3 = 0Now solve.

7. Hint: Take a = 1 + m2, b = 2mc, c = c2 − a2

Use D = b2 − 4 ac = 0.

8. Hint: Let roots ofAx2 + 2Bx + C = 0 be α′ and β′∴ α′ = α + δ; β′ = β + δ

∴ α′ − β′ = α − β;∴ (α′ − β′)2 = (α − β)2

∴ (α′ + β′)2 − 4α′ β′ = (α + β)2 − 4αβ

⇒ 2

24B C.4

AA− =

2

24 .4

b caa

−

⇒ −2

2b ac

a=

−2

2B AC

A

⇒ −−

2

2B ACb ac

=

2

Aa .

9. Let ∆ABC is a right-angledtriangle such that ∠C = 90° andb > a.∴ c2 = a2 + b2

⇒ a2 + b2 = (3 5 )2 = 45

⇒ 4a2 + 4b2 = 180 ... (i)Let the new corresponding sides be a', b'and c' such thata' = 3a, b' = 2b and c' = 15 cmThen, (3a)2 + (2b)2 = (15)2

⇒ 9a2 + 4b2 = 225 ... (ii)Subtracting equation (i) from equation (ii),we have

5a2 = 45 ⇒ a = 3Substituting a = 3 in equation (ii), we have

9 × 9 + 4b2 = 225⇒ 4b2 = 225 – 81 = 144 ⇒ b = 6Hence, the original length of sides are 3 cm,

5 cm and 3 5 cm.

ORAccording to the question, the two times:(i) t minutes past 2 p.m. and

(ii) 2

60 34t

− − minutes past 2 p.m. are

equal. It means

t =2

60 34t

− −

⇒ t +2

34t

− = 60

⇒ t2 + 4t – 252 = 0⇒ t2 + 18t – 14t – 252 = 0


⇒ t(t + 18) – 14(t + 18) = 0⇒ (t + 18)(t – 14) = 0⇒ t = – 18 (rejected), t = 14.

WORKSHEET– 201. (A) The wrong equation is x2 + 17x + q = 0

∴ q = (– 2) × (– 15) = 30Now, the original equation will bex2 + 13x + 30 = 0. Its roots are – 10, – 3.

2. (D) x = 23

must satisfy kx2 – x – 2 = 0

∴ k × 4 29 3

− – 2 = 0 ⇒ k = 6.

3. (C)Hint: For no real root D < 0.

4. x2 + p(2x + 4) + 12 = 0⇒ x2 + 2px + 4p + 12 = 0For real and equal roots, D = 0∴ 4p2 – 4 × (4p + 12) = 0⇒ 4p2 – 16p – 48 = 0⇒ 4(p – 6) (p + 2) = 0⇒ p – 6 = 0 or p + 2 = 0⇒ p = 6 or – 2.

5. Product of roots =ca

⇒ 1× ( 2)

2− =

1q

p−+

⇒ − p −1 = − q⇒ q − p = 1 ... (i)

Also sum of roots = −12

2 =

+3

1p

⇒ 12

− =+1

1p ⇒ p = − 3 ... (ii)

∴ From (i), q = − 2∴ p + q + 5 = − 3 − 2 + 5 = 0.

6. Hint: Use D ≥ 0 for both the equation.

7. Hint: Use sin α + cos α = –ba

... (i)

and sin α . cos α = ca

... (ii)

Squaring both sides of (i) and using (ii)you will get the result.

8. The given equation is1 32 4

x xx x

− −+− −

= 13

3

⇒( 1)( 4) ( 2)( 3)

( 2)( 4)x x x x

x x− − + − −

− −= 10

3

⇒2 2

2

5 4 5 66 8

x x x xx x

− + + − +− +

= 103

⇒2

2

5 56 8

x xx x

− +− +

=53

⇒ 5x2 – 30x + 40 = 3x2 – 15x + 15⇒ 2x2 – 15x + 25 = 0Let us use quadratic formula.

x = 2 4

2b b ac

a− ± −

= 15 225 4×2×25

2×2± −

⇒ x =15 5

4± ⇒ x = 5,

52

.

9. Yes, 25 m and 16 mHint: Let the two adjacent sides of the fieldbe a and b.Then 2(a + b) = 82 ⇒ a + b = 41And ab = 400.

OR3 cm and 9 cmHint: Let smaller leg = xFrom figure,

x2 + y2 = (3 10 )2 = 90⇒ y2 = 90 − x2 ... (i)

Also (3x)2 + (2y)2 = (9 5 )2

⇒ 9x2 + 4y2 = 405 ... (ii)Use (i) and (ii) and then solve.

ASSESSMENT SHEET – 1

1. (C) D = ( )25− – 4 × 2 × 1 = 5 – 8 = – 3 < 0.

Given equation has no real roots.

2. (C) 3x2 + 2 6x + 2 + x2 = 4x2 – 4x

i.e., ( )6 2+ x + 1 = 0which is not a quadratic equation.

A

B C

y

x

3 10 cm


3. (x – 1)(x + 2) + 2 = 0 ⇒ x2 + x = 0⇒ x(x + 1) = 0 ⇒ x = 0 or –1So, roots are 0 and –1.

4. False, because a quadratic equation havingnegative discriminant has no real root.

5. 3x2 + 5 5x – 10 = 0 ... (i)

We divide 5 5x into two parts such thatsum and product of them are 5 5x and– 30x2 respectively.

Such parts are 6 5x and 5x− so equation (i)forms as follows:

3x2 + 6 5x – 5x – 10 = 0

or 3x ( )2 5x + – ( )5 2 5x + = 0

or ( )( )2 5 3 5x x+ − = 0

i.e., x + 2 5 = 0 or 3x – 5 = 0

i.e., x = – 2 5 or x =5

3Hence, the roots of equation

3x2 + 5 5 10x − = 0 are 2 5− and 5

3.

6. If a quadratic equation has equal real roots,then its discriminant vanishes.i.e., D = 0or b2 – 4ac = 0From the given equation,a = k – 12; b = 2(k – 12); c = 2So, b2 – 4ac = 0 provides

{2(k – 12)}2 – 4(k – 12) × 2 = 0i.e., 4(k2 – 24k + 144) – 8k + 96 = 0i.e., k2 – 24k + 144 – 2k + 24 = 0i.e., k2 – 26k + 168 = 0i.e., k2 – 14k – 12k + 168 = 0i.e., k(k – 14) –12(k – 14) = 0i.e., (k – 14)(k – 12) = 0i.e., k – 14 = 0 or k – 12 = 0i.e., k = 14 or 12.But k ≠ 12. ∴ k = 14.

7. a2x2 – 3abx + 2b2 = 0Dividing throughout by a2, we get

x2 – 3bx

a+

2

22ba

= 0

Here, 12

of coefficient of x is – 32

ba

Adding both the sides to 23

2ba

− , we get

x2 –3bx

a+

232

ba

− +

2

2

2ba

= 23

2ba

−

i.e., 2 2 2

2 23 9 22 4

b b bx

a a a − = −

=2

24ba

i.e., 23

2b

xa

− =

2

2ba

i.e., 32

bx

a− = ±

2ba

i.e., 32 2

b bx

a a− = or 3

2 2b b

xa a

− = −

i.e., 2bx

a= or x =

ba

Hence, the roots of the given equation are

2ba and

ba .

8. Let the speed of the faster train be x km/hrand that of the slower train be y km/hr.

Time =Distance

Speed

Time taken by the slower train – Time takenby the faster train = 3 hrs.

⇒ 600y

– 600

x = 3 ... (i)

The speed of the slower train is 10 km/hrless than that of the faster traini.e., x – y = 10or x = y + 10 ... (ii)Substitute x = y + 10 from equation (ii ) inequation (i) to get

600 600

10y y−

+ = 3

⇒ 600(y + 10) – 600y = 3(y + 10)y⇒ 200y + 2000 – 200y = y2 + 10y


⇒ y2 + 10y – 2000 = 0⇒ y2 + 50y – 40y – 2000 = 0⇒ y(y + 50) – 40 (y + 50) = 0⇒ (y + 50)(y – 40) = 0⇒ y = 40, – 50Neglecting y = –50 because speed is a non-negative quantity, we get

y = 40Substitute this value of y, i.e., y = 40in equation (ii) we get

x = 40 + 10 = 50Hence, speed of the faster train = 50 km/hrand speed of the slower train = 40 km/hr.

ASSESSMENT SHEET–2

1. (B) 722

3

+ t 23

– 3 = 0

⇒28 29 3

+ t – 3 = 0

⇒ t = 3 28

32 9

−

⇒ t = – 3 12 9

× = 16

− .

2. (A) Let us find the discriminant of:

x2 + 2 3x – 1 = 0

D = ( )22 3x – 4 × 1 × (–1) = 12 + 4 = 16 > 0

⇒ x2 + 2 3x – 1 = 0 has real roots.

3. Consider x2 + 5px + 16 has no real roots.⇒ D < 0⇒ (5p)2 – 4 × 1 × 16 < 0

⇒ 25p2 < 64 ⇒ p < ± 6425

⇒ 85

− < p < 85

.

4. True.Let equation is ax2 + bx + c = 0Case I. a > 0 and c < 0 ⇒ ac < 0 ⇒ – ac > 0

∴ D = b2 – 4ac > 0 ∴ b2 ≥ 0Case II. a < 0 and c > 0 ⇒ ac < 0 ⇒ – ac > 0

∴ D = b2 – 4ac > 0.

5. (a – b) x2 + (b – c) x + (c – a) = 0As this equation has equal roots, thediscriminant of it vanishes.i.e., D = 0⇒ (b – c)2 – 4 × (a – b) × (c – a) = 0⇒ b2 – 2bc + c2 – 4ac + 4a2

+ 4bc – 4ab = 0⇒ 4a2 + b2 + c2 – 4ab + 2bc – 4ac = 0⇒ (2a – b – c)2 = 0

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy+ 2yz + 2zx]

⇒ 2a – b – c = 0⇒ 2a = b + c.

Hence proved.

6. x2 – ( )3 1 3 0x+ + = ... (i)

Coefficient of x = – ( )3 1+

Half of coefficient of x = – 3 12+

Adding 2

3 12

+− to both the sides of

equation (i), we get

x2 – ( )3 1+ x +2

3 12

+− + 3

= 2

3 12

+−

⇒ 2

3 12

x +− =

3 1 2 34

+ +– 3

⇒ 2

3 12

x +− =

3 1 2 34

+ −

⇒2

3 12

x +− =

23 12

−

⇒ x – 3 12+

= ± 3 12−

i.e., x = 3 12+

+3 12−

,3 12+

–3 12−

i.e., x = 3 or x = 1.

Hence, x = 3 , 1.


7. Given equation isa2b2x2 – (4b4 – 3a4)x – 12a2b2 = 0 ... (i)General quadratic equation isAx2 + Bx + C = 0 ... (ii)Comparing the coefficients of like powers ofx of equations (i) and (ii), we getA = a2b2; B = – (4b4 – 3a4); C = –12a2b2

Now,D = B2 – 4AC

= {– (4b4 – 3a4)}2 – 4a2b2 (– 12a2b2)= 16b8 – 24b4a4 + 9a8 + 48a4b4

= 16b8 + 9a8 + 24a4b4

= (4b4 + 3a4)2.Using the quadratic formula to solveequation (i), we have

x =B D

2A− ±

=4 4 4 4 2

2 2

4 3 (4 3 )

2

b a b a

a b

− ± +

=4 4 4 4

2 2

4 3 (4 3 )2

b b b aa b

− ± +

i.e., x =4 4 4 4

2 2

4 3 4 32

b a b aa b

− + +,

4 4 4 4

2 2

4 3 4 32

b a b aa b

− − −

i.e., x =4

2 2

82

ba b

, 4

2 2

6

2

a

a b

−

i.e., x =4

2

4ba

, – 2

2

3ab

.

8. Let the present age of Sumita be x yearsand that of her daughter Riya be y years.According to the first given condition,

x = 2 + y2 ... (i)Riya would reach at the age of x years after(x – y) years.After (x – y) years, Sumita’s age

= x – y + x= (2x – y) years

According to the second given condition2x – y = 10y – 1

2x – 11y + 1 = 0 ... (ii)

Putting x = 2 + y2 from (i) in equation (ii),we get

2(2 + y)2 – 11y + 1 = 0⇒ 4 + 2y2 – 11y + 1 = 0⇒ 2y2 – 11y + 5 = 0⇒ 2y2 – 10y – y + 5 = 0⇒ 2y (y – 5) – 1 (y – 5) = 0⇒ (y – 5)(2y – 1) = 0

⇒ y = 5 or 12

Let us use equation (i).

When y = 12

, x = 2 + 21

2

= 94

= 2.25

But at the age of 2.25 years, Sumita cannot

be mother. So, we must reject y = 12

years.

When y = 5, x = 2 + 52 = 27Hence, present age of Sumita is 27 yearsand Riya is 5 years.

CHAPTER TEST1. (C) Let us consider option (C).

⇒ 2x2 + 3 + 2 6x + x2 = 3x2 – 5x

⇒ (5 + 2 6 ) x + 3 = 0which is not a quadratic equation.

2. (B) 9x2 + 34

x – 2 = 0

Let us add and subtract 1

64.

9x2 + 34

x + 1 1

264 64

− − = 0

⇒ 21

38

x + −

21 64 2

8

+ = 0

Clearly, the required number is 164

.


3. (C) The given equation can be written asx4 + x2 + 1 = 0

Here, D = 12 – 4 × 1 × 1 = – 3 < 0As D < 0, there is no real root.

4. 2x2 – kx + k = 0 has equal roots, if discrimi-nant = 0.⇒ (– k)2 – 4 × 2 × k = 0 ⇒ k(k – 8) = 0⇒ k = 0 or 8.

5. True.Let us consider a quadratic equation

23 7 3 12 3 0x x− + =

Here, D = ( )2– 7 3 – 4 × 3 × 12 3

⇒ D = 147 – 144 = 3⇒ D > 0⇒ Roots are real and distinct.

So, x = 7 3 3

2 3±

∴ x = 4, 3 which are

rationals.OR

No.(x – 1)2 + (2x + 1) = 0

⇒ x2 – 2x + 1 + 2x + 1 = 0⇒ x2 + 2 = 0Here, D = 02 – 4 × 1 × 2 = – 8 < 0.Hence, the given equation has no real root.

6. 2111

2x x− + 1 = 0

Here, 1

= 2

a , b = 11− , c = 1

D = b2 – 4ac

= ( )211− – 4 × 1

2 × 1 = 9

x = D2

ba

− ± =

( )11 3

2 × 1

− − ±

α = 11 32

+, β =

11 32

−.

7. Let the required natural number be N.N2 – 84 = (N + 8) × 3 ⇒ N2 – 3N – 108 = 0⇒ (N – 12)(N + 9) = 0 ⇒ N = 12 or –9

But –9 is not a natural number.So, N = 12 is the required natural number.

8. (b – c)x2 + (c – a) x + (a – b) = 0A quadratic equation is a perfect square, ifits discriminant (D) is equal to zero.Here, A = b – c, B = c – a and C = a – bNow, D = 0⇒ D = B2 – 4AC = (c – a)2 – 4(b – c)(a – b)⇒ c2 + a2 – 2ca – 4 (ab – b2 – ca + bc) = 0⇒ c2 + a2 + 4b2 + 2ca – 4bc – 4ab = 0⇒ (c + a – 2b)2 = 0[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy

+ 2yz + 2zx]

⇒ c + a – 2b = 0 ⇒ b = 2

a c+.

Hence proved.9. Let the thickness of the region having the

grass be x metres.From the adjoining figure, we haveAB = PQ = DC = SR = 50 mAP = BQ = KQ = x mKL = QR = JM = PS = (40 – 2x) m

Now, area of the grass field = 1184⇒ AB × AP × 2 + QR × KQ × 2 =1184⇒ 50 × x × 2 + (40 – 2x) × x × 2 = 1184⇒ 180x – 4x2 = 1184⇒ x2 – 45x + 296 = 0⇒ x2 – 37x – 8x + 296 = 0⇒ (x – 37)(x – 8) = 0⇒ x = 8 or x = 37But x = 37 is not possible as 2x > 50∴ x = 8 m∴ Length of the pond

= JK = 50 – 16 = 34 m And breadth of the pond = JM

= 40 – 16 = 24 m.

❑❑


2Chapter

ARITHMETIC PROGRESSIONS

WORKSHEET– 241. (D) Let us consider option (D).

2nd term – 1st term = – 6 – (–10) = 43rd term – 2nd term = – 2 – (– 6) = 44th term – 3rd term = 2 – (– 2) = 4∴ – 10, – 6, – 2, 2,........ is an A.P.

2. (B) 11th term of the A.P. – 62, – 59, – 56, ..., 7,10 is – 62 + (11 – 1) × 3, i.e., – 32.

3. (A) nth term = – 15 + (n – 1) × (– 3)= – 15 – 3n + 3 = – 3n – 12.

4. ∵ Tn = 7 – 4n∴ Tn – 1 = 7 – 4(n – 1) = 11 – 4nNow, d = Tn – Tn – 1 = 7 – 4n – 11 + 4n

= – 4.

5. FalseLet first term be a and common differencebe d of an A.P.Then

3rd term, a + 2d = 4 ... (i)9th term, a + 8d = – 8 ... (ii)

Subtracting equation (ii) from (i), we get⇒ – 6d = 12⇒ d = – 2Putting d = – 2 in (i), we get... a = 8... Given a + (n – 1)d = 0⇒ 8 + (n – 1) (– 2) = 0⇒ – 2n = – 10 ∴ n = 5

5th term = a + (n – 1)d= 8 + (5 – 1) (– 2)= 0

Thus its 5th term is 0. So the given statementis false.

6. First term = a = – 1

Second term = a + d = – 1 + 12

= – 12

Third term = a + 2d = – 1 + 2 × 12

= 0

Fourth term = a + 3d = – 1 + 3 × 12

= 12

Hence, the first four terms are: – 1, – 12

, 0, 12

.

7. Let the first term be a and the commondifference be d.A.P. = a, a + d, a + 2d,........According to question,T3 = 16 and T7 = 12 + T5

⇒ a + 2d = 16 and a + 6d = 12 + a + 4d⇒ a + 2d = 16 and d = 6⇒ a = 4 and d = 6So, the required A.P. will be 4, 10, 16, ......

8. (i) Here, a = 9, d = 17 – 9 = 8Let 636 be the sum of n term of this A.P.where (n ∈ N)

Let’s use Sn = 2n

[2a + (n – 1)d]

636 = 2n [2 × 9 + (n – 1)8]

⇒ 636 = 2n

[18 + 8n – 8]

⇒ 636 = 2n

[10 + 8n]

⇒ 636 = 2n × 2[5 + 4n]

⇒ 4n2 + 5n – 636 = 0

Using quadratic formula

n = 2 . .– 5 5 – 4 4 (– 636)

.2 4

±

= – 5 10201

8±

163TIRA MH ITE GORP ER SSC I NO S

= – 5 101

8±

= –10696

,8 8

∴ n = 12, – 13.25Rejecting n = –13.25 due to negativevalue.

∴ n = 12Thus the number of terms will be 12.

(ii) 3320

Hint : d =160

∴ an =1 1

+ ( 1)15 60

n −

∴ an =1 1

+ 10 ×15 60

=4 10

60+

=1460

= 730

∴ S11 = { }11 1 72 15 30

+ =11 92 30

××

=3320

.

9. ` 80n where n = 1, 2, 3, ..., 30; yes; ` 2400

Hint: S.I. =PR100

n ⇒ S.I. = 80n

Interest at the end of 30 years = ` 2400.

OR

Sn = 55350Hint: Sequence is:

108, 117, 126, ......, 999∴ an = 108 + (n – 1) . 9

⇒ 999 – 108 = 9(n – 1)⇒ 891 = 9(n – 1) ⇒ n = 100

∴ Sn = 100

2(108 + 999) = 50 × 1107

= 55350.

WORKSHEET– 25

1. (A) S15 = 15 3 3

2 × + (15 – 1) × 5 –2 5 5

= 17 1515 6 28

× + =2 5 5 5

×

= 51 5 .

2. (C)

Hint: Let an = –22

... 5 + (n – 1)(– 3) = – 22

⇒ – 3n = – 30 ⇒ n = 10 ∴ a10 = – 22.

3. (B) Common difference = – 2 – 1 = – 5 – (– 2)= – 3.

4. p = 4Hint: Use: if a, b, c are in A.P. ⇒ 2b = a + c.

5. FalseIn an A.P., having nth term an = a + (n – 1)d,we know that an is a linear polynomial in n.Here an = n2 + n + 1 is not a linearpolynomial in n. So it can't be nth term ofan A.P.

6. Let the first term be a and the commondifference be d.

Middle term = th21 1

2+

term = 11th term.

Sum of three middle terms:a10 + a11 + a12 = 129

⇒ (a + 9d) + (a + 10d) + (a + 11d) = 129⇒ 3a + 30d = 129 ... (i)Sum of last three terms:

a19 + a20 + a21 = 237

⇒ (a + 18d) + (a + 19d) + (a + 20d) = 2373a + 57d = 237

...(ii)Subtracting equation (i) from equation (ii),we have

27d = 108 ⇒ d = 4Substituting d = 4 in equation (i), we have

a = 129 – 120

3 = 3

So the required A.P. will be 3, 7, 11, ...OR

a5 + a9 = 72 ⇒ 2a + 12d = 72 ... (i)also a7 + a12 = 97 ⇒ 2a + 17d = 97 ... (ii)

(i) – (ii) ⇒ − 5d = − 25


⇒ d = 5From (i), ⇒ a = 6∴ A.P. is 6, 11, 16,....

7. Sn = 2n – 3n2

Hint: a1 = 5 – 6 = – 1a2 = 5 – 12 = – 7a3 = 5 – 18 = – 13

... d = a2 – a1 = – 6

... Sn = ( )( ){ }2 + 1 62n

n− − −

= { }−4 62n

n

= n(2 – 3n) = 2n – 3n2.

8. a = 7d = 11 – 7 = 4.

Let an = 111⇒ a + (n – 1) × d = 111⇒ 7 + (n – 1) × 4 = 111⇒ 7 + 4n – 4 = 111⇒ 4n = 108⇒ n = 27

... Sn = S27 = ( )277 111

2+

=27

1182

×

= 27 × 59= 1593.

9. 60°, 80°, 100°, 120°Hint: Let the angles be:

a − 3d, a − d, a + d, a + 3d.

OR∵ Sn = 5n2 – 3n∴ Sn–1 = 5(n – 1)2 – 3(n – 1)

= 5n2 + 5 – 10n – 3n + 3= 5n2 – 13n + 8

nth term (an) = Sn – Sn – 1= 5n2 – 3n – (5n2 – 13n + 8)= 10n – 8

a1 = 10 × 1 – 8 = 2a2 = 10 × 2 – 8 = 12a3 = 10 × 3 – 8 = 22

Therefore, the A.P. is 2, 12, 22,.......

Substituting n = 10 in an = 10n – 8, we get10th term = 10 × 10 – 8 = 92.

WORKSHEET– 26

1. (A) a = – 3, d = –12

– (– 3) =52

∴ a11 = a + (11 – 1)d = – 3 + 10 ×52

= 22.

2. (B) Let nth term be – 81.∴ 21 + (n – 1)(– 3) = – 81⇒ 24 – 3n = – 81⇒ n = 35.

3. (C) a18 – a13 = a + 17d – a – 12d = 5d= 5 × 5 = 25.

4. The given A.P. is 17, 14, ......, – 34, – 37, – 40.5th term from the end = – 40 + (5 – 1) × 3

= – 40 + 12 = – 28.5. 178

Hint: Let the first term be a and the commondifference be d.a + 10d = 38 and a + 15d = 73.

OR

The given A.P. is 8 , 18 , 32, ......or 2 2 , 3 2 , 4 2 , .........

Here d = 3 2 – 2 2 = 2Hence, the next two terms will be 4 2 + 2

and 4 2 + 2 2 or 5 2 and 6 2 .

6. As 3k + 4, 7k + 1, 12k − 5 are in A.P.⇒ 2 × (7k + 1) = 3k + 4 + 12k − 5⇒ 14k + 2 = 15k − 1⇒ 2 + 1 = 15k − 14k⇒ 3 = kThus, value of k = 3.

7. n = 4 or 13

Hint: Use ( ){ }S 2 12nn

a n d= + − × .

8. Let the first term and the common differencebe a and d respectively.

Second term = 14⇒ a + (2 – 1) × d = 14 [∵ an = a + (n – 1)d]⇒ a + d = 14 ...(i)


Third term = 18⇒ a + (3 – 1)d = 18⇒ a + 2d = 18 ...(ii)Solving equations (i) and (ii), we obtain

d = 4 and a = 10.Sum of first 51 terms

S51 = 512

[2a + (51 – 1)d]

[ ]{ }S = 2 ( –1)2nn

a n d+

= 512

[2 × 10 + 50 × 4]

= 512

× 220 = 5610.

9. The sequence of all months' savings (in rupees)is 320, 360, 400, .......Let Sn = 20000

⇒ 20000 = ( ){ }2 12n

a n d+ − ×

Using a = 320d = 40,

we get

20000 = ( ){ }640 1 402n

n+ − ×

⇒ 40000 = { }40 600n n +⇒ 40n2 + 600n − 40000 = 0⇒ 4n2 + 60n − 4000 = 0⇒ n2 + 15n − 1000 = 0⇒ n2 + 40n − 25n − 1000 = 0⇒ n(n + 40) − 25(n + 40) = 0⇒ n = 25 or − 40 (Reject)... n = 25 months.

WORKSHEET– 27

1. (D) 5a5 = 10a10

⇒ 5(a + 4d) = 10(a + 9d)⇒ 5a = – 70d ⇒ a = – 14dNow a15 = – 14d + 14d = 0.

2. (B) an = 505 ⇒ a + (n – 1) × d = 505⇒ 1 + 7n – 7 = 505

⇒ n = 5117

= 73

∴ Middle term is th1

2n +

= 37th

∴ a37 = 1 + 36 × 7 = 253.

3. (A) 2 (p +10) = 2p + 3p + 2⇒ 2p + 20 = 5p + 2⇒ 18 = 3p⇒ p = 6.

4. a4 = 1

92

= 192

⇒ a + 3d = 192

⇒ 5 + 3d = 192

⇒ d = 32

∴ First box =1

62

= 132

Second box = 8.

5. 15th term from end of – 10, – 20, – 30, ............,– 980, – 990, – 1000

= 15th term of – 1000, – 990, – 980, .............,– 20, – 10

= –1000 + (15 – 1) × (– 990 + 1000)= –1000 + 140 = – 860.

6. 6n – 1Hint: Use

an = Sn − Sn–1

7. Hint: Use S20 = S30 and show that S50 = 0.

8. A.P.: 63, 65, 67, ...........a = 63, d = 65 – 63 = 2∴ nth term of 63, 65, 67,.....= a + (n – 1) × d = 63 + (n – 1) × 2 ... (i)A.P.: 3, 10, 17,.........a′ = 3, d′ = 10 – 3 = 7

∴ nth term of 3, 10, 17,.......

= a′ + (n – 1)d′ = 3 + (n – 1) × 7 ... (ii)


According to the question,63 + (n – 1) × 2 = 3 + (n – 1) × 7

[Using (i) and (ii)]⇒ 2n – 2 – 7n + 7 = 3 – 63⇒ – 5n + 5 = – 60⇒ – 5n = – 65 ⇒ n = 13.

9. The sequence of savings (in rupees) is4, 5.75, 7.5, ........, 19.75Here, a = 4

d = 1.75an = 19.75

⇒ a + (n − 1) × d = 19.75⇒ 4 + (n − 1) × (1.75) = 19.75⇒ 4 + 1.75n − 1.75 = 19.75⇒ 1.75n = 19.75 − 2.25⇒ 1.75n = 17.50⇒ n = 10... In 10th week her saving will be ` 19.75.

ORLet first term be a and common differencebe d.According to question,

a4 + a8 = 24⇒ a + 3d + a + 7d = 24

a + 5d = 12 ... (i)and a6 + a10 = 44⇒ a + 5d + a + 9d = 44

a + 7d = 22 ... (ii)Subtracting (i) from (ii), we get

2d = 10∴ d = 5Putting d = 5 in (i), we get

a = –13∴ First three terms of this A.P. will be – 13,– 8, – 3.

WORKSHEET– 28

1. (B) 3 2 – 72

a a+= a + 4

⇒ 5a – 7 = 2a + 8⇒ 3a = 15⇒ a = 5.

2. (A) We need to obtain the sum:1 + 3 + 5 +............+ 49

an = a + (n – 1)d ⇒ 49 = 1 + (n – 1) × 2⇒ n = 25

∴ Sn = 2n

(a + an) =252

× (1 + 49) = 625.

3. (D) an = a + (n – 1) × d⇒ 50 = 5 + (n – 1) × 3

⇒ n – 1 = 50 – 5

3 ⇒ n = 16.

4. Let missing terms be x and y such that 2, x,26, y are in A.P.Common difference = x – 2 = 26 – x = y – 26

∴ x = 282

= 14 and

y = 26 + 26 – 14 = 38.

5. a9 = –10 ⇒ a + 8d = – 10

⇒ a + 8 ×54

= –10 ⇒ a = – 20

Now, a27 = a + 26d = –20 + 26

54

= – 20 + 652

= 252

= 1

122

.

6. 5, 7, 9Hint: Let the three numbers be

a − d, a, a + d.

7. The sequence of such numbers is:101, 108, 115, .............., 997

∴ an = 997 ⇒ 101 + (n − 1) × 7 = 997

⇒ n = 129 ∴ S129 = { }129101 997

2+

= 129 × 549 = 70821.8. S24 = 672

Hint: a1 = 5,a2 = 7,

... d = 2

... S24 = 12(10 + 46) = 672.

9. ` 40, ` 60, ` 80, ` 100Hint: Let the prizes (in `) be x, x + 20,x + 40, and x + 60.


ORAccording to the given question, we have

a = 8 ...(i)an = 33 ⇒ a + (n – 1)d = 33 ...(ii)

Sn = 123 ⇒ 2n

(a + an) = 123 ...(iii)

Substituting the value of a from equation (i),and that of an from equation (ii) in equation(iii), we get

2n

(8 + 33) = 123 ⇒ n = ×123 2

41 ⇒ n = 6

Substituting a = 8 and n = 6 in equation (ii),we get

8 + (6 – 1) × d = 33 ⇒ d = 33 – 8

5 ⇒ d = 5

Thus, n = 6 and d = 5.

WORKSHEET– 29

1. (A) Hint: 12, 16, 20,...., 248... 248 = 12 + (n – 1)4

⇒2364

= n – 1

⇒ n = 60.

2. (B) Sn = 90 ⇒ 90 = ( ){ }4 1 82n

n+ − ×

⇒ 180 = n (8n − 4)⇒ 2n2 − n − 45 = 0⇒ (2n + 9) (n − 5) = 0⇒ n = 5∴ an = a5 = a + 4 × d = 2 + 4 × 8 = 34.

3. (C) an = na′

⇒ 63 + (n − 1) × 2 = 3 + (n − 1) × 7⇒ 2n + 61 = 7n − 4⇒ 5n = 65⇒ n = 13.

4. Common difference = 2p – 1 – p= 7 – (2p – 1)

∴ 2p – p + 2p = 7 + 1 + 1⇒ 3p = 9 ⇒ p = 3.

5. a = 103; d = 101 − 103 = −2... an = 49

⇒ 103 + (n − 1) × ( −2) = 49⇒ − 2n = 49 − 105⇒ n = 28

S = 282

{103 + 49)

= 14 × 152 = 2128.

6. –1, 4, 740Hint: a3 = 7

a7 = 3 × a3 + 2.


∵ Sn = 2n

[2a + (n – 1)d] ∴ 42 = 62

[2a + 5d]

⇒ 2a + 5d = 14 ...(i)

∵ an = a + (n – 1)d ∴ +

+9

29a d

a d =

13

⇒ 3a + 27d = a + 29d ⇒ 2a – 2d = 0 ⇒ a = dSubstituting a = d in equation (i), we have

d = 2 and so a = 2Now, a18 = 2 + 17d = 2 + 17 × 2 = 36.Hence the first term is 2 and 18th term is 36.

8. Hint: mam = nan⇒ m{a + (m – 1)d} = n{a + (n – 1)d}⇒ m{a + (m – 1)d} – n{a + (n – 1) d} = 0⇒ a(m – n) + {m(m – 1) – n(n – 1)} d = 0⇒ a(m – n) + {(m2 – n2) – (m – n} d = 0⇒ a(m – n) + {(m – n) (m + n – 1) d = 0⇒ a + (m + n – 1) d = 0 {... m ≠ n}⇒ am + n = 0.

9. The penalty (in `) is in A.P. as follows:200, 250, 300, ........... 30 termsSo, the contractor has to pay the penalty(in `) as the sum of the following series:200 + 250 + 300 + ............30 termsHere, a = 200, d = 250 – 200 = 50, n = 30

∴ Sum = 2n

[ 2a + (n – 1)d]

= 302

[ 2 × 200 + 29 × 50]

= 15(400 + 1450) = 27750Hence, the required penalty is ` 27750.


WORKSHEET–30

1. (B)Hint: The sequence is:

3, 9, 15, ......., 99.

Use: Sn = { }2n

a l+ .

2. (A) an = Sn – Sn–1

= (2n2 + 5n) – [2(n – 1)2 + 5(n – 1)]

= 2n2 + 5n – 2(n2 + 1 – 2n) – 5n + 5

= 2n2 + 5n – 2n2 – 2 + 4n – 5n + 5

= 4n + 3.

3. Here, –56

– (–1), i.e. 16

and –23

–

5–

6,

i.e., 16

are equal. So, the given sequence is

in A.P. with common difference 16

.

Therefore, the next three terms will be

2 1 2 1– + , – + 2 ×

3 6 3 6 and + ×2 1

– 33 6

These are 1 1 1– , – and –

2 3 6.

4. ∵ 8x + 4, 6x – 2 and 2x + 7 are in A.P.

∴ 6x – 2 – (8x + 4) = 2x + 7 – (6x – 2)

⇒ – 2x – 6 = – 4x + 9

⇒ 2x = 15

⇒ x = 152

.

5. 3, 9, 15Hint: Let the three numbers be

a − d, a, a + d.

OR

a17 = a10 + 7⇒ a + 16d = a + 9d + 7⇒ 7d = 7⇒ d = 1.

6. 30Hint: Let Sn < 0.

7.1 5

,2 2

−

Hint: a4 + a8 = 24⇒ 2a + 10d = 24 ...(i)and a6 + a10 = 34⇒ 2a + 14d = 34 ...(ii)Solve (i) and (ii).


a3 = 7⇒ a + 2d = 7 ...(i)∵ a7 = 3a3 + 2⇒ a + 6d = 3 × 7 + 2

a + 6d = 23 ...(ii)Using equations (i) and (ii), we get

4d = 16 ⇒ d = 4 ...(iii)Substituting d = 4 in (i), we have

a + 2 × 4 = 7 ⇒ a = – 1 ...(iv)Now, using the formula of sum of n terms:

Sn = 2n

[2a + (n – 1)d]

∴ S20 = 202

[2 × (– 1) + (20 – 1) × 4]

= 10(– 2 + 76) [Using (iii) and (iv)]

= 740.

9.SS

n

n′=

2 33 2

nn

++

⇒( )( )

2 1

2 1

a n d

a n d

+ − ×′ ′+ − ×

=2 33 2

nn

++

⇒

12

12

na d

na d

− + × − ′ ′+ ×

=2 33 2

nn

++

...(i)

... To obtain 7

7

aa′

, i.e., 66

a da d

+′ ′+

replace 1

2n −

by 6 i.e. n by 13


We get from (i),

66

a da d

+′ ′+

= ( )( )

2 3 133 2 13

++

⇒ 7

7

aa′

= 4129

Thus, the ratio of 7th terms is 41 : 29.

WORKSHEET – 31

1. (C) a1 = x; a2 = y; l = 2x∴ d = y – x∴ 2x = x + (n – 1).(y – x)

⇒x

y x−= n – 1

n = –

x y xy x+ −

= –y

y x

∴ Sn = ( )1× + 2

2y

x xy x−

= 3

2( )xy

y x−.

2. (B) 10th term from end of 4, 9,....., 244, 249,254

= 10th term from begining of 254, 249, 244,...., 9, 4.

= 254 + 9 × (–5) = 254 – 45 = 209.

3. (D) 105, 112, 119, ........., 994an = a + (n – 1)d

⇒ 994 = 105 + (n – 1) × 7⇒ 994 = 105 + 7n – 7⇒ 994 = 98 + 7n

∴ n = 994 – 98

7 = 128.

4. Sn = 2n

(a + l) ⇒ 144 = 92

(a + 28)

⇒ a + 28 = 32 ⇒ a = 4.5. Yes,

Hint: a30 – a20 = a + 29d – a – 19d= 10d = – 40.

6. Let the first term and common difference offirst A.P. be A and D respectively and thatof the second A.P. be a and d respectively.

[ ]

[ ]

+

+

2A ( – 1)D2

2 ( – 1)2

nn

na n d

= +

+7 1

4 27n

n

⇒++

2A ( – 1)D2 ( – 1)

na n d

= +

+7 1

4 27n

n

⇒

– 1A + D

2– 1

+2

n

na d

= +

+7 1

4 27n

n

To prepare the 5th term in numerator anddenominator of LHS of this last equation,

we should put – 12

n = 4, i.e. n = 9.

Therefore,

A + 4D+ 4a d

= +7 × 9 1

4 × 9 + 27 ⇒ 5

5

Aa

= 6463

Hence, the required ratio is 64 : 63.

7. Hint: Use a′ + (p – 1)d = aa′ + (q – 1)d = ba′ + (r – 1)d = c.

ORLet the first term be a and the commondifference be d.Now, a19 = 3 × a6 ⇒ a + 18d = 3 (a + 5d)

2a = 3d ...(i)Also, a9 = 19 ⇒ a + 8d = 19 ...(ii)From equations (i) and (ii),

32

d + 8d = 19 ⇒ 19d = 38

⇒ d = 2 and so a = 3.[From equation (i)]

Hence the A.P. is a, a + d, a + 2d, ..........i.e., 3, 5, 7, .........

8. As given numbers are in A.P....2 × [2k2 + 3k + 6] = 4k + 8 + 3k2 + 4k + 4⇒ 4k2 + 6k + 12 = 3k2 + 8k + 12⇒ k2 − 2k = 0 ⇒ k (k − 2) = 0⇒ k = 0 or k = 2.

9. ` 7250Hint: The sequence is 125, 150, 175, .....

Here, a = 125, d = 25, n = 20

Use Sn =2n

[2a + (n – 1)d].


WORKSHEET – 321. (A) Sn = 3n2 − n

Put n = 1 ∴ S1 = a1 = 2Put n = 2 ∴ S2 = a1 + a2 = 10

⇒ a2 = 8∴ d = a2 − a1 = 6.

2. (B)Hint: a + 2d = 4

a + 8d = – 8– – +– 6d = 12 d = – 2

a = 8 a10 = a + 9d

= 8 – 18 = – 10.

3. (D)Hint: an= Sn − Sn−1.

4. Let the nth term be – 44.∴ an = – 44⇒ a +(n – 1)d = – 44⇒ 40 + (n – 1)(– 4) = – 44⇒ (n – 1) = 21⇒ n = 22.

5. – 8930Hint: a = − 5; d = − 8 − (− 5) = − 3.

an = − 230. Find n.

Then use Sn = { }2n

a l+ .

6. nth term is an = 5n – 3 Substituting n = n – 1, we have(n – 1)th term is an –1 = 5 (n – 1) –3 = 5n – 8∴Common difference is d = an – an – 1

= 5n – 3 – 5n + 8= 5

Substitute n = 1 in an = 5n – 3 to get firstterm

a1 = 5 × 1 – 3 ⇒ a1 = 2Now, using

Sn = 2n

[ 2a1 + (n – 1)d]

The sum of first 20 terms is

S20 = 202

[2 × 2 + (20 – 1) × 5] = 10 × 99

= 990.

7. Let the same common difference be d.30th term of one A.P. = 3 + (30 – 1) × d

= 3 + 29d ...(i)30th term of other A.P. = 8 + (30 – 1)d

= 8 + 29d ...(ii)Now, the required difference

= (8 + 29d) – (3 + 29d)[Using equations (i) and (ii)]

= 8 + 29d – 3 – 29d = 5.

8. Hint:

( ){ }2 12p

a p d′ + − = a

( ){ }2 12q

a q d′ + − = b

( ){ }2 12r

a r d′ + − = c

OR

Let the first term and the common differenceof the given A.P. be a and d respectively.

5th term = 0 ⇒ a + 4d = 0⇒ a = – 4d ...(i)

23rd term: a23 = a + 22d⇒ a23 = – 4d + 22d

[From equation (i)]⇒ a23 = 18d ...(ii)

11th term: a11 = a + 10d⇒ a11 = – 4d + 10d

[From equation (i)]

⇒ a11 = 6d ⇒ a11 = 6 × 23

18a

[From equation (ii)]⇒ a23 = 3a11

⇒ 23rd term = 3 × 11th termHence proved.

9. Let the digits of the number be a – d, a anda + d such the required number is100(a – d) + 10a + a + d as the digits are inA.P.


So, the required number = 111a – 99d ...(i)Sum of the digits = 15

⇒ a – d + a + a + d = 15⇒ a = 5 ...(ii)The number obtained by reversing the digits

= 100(a + d) + 10a + a – d= 111a + 99d ...(iii)

According the given condition, we have111a – 99d = 594 + 111a + 99d

[Using equation (i) and (iii)]⇒ –2 × 99d = 594⇒ d = –3 ...(iv)Using equations (i), (ii) and (iv), we arrivethat the original number is111 × 5 – 99 × (–3), that is 852.

OR

16 rows, 5 logs are placed in top row.Hint: Put Sn = 200, a = 20, d = –1

in formula Sn =2n

[2a + (n – 1)d]

So 41n – n2 = 400⇒ n = 16, 25... n = 25 not possiblebecause if n = 25 then the number of logs intop row

= – 4... n = 16 and a16 = 5.

WORKSHEET – 33

1. (A) a = 10, d = 7 – 10 = – 3a30 = a + 29d = 10 + 29(–3) = – 77.

2. (C) x + 10 – 2x = 3x + 2 – (x + 10)⇒ x – 2x – 3x + x = 2 – 10 – 10

⇒ x = 183

= 6.

3. (C) an = 2n + 1 ∴ a1 = 2 × 1 + 1 = 3

Now, Sn = 2n

(a1 + an) = 2n

(3 + 2n +1)

= n(n + 2)

4. No.Let an = 68

⇒ a + (n − 1) × d = 68

⇒ 7 + (n − 1) × 3 = 68⇒ 3n = 64

⇒ n =643

which is not a whole number so an = 68 notpossible.

5. General term is an = (– 1)n 3n + 1

Substituting n = 1, 2, 3, 4 successively we geta1 = (–1) 32 = – 9, a2 = (–1)2 33 = 27

a3 = (–1)3 34 = – 81, a4 = (–1)4 35 = 243.Therefore, first four terms are – 9, 27, – 81,243.

6. The sequence is 23, 21, 19, ............, 5∴ a = 23

d = 21 − (23)= − 2

∴ an = 5⇒ a + (n − 1) × d = 5⇒ 23 + (n − 1) × (− 2) = 5⇒ 23 − 2n + 2 = 5⇒ − 2n = − 20⇒ n = 10.Hence, number of rows is 10.

7. In the series (–5) + (– 8) + (–11) + .....+ (– 230),a = – 5, d = – 8 – (–5) = – 3.Let the number of terms be n, then–230 = – 5 + (n – 1) (– 3)

[∵ an = a + (n – 1)d]

⇒ n – 1 = – 225

– 3 ⇒ n = 76

Now, sum of first n terms is given by

Sn = 2n

(a + an)

= 762

(–5 – 230) (∵ an = – 230)

= – 38 × 235 = – 8930.

8. n2

Hint: S7 = 49S17 = 289

Find a and d and then

find Sn =2n

[2a + (n – 1)d].


9. To pick up the first potato, distance run= 2(5) m

To pick up the second potato, distance run= 2(5 + 3) = 2 (8) m

To pick up the third potato, distance run= 2(8 + 3) = 2 (11) m

.........................................................................

.........................................................................

∴ Sequence of the distance run is:2 (5), 2 (8), 2 (11), ............, till 10 terms.

∴ Total distance covered= 2 [5 + 8 + 11 + ........ + 10 terms]

= ( ) ( ) ( ){ }102 2 5 10 1 3

2 + −

= ( )2 5 37 = 10 × 37= 370 m.

WORKSHEET– 34

1. (A) 2 × (6x – 2) = 8x + 4 + 2x + 7

⇒ 12x – 4 = 10x + 11 ⇒ x = 152

.

2. (B) Sn = 1 + 3 + 5 + ......... (n terms)

∴ a = 1, d = 2

∴ Sn =2n [2a + (n – 1)d]

⇒ Sn =2n

[2 + (n – 1) × 2] = n2.

3. (D) an = Sn − Sn − 1

= n2 + 2n − (n − 1)2 − 2(n − 1)

= n2 + 2n − n2 − 1 + 2n − 2n + 2

= 2n + 1.

4. Let the common difference and the first termbe d and a respectively.

Now, a18 – a14 = 32

⇒ a + 17d – a –13d = 32

⇒ 4d = 32

⇒ d = 8.

5. an = Sn − Sn − 1

= ( ) ( )

22 3 13 13 131

2 2 2 2

nnn n

− + − + −

= ( ) + − + − + −

223 13 3 13 13

1 22 2 2 2 2n

n n n n

= + − − + − +2

23 13 3 3 13 133

2 2 2 2 2 2n

n n n n .

= 3n + 5∴∴∴∴∴ a25 = 3 × 25 + 5 = 80.

6. Hint: Let a = first termand d = common difference

am =1n

⇒ a + (m – 1)d = 1n

an =1m

⇒ a + (n – 1)d =1m

Solving d =1

mn and a =

1mn

∴ Smn = ( ){ }2 12

mna mn d+ −

= ( )11

2mn + .

7. Let a3 = x; a7 = y

∴ x + y = 6x y = 8

⇒ x2 – 6x + 8 = 0⇒ (x – 4) (x – 2) = 0

⇒ x = 4 or x = 2⇒ x = 4, y = 2; x = 2, y = 4

⇒ a3 = 4 and a7 = 2

⇒ a = 5; d = –12

⇒ S16 = 20

and a3 = 2 and a7 = 4

⇒ a = 1; d =12

⇒ S16 = 76.

8. (i) 12 (ii) 62

Hint: (i) Sn = 636.(ii) Let an = a51 + 132.


9. l1 = Length of 1st semicircle = πr= π (0.5)

l2 = Length of 2nd semicircle = π (1)

l3 = Length of 3rd semicircle = π (1.5)

l4 = Length of 4th semicircle = π (2)............................................................................................................................

∴Sequence is:

π (0.5), π (1), π (1.5), π (2), ......... till 13th term

We will find

S13 = π [0.5 + 1 + 1.5 + 2 +....... + 13th term]

= π ( ) ( ) ( ){ }132 0.5 13 1 0.5

2 + − ×

= ( )137

2π =

13 227

2 7× ×

= 143 cm.

WORKSHEET– 351. (A)

Hint: a + 6d = 34and a + 12d = 64

⇒ a = 4, d = 5.

2. (C)Hint: d = 8 2− = 2 2 2− = 2 .

a = 2

Use Sn =2n

[2a + (n – 1)d].

3. (D) ∵ 21 is an odd number ∴ a21 = 1∴ 40 is an even number ∴ a40 = –1

4. Let nth term be 181a = 5, d = 8, an = 181Now, 5 + (n – 1) × 8 = 181

⇒ n – 1 = 176

8

⇒ n = 23.

5. 28th termHint: Let an < 0

∴83 3

– 04 4

n <

⇒ 83 – 3n < 0⇒ 3n > 83

⇒ 2

273

n >

⇒ n = 28.

6. Hint: Use 2

2SS

m

n

mn

= .

7. 6 or 12Hint: Let Sn = 72

⇒ 2n

[2a + (n – 1)d] = 72.

8. n = 9; angle = 32°.

Hint: Sum of all angles = 360°.

9. Volume of concrete required to build the

1st step = 31 150 m

4 2× ×

Volume of concrete required for

2nd step = 32 150 m

4 2 × ×

Volume of concrete required for

3rd step = 33 150 m

4 2 × ×

..............................................................

..............................................................Volume of concrete required for 15th step.

= 315 150 m

4 2× ×

∴ Total volume of concrete required:

S15 =508

[ 1 + 2 + 3 + ............ + 15]

= ( ) { }25 151 15

4 2 2n

nS a l × × + = +

∵

=25 15 16

8× × = 750 m3.

WORKSHEET– 36

1. (A)

Hint: Use: Sn = 2n

[2a + (n – 1)d].


2. (B)Hint: Use: an = a + (n – 1)d.

3. (B)Hint: Common difference = – 2 – 1 or – 5 + 2,i.e., – 3.

4. 10n – 2Hint: an = Sn– Sn – 1

= 5n2 + 3n – 5 (n – 1)2 – 3 (n – 1)= 5n2 + 3n – 5 (n2 + 1 – 2n) – 3n + 3= 5n2 + 3n – 5n2 – 5 + 10n – 3n + 3

an = 10n – 2.

5. a = 2

S5 = ( )10 51

S – S4

⇒ { }+54 4

2d = ( ) ( ) × + − +

1 55 4 9 4 4

4 2d d

⇒ { }54 1

2d× + =

54 9 2 2

4d d+ − −

⇒ 8 + 8d = 4 + 9d − 2 − 2d⇒ d = −6∴ S30 = 15 {4 + 29 (−6)}

= 15 × (−170) = − 2550.

6. Hint: S1 = ( ){ }2 12n

a n d+ −

S2 = ( ){ }22 2 1

2n

a n d+ −

S3 = ( ){ }32 3 1

2n

a n d+ −

Calculate 3(S2 – S1).

7. 900Hint: S24 = 12(a1 + a24)Also note:

a5 + a20 = a1 + a24a10 + a15 = a1 + a24

Hence given relation gives:3(a1 + a24) = 225

a1 + a24 = 75∴ S24 = 900.

8. 7, 8, 9Hint: Let the three numbers be:

a − d, a, a + d.

9. We have 1 1

n n

n na b

a b− −++

=2

a b+

⇒ 2an + 2bn = an + abn–1 + ban–1 + bn

⇒ an + bn – abn–1 – ban–1 = 0⇒ an–1 (a – b) – bn–1 (a – b) = 0⇒ (a – b) (an–1 – bn–1) = 0

⇒ a = b or 1na

b

−

= 1

Taking 1na

b

−

= 1

⇒1na

b

−

= 0a

b

⇒ n – 1 = 0... n = 1.

ORan = x

⇒ a + (n − 1) × d = x⇒ 1 + (n − 1) × 3 = x⇒ 3n − 2 = x ...(i)

∴ S = ( )2n

a l+

590 = ( )12n

x+

1180 = n (1 + 3n − 2)1180 = 3n2 − n

⇒ 3n2 − n − 1180 = 0⇒ 3n2 − 60n + 59n − 1180 = 0⇒ 3n(n − 20) + 59 (n − 20) = 0

⇒ n = 593

−(Reject) or n = 20

∴ From (i) ⇒ x = 3 × 20 − 2x = 60 − 2 ⇒ x = 58.

WORKSHEET – 371. (C) Let x = nth term

∴ x = 2 + (n – 1) 3 ⇒ x = 3n – 1

∴ Sn = { }22n

x+ ⇒ 155 = { }+ −2 3 12n

n

⇒ 310 = 3n2 + n ⇒ 3n2 + n – 310 = 0⇒ n(3n+31) – 10 (3n + 31) = 0 ⇒ n = 10... x = 29.


2. (A)Hint: Use an = a + (n − 1). d.

3. 19668Hint: The sequence is:

103, 119, 135, ......., 791.

4. n = 6Hint: Let a = 3, b = 17∴ a, x1, x2, ..........., xn, b are in A.P.

∴ d = 1

b an

−+

= 14

1n +

∴ x1 = a + d = 3 + 14

1n + =

3 171

nn

++

...(i)

xn = a + nd = 3 + 14

1n

n + =

17 31

nn

++

...(ii)

1

n

xx

= 13

⇒ 3x1 = xn

∴ Using (i) and (ii), we getn = 6.

5. Hint: S1 =2n

(n + 1); a = 1, d = 1

S2 = n2 ; a = 1, d = 2

S3 =2n (3n – 1); a = 1, d = 3

∴S1 + S3= ( ) ( )1 3 12 2n n

n n+ + −

S1 + S3 = 2n2

∴S1 + S3= 2S2.

6. Let a1 = a and common difference = d.Now, a1 + a7 + a10 + a21 + a24 + a30 = 540a + a + 6d + a + 9d + a + 20d + a + 23d + a + 29d

= 540⇒ 6a + 87d = 540⇒ 2a + 29d = 180 ...(i)Further, the required sum

S30 = 302

(a1 + a30) = 15(a + a + 29d)

= 15(2a + 29d)= 15 × 180 [Using equation (i)]= 2700.

7. Hint: Let the first term and the commondifference be a and d respectively.

a9 = 0 ⇒ a + 8d = 0 ⇒ a = – 8da29 = a + 28d = – 8d + 28d = 20da19 = a + 18d = – 8d + 18d = 10d.

8. 2, 6, 10, 14.Hint: Let the four parts be:

a − 3d, a − d, a + d, a + 3d.

OR

The sequence is: 150, 146, 142, ..........∴ Total number of workers who worked

all the n days

= 150 + 146 + ....... + n terms.∴ = n(152 − 2n)

Now had the workers not dropped thenthe work would have finished in (n − 8)days with 150 workers working on eachday.∴ Total number of workers who would

have worked all the n days is 150(n −8).∴ n(152 − 2n) = 150(n − 8)

⇒ n2 − n – 600 = 0⇒ (n − 25) (n + 24) = 0

⇒ n = 25 or n = – 24 (Reject)∴ n = 25.


1. (B) an = a + (n – 1)d

⇒ 210 = 21 + (n – 1) × 21

⇒ n – 1 = 210 – 21

21⇒ n = 10.

2. (A) a18 – a14 = 32 ⇒ a + 17d – (a + 13d) = 32

⇒ d = 324

⇒ d = 8.

3. Sn = 2n2 + 5n

⇒ Sn–1 = 2(n – 1)2 + 5(n – 1)

= 2(n2 + 1 – 2n) + 5n – 5

= 2n2 + n – 3


nth term = Sn – Sn – 1

= 2n2 + 5n – (2n2 + n – 3)= 4n + 3.

4. False, because nth term of an A.P. is alwaysa linear polynomial in n.

5. First term of the A.P. = a = 4

–3

Common difference of the A.P. = d

= – 1 –

4–

3=

13

Let the A.P. consists n terms.

∴ nth term = 14

3 = 13

3...(i)

But nth term is given byan = a + (n – 1)d

= 4

–3

+ (n – 1) 13

= 3n

– 53

...(ii)

From equations (i) and (ii), we get

3n

– 53

= 133

⇒ n = 18

Since n = 18 is even number so, the middle

most terms will be th18

2

and th18

12

+ terms, i.e., 9th and 10th terms.

Now, a9 = –43

+ (9 – 1) ×13

= –43

+83

=43

And a10 = –43

+ (10 – 1)13

= –43

+ 93

= 53

Therefore, the required sum

= a9 + a10 = 43

+53

=93

= 3.

6. First term = a = +–x y

x y

Common difference = d = +

3 – 2x yx y

– +–x y

x y

= +

2 –x yx y

Now, sum of n terms

=2n

[2a + (n – 1)d]

= 2n – 2 –

2 ( – 1)x y x y

nx y x y

× + + +

= 2 – 2 2 – – 2

2x y nx ny x yn

x y

+ + +

= 2 – –

2nx ny yn

x y +

= ( )+2n

x y {n(2x – y) – y}.

7. Let the first term and the common differenceof the given A.P. be a and d respectively.According to the given condition,

11

18

aa

= 102 2

3 17 3a da d

+⇒ =

+[Using an = a + (n – 1)d]

⇒ 3a + 30d = 2a + 34d⇒ a = 4d

Now, 5

21

aa =

++

420

a da d

= 4 44 20

d dd d

++

(Substituting a = 4d)

= 8 124 3

dd

=

i.e., a5 : a21 = 1 : 3.

Now, 5

21

SS

= +

+

52 4

221

2 202

a d

a d

( ){ }Using S 2 – 12nn

a n d = +


= ( )( )

5 8 4

21 8 20

d d

d d

++[Substituting a = 4d]

= 60588

dd

= 549

i.e., S5 : S21 = 5 : 49.

8. Distances covered by a girl during 1st

minute, 2nd minute, 3rd minute,..... arerespectively 20 m, 18 m, 16 m,.......whichform an A.P. with first term (a) = 20 m andcommon difference (d) = – 2 m.(i) Distance covered in 10th minute

= 10th term of the A.P.= a + (10 – 1)d = 20 + 9 × (– 2) = 20 – 18= 2 m.

(ii) Distance covered in 10 minutes= sum of first 10 terms

= 102

[2a + (10 – 1)d] = 5[2 × 20 + 9 (– 2)]

= 5 × 22 = 110 m.

ASSESSMENT SHEET – 41. (D) 2 + 4 + 6 +....... n terms

= k(1 + 3 + 5 +..... n terms)⇒ 2(1 + 2 + 3 +........ n terms)= k(1 + 3 + 5 + ...... n terms)

⇒ 2+( 1)

2n n

= k ×2n

× {2 + (n – 1) × 2}

⇒ n2 + n = kn2

⇒ k = 1n

n+

.

2. (D) a + 8d = 449 and a + 448d = 9⇒ a = 457; d = – 1Further, 0 = 457 + (n – 1) (– 1)⇒ n = 458.

3. Let the A.P. has n terms. So, an = 2a.2a = a + (n – 1) × (b – a)

⇒ 2a = a + n(b – a) – b + a

⇒ n = –b

b a

Sn = ( )2 –b

b a (a + 2a)

⇒ Sn = ( )3

2 –ab

b a.

4. Yes; the first term is ` 2000 of an A.P. andthe common difference is ` 200 which isinterest per year.

5. The integers between 100 and 200, whichare divisible by 9 are:108, 117, 126,........, 198.Which is an A.P. with first term(a) = 108and common difference (d) = 9.Let this A.P. has n terms nth term is given by

an = a + (n – 1)d ⇒ 198 = 108 + (n – 1) × 9

⇒ n – 1 = 198 –108

9 ⇒ n = 1 +

909

⇒ n = 11

Now,

Sn = 2n

(a + l), l being last term

S11 = 112

(108 + 198) = 112

× 306

= 1683.Required sum

= 101 + 102 +....+ 199 – S11

= 992

(101 + 199) – 1683 = 14850 – 1683

= 13167.

6. We need to prove(x + 2y – z) (2y + z – x) (z + x – y) = 4xyz

...(i)If x, y, z are in A.P., then

y – x = z – y ⇒ y = 2

z x+...(ii)

Let us take LHS of (i)(x + 2y – z) (2y + z – x) (z + x – y)

= (x + z + x – z) (z + x + z – x)

–2

z xz x

+ + [Using (ii)]

= 2x × 2z × 2

z x+ = 2xz(z + x)

= 2xz × 2y [Using (ii)]

= 4xyz = RHS. Hence proved.


7. Let first term = a,Common difference = d

Sum of first n terms = 2n

[2a + (n – 1)d]

Sum of first 20 terms = 400

⇒ 202

[2a + (20 – 1)d] = 400

2a + 19d = 40 ...(i)Sum of first 40 terms = 1600

402

[2a + (40 – 1)d] = 1600

2a + 39d = 80 ...(ii)Subtracting equation (i) from equation (ii),we get

20d = 40 ⇒ d = 2Substituting d = 2 in equation (i), we get

2a + 38 = 40 ⇒ a = 1Now,

sum of first 10 terms = 102

[2a + (10 – 1)d]

= 5(2 + 9 × 2)= 100.

8. See Worksheet – 37, Sol. 8 (OR part).

CHAPTER TEST

1. (C) Sn = 2n

(a + an) ⇒ 399 = 2n

(1 + 20)

⇒ 21n = 2 × 399 ⇒ n = 38.

2. (B) a + d = 13 and a + 4d = 25⇒ a = 9, d = 4Now, a7 = a + 6d = 9 + 24 = 33.

3. (A) ∵ ap = 3 –1

6p

∴ an = 3 – 1

6n

and a1 = 13

Now, Sn = 2n +

1 3 – 13 6

n =

12n

(3n + 1).

4. Let the nth term be requireda9 = 449 ⇒ a + 8d = 449

a449 = 9 ⇒ a + 448d = 9

So, a = 457 and d = – 1Now, an = 300 ⇒ 457 + (n – 1) (–1) = 300⇒ (n – 1) = 157 ⇒ n = 158.

5. True, the reason is:d = 14 – 8 = 6, a = 8

a53 = a + 52d = 8 + 52 × 6 = 320a41 = a + 40d = 8 + 40 × 6 = 248

Now, a53 – a41 = 72.

6. The first two digit number divisible by 4 is12 and the others are: 16, 20, 24,............., 96.Now, we have to find the sum of thefollowing series:12 + 16 + 20 +.............+ 96Here, a = 12, d = 16 – 12 = 20 – 16 = 4Let the number of terms in the series be n.nth term is given by

a + (n – 1)d = 96⇒ 12 + (n – 1) × 4 = 96⇒ (n – 1) = 21⇒ n = 22

Now, S22 = 222

(12 + 96)

S ( )2nn

a l = + ∵

= 11 × 108 = 1188.

7. Let the profit to be ceased at nth day.Sale on first day = ` 8100

Sale on second day = ` (8100 – 150)= ` 7950

So, the sale (in `) will be day by day asfollows:8100, 7950, 7800,..........n termsHere, a = 8100, d = –150The profit will be ceased when it is equalto or less than ` 1500.Therefore, 8100 + (n – 1) × (– 150) ≤ 1500

[∵ an = a + (n – 1)d]

⇒ 8100 – 150n + 150 ≤ 1500⇒ 150n ≥ 6750 ⇒ n ≥ 45Hence, the profit to be ceased at 45th day.


8. If x, y and z are in A.P., then

y = +2

x z...(i)

Now, in the given equation,LHS = (x + 2y – z) (2y + z – x) (z + x – y)

= (x + x + z – z) (x + z + z – x) (2y – y)[Using equation (i)]

= 2x × 2z × y= 4xyz= RHS. Hence proved.

9. Total number of rungs =

+

2.5 m1

25cm

= 25025

+ 1 = 11

Length of the largest rung = 45 cmLength of the shortest rung = 25 cm

Let the length of each rung decreases byx cm from bottom to top.So, lengths (in cm) of all rungs frombottom to top are respectively45, 45 – x, 45 – 2x, .......25.This is an A.P. of 11 terms∴ 45 + (11 – 1) × (– x) = 25

[Using a + (n – 1)d = an]⇒ – 10x = – 20 ⇒ x = 2 cmSo, the A.P. now becomes45, 43, 41,........., 25Now, required length of wood

= Sum of this A.P.

= 112

(45 + 25) = 112

× 70

= 385 cmHence, the required length of the wood is3.85 m.

❑❑


3Chapter

CIRCLES

6. See solved example 4.

7. AB = 13 cm, AC = 15 cmHint:Usear(∆OBC + ∆OAC + ∆OAB) = ar(∆ABC).

8. Hint: Join ABLet OA = r ⇒ OP = 2rIn ∆OAP,

sinθ = OA 1OP 2 2

rr= =

θ = 30°∴ ∠APB = 2 × 30° = 60°

In ∆ABP, AP = BP∴ ∠A = ∠B

But, ∠A + ∠B = 180° – ∠APB = 120°⇒ ∠A = ∠B = 60°∴ ∆APB is an equilateral triangle.

WORKSHEET– 421. (B) As ∠OQP = 90º∴ x = 90º – 30º = 60º.

2. (Α) ∠QOR = 180º – 30º = 150º

∠PRQ = 12

× 150º = 75º.

3. (C) BC = BP + PC= BR + CQ= 3 + [AC – AQ]= 3 + [11 – 4]= 10 cm.

WORKSHEET– 411. (B) Join OX.

In ∆XOY,∠OXY= 90°,

∴ XY = 2 2OY – OX

= 2 213 – 5 = 12 cm.

2. (D) OT2 + TQ2 = OQ2

⇒ OT2 + (24)2 = (25)2

⇒ OT2 = 625 – 576 = 49⇒ OT = 7 cm.

3. (B) As AB || PR⇒ ∠BQR = ∠ABQ = 70°

(Alternate interiar angles)

Also ∠ABQ = ∠BAQ = 70° {∵ ∆AMQ ≅ ∆BMQ}

∴ In ∆AQB, using Angle sum property∠AQB = 180° – 70° – 70° = 40°.

4. False,Perimeter of ∆ABC = AB + BC + AC= AB + BQ + CQ + CR + AR= AB + BP + CQ + CQ + AP= AB + (BP + AP) + 2CQ= 2(AB + CQ)= 2(8) = 16 cm.

5. LHS= AB + CD= (AP + PB) + (CR + RD)= AS + BQ + CQ + DS= (AS + DS) + (BQ + CQ)= AD + BC= RHS.

181CRIC SEL

4. True.Let M be the point ofcontact and O be thecentre of the circle.

∠ABM = ∠ACM(∵ AB = AC)

12

∠ABM = 12

∠ACM

∠OBM = ∠OCM ... (i)∠BMO = ∠CMO ... (ii) (Each 90°)

OM = OM ... (iii) (Common)Using equations (i), (ii) and (iii) in ∆BMOand ∆CMO, we have

∆BMO ≅ ∆CMO (AAS corollory)∴ BM = CM (CPCT)⇒ BC is bisected at the point of contact.

5. As AD + BC = AB + CD⇒ AD + 7 = 6 + 4⇒ AD = 3 cm.

6. 11 cmHint: OQBP is a square∴ OQ = BP = 11 cm.

7. Let the tangents be PQ andPR corresponding to thechord QR of the circle withcentre O.Join OQ, OR and OP.In ∆PQO and ∆PRO,∠PQO = ∠PRO = 90°(Angles formed betweentangent and correspondingradius)

PO = PO (Common)QO = RO (Radii of same circle)

Therefore, we arrive at∆PQO ≅ ∆PRO(RHS axiom of congruence)So, PQ = PR (∴ CPCT)Thus, ∆PQR is an isosceles triangle.∴ ∠PQR = ∠PRQ. Hence proved.

8. Let the given parallelogrambe ABCD whose sidestouches a circles at P, Q, Rand S as shown in theadjoining figure.Since, length of twotangents drawn from anexternal point to a circle are equal.∴ AP = AS ...(i)Similarly, we have

PB = BQ ...(ii)DR = SD ...(iii)RC = QC ...(iv)

Adding these four equations, we haveAP + PB + DR + RC = AS + BQ + SD + QC

⇒ (AP + PB) + (DR + RC)= (AS + SD) + (BQ+ QC)

⇒ AB + DC = AD + BC∵ AB = DC and AD = BC

(ABCD is a parallelogram)∴ AB = BCThus, AB = BC = CD = DAHence, ABCD is a rhombus.

9. Let line l be thetangent at apoint P to thecircle with centreO. Let us takeany point Q onthe tangent l as shown in the figure.Join OQ to meet the circle at M.We know that if a point is met with thedifferent points of a line, then the shortestline segment is the perpendicular on thatline. Consider the adjoining figure:

OM = OP (Radii of same circle)OQ = OM + MQ

⇒ OQ = OP + MQ⇒ OQ > OPi.e., OP < OQClearly, OP is the shorter than OQ. Similarly,we can prove that OP is the shortest all OV,V being a variable point on the line otherthan P. Therefore, OP is the perpendicular toline l.


Hence, tangent l ⊥ radius OP.2nd Part: Join OY

∠OYX = 90ºand ∠OAY = b + a = ∠OYA

[∵ OA = OY = radius]⇒ b + a = 90º – a⇒ b + 2a = 90º.

WORKSHEET – 431. (B) ∠Q = ∠R = 90°

In quadrilateral PQOR,

∠P = 360° – (90° + 130° + 90°) = 50°

2. (C)Hint: AC = 10 cm∴ ar(∆ABC)= ar(∆AOB) + ar(∆BOC)

+ ar(∆AOC)

⇒ 24 =12

× (8 × r + 6 × r + 10 × r)

⇒ 48 = r × 24 ⇒ r = 2 cm.

3. ∠PTQ + ∠POQ = 180º⇒ ∠PTQ = 180º – 110º

= 70º.

4. CP = CQ = 11 cmBQ = CQ – CB

= 11 – 7 = 4 cm∴ BR = QB = 4 cm.

5. See Worksheet – 42, Sol. 8.

6. Hint: XP = XQ⇒ XA + AP = XB + BQ⇒ XA + AR = XB + BR

AP AR

and BQ BR=

=

∵.

7. Hint:As PA = AC... ∆PAO ≅ ∆CAO (SSS)⇒ ∠PAO = ∠CAO (CPCT)

⇒ ∠PAC = 2∠CAO

Similarly,∠CBQ = 2∠CBO

As ∠PAC + ∠CBQ = 180°

⇒12

∠PAC + 12

∠CBQ = 12

× 180°

⇒ ∠CAO + ∠CBO = 90°... ∠AOB = 90°.

8. 9 cm

Hint: ∠AMP = 90°

cos 60° = AMAP

⇒ 12

=AM

9

⇒ AM = 92

∴ AB = 2AM ⇒ AB = 9 cm.

WORKSHEET – 441. (A)

See Worksheet – 41, Sol. 2.

2. (C) ∠PTQ + ∠POQ = 180º

⇒ ∠PTQ = 180º – 115º = 65º.

3. False.∵ ∠OQL = 90°∴ ∠OQS = 90° – ∠SQL = 40°

183CRIC SEL

Similarly,∠ORS = 90° – ∠SRM = 30°

∵ In ∆SOQ, ∠OSQ = ∠OQS = 40°And in ∆SOR, ∠OSR = ∠ORS = 30°

∴ ∠QSR = OSQ + ∠OSR= 40° + 30° = 70°.

4. Perimeter of ∆ABC= AB + BC + AC= (AQ – BQ) + BC + (AR – CR)= AQ + AR + BC – (BP + PC)= 2AQ [∵ AQ = AR]

∴ AQ = AR = 12

(Perimeter of ∆ABC).

5. Let OT intersects PQ at M.

OM = 2 27 – 5

= 24 = 2 6 cm ... (i)

Let ∠PTM = θ ... (ii)Then ∠TPM = 90° – θ ... (iii)And ∠OPM = θ ... (iv)

(∵ OP ⊥ PT)And ∠MOP = 90° – θ ... (v)In ∆TPM and ∆POM,

∠TPM = ∠POM = 90° – θ[From (iii) and (v)]

∠PTO = ∠MPO = θ[From (ii) and (iv)]

∠TMP = ∠PMO (Each 90°)

So, by AAA criterion of similarity, we have ∆TPM ~ ∆POM

⇒TPPO

= PMOM

⇒ TP7

= 5

2 6

⇒ TP = 35

2 6×

66

=35 6

12cm.

6. See Worksheet – 41, Sol. 5.7. Let the given two

tangents be PA and PBto the circle withcentre O.We need to prove∠APB + ∠AOB = 180°.We know that the angle formed by atangent to the circle and the radius passingthrough the point of contact is 90°.

∴ ∠PAO = ∠PBO = 90°Applying angle sum property in thequadrilateral AOBP, we get

∠PAO + ∠AOB + ∠PBO + ∠APB = 360°⇒ 90° + ∠AOB + 90° + ∠APB = 360°⇒ ∠AOB + ∠APB = 180°.

Hence proved.

8. We have givenl || m to a circle. DEis intercept madeby tangent at C,between l and m.

We have to prove ∠DEF = 90°Construction: Join A to F, F to B and F to C.Proof: In a triangles ADF and DFC, we have

DA = DC(Tangents drawn from an external point

are equal in length)DF = DF (Common)AF = CF (Radii of the same circle)

∴ ∆ADF ≅ ∆CDF (SSS)⇒ ∠ADF = ∠CDF (CPCT)⇒ ∠ADC = 2∠CDF ... (i)Similarly, ∠CEB = 2∠CEF ... (ii)Now, ∠ADC + ∠CEB = 180°Sum of the interior angles on the same sideof transversal is 180°.⇒ 2∠CDF + 2∠CEF = 180°


⇒ ∠CDF + ∠CEF = 90° ... (iii)In ∆DEF,∠DEF + ∠EDF + DFE = 180°

⇒ 90° + ∠DFE = 180° [From (iii)]⇒ ∠DFE = 90°.

Hence proved.

WORKSHEET– 451. (D) Let radius = r

∴ PO = PB = BQ = r

AB = 2 217 – 15 = 8 cm

Now,AC = AR + RC

⇒ 17 = 8 – r + 15 – r⇒ 2r = 23 – 17 = 6⇒ r = 3 cm.

2. (B) AB = 2 2OB – OA

= 169 – 25

= 144 = 12 cm.

3. True, because in right angled isoscelestriangle AOB,

OP = 2 2+a a = 2a

4. In ∆APQ,∠PAQ = ∠AQP = θ (say)

(∵ AP = PQ = radius)∠RPB = ∠QAP = θ (Corresponding angles)∠RPQ = ∠AQP = θ (Alternate angles)Now, in ∆RPQ and ∆RPB,

RP = RP (Common)∠RPQ = ∠RPB (Each θ)

PQ = PB (Each radius)So, by SAS creterion of similarity, we have

∆RPQ ~ ∆RPB∴ ∠RBP = ∠RQPBut RQ ⊥ PQ,∴ ∠RQP = 90°∴ ∠RBP = 90°⇒ BR is tangent at B. Hence proved.

5. Join OR and OS.

Let AP = x ∴ AS = xIn quadrilateral OSDR,

∠O + ∠S + ∠D + ∠R = 360°⇒ ∠O + 90° + 90° + 90° = 360°

(∵ OS ⊥ AD and OR ⊥ CD)⇒ ∠O = 90°⇒ OSDR is a square.⇒ DR = DS = rNow, ABCD is a subscribed quadrilateral∴ AB + CD = BC + DA⇒ x + 27 + 25 = 38 + r + x⇒ r = 14 cm.

6. See Worksheet – 44, Sol. 4.OR

We know that twotangents drawn froman external point to acircle are equal inlength.∴ PA = PB⇒ ∠PBA = ∠PAB = θ (say) ...(i)⇒ ∠APB = 180° – 2θ ...(ii)

(Using ∆APB)Further, PA is tangent and AO is corres-ponding radius⇒ PA ⊥ AO⇒ ∠PAO = 90°

185CRIC SEL

⇒ ∠OAB = 90° – ∠PAB⇒ ∠OAB = 90° – θ ...(iii)

[Using equation (i)]Dividing equation (ii) by equation (iii), weget

APBOAB

∠∠

= 180 – 2

90 –° θ° θ

= ( )2 90 –90 –

° θ° θ

= 2

⇒ ∠APB = 2∠OAB. Hence proved.

7. Let the given quadrilate-ral be ABCD subscribinga circle with centre O. Letthe sides AB, BC, CD andDA touch the circle at P,Q, R and S respectively(see figure).Join OA, OB, OC, OD, OP, OQ, OR and OS.We need to prove∠AOB + ∠COD = ∠BOC + ∠DOA = 180°.Proof: In ∆AOP and ∆AOS,

OP = OS (Radii of same circle)AP =AS (Tangents from external points)AO = AO (Common)

∴∆AOP ≅ ∆AOS(SSS axiom of congruence)∴∠1 = ∠8 ...(i) (CPCT)Similarly, we can prove that∠2 = ∠3, ∠4 = ∠5 and ∠6 = ∠7 ... (ii)As, ∠1, ∠2, ∠3, ∠4, ∠5, ∠6, ∠7 and ∠8 aresubtended at a point∴ ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8

= 360°⇒ ∠1 + ∠1 + ∠2 + ∠2 + ∠5 + ∠5 + ∠6 + ∠6

= 360°Also, ∠8 + ∠8 + ∠3 + ∠3 + ∠4 + ∠4 + ∠7 + ∠7

= 360°[Using results from equations (i) and (ii)]

⇒ 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360°Also, 2(∠3 + ∠4) + 2(∠7 + ∠8) = 360°⇒ 2∠AOB + 2∠COD = 360°Also, 2∠BOC + 2∠DOA = 360°⇒ ∠AOB + ∠COD = ∠BOC + ∠DOA

= 180°Hence proved.

8. See solved example 4 for first part.Second part:

x + y = 12 ...(i)10 – x = 8 – y

⇒ – x + y = – 2 ...(ii)Solving (i) and (ii):

2y = 10⇒ y = 5∴ x = 7AD = 7 cm, BE = 5 cm, CF = 3 cm.

WORKSHEET–461. (A)

As CD = DB = 4 cmAD = CD = 4 cm

∴ AB = 4 + 4= 8 cm.

2. (B) Hint: AB = 2AM

3. Let CQ = x andBQ = y such that x + y = c ... (i)

As CR and CQ aretangents from theexternal point C,∴ CQ = CR⇒ x = a – r ... (ii)Similarly,

y = b – r ... (iii)Add (i) and (ii).

x + y = a + b – 2r⇒ c = a + b – 2r [Using (i)]

⇒ r = –

2+a b c

.

4. 16 cmGiven: AP = 5 cm⇒ BP = 12 – 5 = 7 cmalso AP = 5 cm = AQ∴ QC = 14 – 5 = 9 cm∴ BC = BR + RC

= BP + CQ= 7 + 9 = 16 cm.


5. See Worksheet – 45, Sol. 7.6. See Worksheet – 43, Sol. 7.7. For proof of theorem see solved example 4.

2nd Part.See Worksheet – 41, Sol. 5.

8. Join OB, OG, OA, OH and OC.Radius = OD = OG = OH = 4 cm

HC = DC = 6 cmBG = BD = 8 cm

Let AG = AH = xar(∆OBC)

= 1

base height2

× ×

= 1

14 42

× ×

= 28 cm2

ar(∆OAB) = ( )1× + 8 × 4

2x

= (2x + 16) cm2

ar(∆OBC) = ( )1× + 6 × 4

2x

= (2x + 12) cm2

∴ ar(∆ABC) = 28 + 2x + 16 + 2x + 12⇒ ar(∆ABC) = (4x + 56) cm2 ... (i)In ∆ABC,

s = AB BC CA

2+ +

= 8 14 6

2+ + + +x x

= x + 14

∴ ar(∆ABC) = ( ) ( ) ( )– AB – BC – CAs s s s

⇒ ar(∆ABC) = ( )+ 14 × 6 × × 8x x ... (ii)

Comparing equations (i) and (ii), we get

4x + 56 = ( )+14 × 6 × × 8x x

⇒ 42 (x + 14)2 = (x + 14) × 6 × x × 8(On squaring both sides)

⇒ 16 (x + 14) (x + 14 – 3x) = 0⇒ x = 7 as x ≠ –14 (∵ x > 0)So, AB = x + 8 = 7 + 8 = 15 cmand AC = x + 6 = 7 + 6 = 13 cm.

ASSESSMENT SHEET–51. (B) BC = 2AB

= 2 169 – 25

= 24 cm.

2. (C) If a circle touches the sides of aquadrilateral, then the angles subtended byeach pair of opposite sides are supplementary.∴ ∠AOB + ∠COD = 180°⇒ 130° + ∠COD = 180°⇒ ∠COD = 50°.

3. In right triangle AOP,

tan (∠APO) = Perpendicular

Base

⇒ tan 30° = AOAP

⇒ 13

= 5

AP ⇒ AP = 5 3

∴ AP = BP = 5 3 cm.

4. True, because∠MAB = ∠ACB

and ∠NAC = ∠ABC, But ∠ABC = ∠ACB∴ ∠MAB = ∠ABCand ∠NAC = ∠ACBBut these are alternate pair of interior angles.∴ MN || BC.

5. Note that tangentsdrawn from anexternal point to acircle are equal inlength.Tangents are drawn from A, B and C, so werespectively get

AQ = AR ... (i)BQ = BP ... (ii)

And CR = CP ... (iii)Taking equation (i), we

AQ = AR = AC + CR = AC + CP[From (iii)]

= AC + BC – BP= AC + BC – BQ [From (ii)]= AC + BC – (AQ – AB)

187CRIC SEL

= AC + BC + AB – AQ⇒ 2 AQ = Perimeter of ∆ABC

⇒ AQ = 12

(Perimeter of ∆ABC)

Hence proved.

6. We are given twotangents AP and BPcorresponding tothe chord AB of acircle with centre O.We need to prove

∠PAB = ∠PBAJoin AO, BO and PO.In ∆AOP and ∆BOP,

∠PAO = ∠PBO = 90°(Angle between tangent and correspondingradius)

OP = OP (Common)AO = BO (Radii of same circle)

Therefore,∆AOP ≅ ∆BOP (RHS criterion)

∴ PA = PB (CPCT)Thus, ∆ PAB is an isosceles triangle.⇒ ∠PAB = ∠PBA. Hence proved.

7. Let O be thecentre of thecircle. Join OT tomeet PQ at R.Join OP and OQ.

In ∆PTR and ∆QTR,PT = QT

(Tangents from an external point to thesame circle)

∠PTR = ∠QTR(Line segment joining the centre and the

point of intersection of the tangents bisectsthe angle between the tangents)TR = TR (Common)

Therefore,∆PTR ≅ ∆QTR (SAS criterion)

⇒ PR = QR (CPCT)Also, ∠ORP = ∠ORQ = 90°(Angle between chord and the line segment

joining the centre and mid-point of thechord)

∴ ∠ORP = ∠PRT = 90°

In ∆OPR, OR = 2 2OP – PR

= 25 – 16 = 3 cm

(Pythagoras theorem)Further in ∆ORP and ∆PRT,

∠ORP = ∠PRT = 90°(Proved above)

∠OPR = ∠PTR(∵ ∠OPR = 90° – ∠RPT = 90° – (90° – ∠PTR))Therefore,

∆ORP ∼ ∆PRT (AA criterion)

⇒ ORPR

= OPTP

(Corresponding sides)

⇒ 34

= 5

TP

⇒ TP = 203

cm.

8. Join QO and TOto intersect PQat R.In ∆TOP and∆TOQ,

TP = TQ(Tangents from same external point)

PO = QO (Radii of same circle)TO = TO (Common side)

So, ∆TOP ≅ ∆TOQ(SSS criterion of congruence)

∴ ∠POR = ∠QOR ... (i) (CPCT)Now, in ∆POR and ∆QOR,

PO = QO (Radii of same circle)∠POR = ∠QOR [From (i)]

OR = OR (Common side)So, ∆POR ≅ ∆QOR

(SAS criterion of congruence)∴ ∠PRO = ∠QRO ... (ii)

(Corresponding angles)∠PRO + ∠QRO = 180º ... (iii)

(Linear pair axiom)From (ii) and (iii),

∠PRO = 90° ... (iv)In ∆TPO,

∠TPO = 90° ... (v)(Angle between tangent and

corresponding radius)


Further, in ∆PRO and ∆TPO,∠PRO = ∠TPO = 90°

[From (iv) and (v)]∠POR = ∠TOP (Common angle)

So, ∆PRO ~ ∆TPO(AA criterion of similarity)

∴ ∠OPR = ∠OTP

⇒ ∠OPQ = 12

∠PTQ

(∴ ∠OTP = ∠OTQ)i.e., ∠PTQ = 2∠OPQ Hence proved.

ASSESSMENT SHEET– 6

1. (D) Join OA

cos 30° = ATOT

⇒ AT = 6 ×3

2= 3 3 cm.

2. (D) ∠BAT = ∠ACB = 55°.

3. AD = DC = 4⇒ AD = 4 cm

Similarly CD = DB = 4 cm⇒ DB = 4 cm

∴ AB = AD + DB = 4 + 4 = 8 cm.4. False, because the centres of the circles lie on

the perpendicular of PQ, which passesthrough A.

5. Let the sides AB,BC and CA of the∆ABC touch thecircle with centre Oat the point P, Qand R respectively.

In quadrilateral OQCR,∠OQC = ∠ORC = 90°

(Angles between tangent andcorresponding radius)

and ∠QCR = 90° (Given)⇒ ∠QOR = 90°⇒ OQCR is a square⇒ CQ = CR = r⇒ BQ = a – r, AR = b – r,⇒ AP = b – r, PB = a – rBut AB = c∴ b – r + a – r = c⇒ 2r = a + b – c

⇒ r = – c

2a b+

. Hence proved.

6. Draw a line QTpassing through Qand perpendicular toQP to meet SR at T.In ∆PQR,

PQ = PR(Tangents from an external point)

∴ ∠PRQ = ∠PQR ... (i)(Angles opposite to equal sides)

∠PQR + ∠PRQ + ∠QPR = 180° ... (ii)(Angle sum property for a triangle)

From equations (i) and (ii),∠PQR + ∠PQR + 30° = 180°

⇒ ∠PQR = ∠PRQ = 75° ... (iii)Now, ∠TQR + ∠PQR = 90°

(Angle between tangent)⇒ TQR = 15° ... (iv) [Using (iii)]∴ SR || QP and QT ⊥ QP∴ QT ⊥ SR⇒ ST = TR ... (v)(∵ TQ passess through the centre of thecircle)In ∆STQ and RTQ,

ST = TR [From (v)]∠STQ = ∠RTQ (∵ QT ⊥ SR)

TQ = TQ (Common)∴ ∆STQ ≅ ∆RTQ (SAS criterion)⇒ ∠SQT = ∠RQT = 15° [Using (iv)]∠SQT + ∠TQR = 15° + 15°

⇒ ∠RQS = 30°.

189CRIC SEL

OR∠APB = 120°

∴ ∠APO = ∠OPB = 60°

In right-angled ∆OAP

cos 60° = APOP

⇒12

= APOP

⇒ OP = 2AP

7. Let O be the centreof the circle withradius r.

Join SO and RO.

In quadrilateralDROS,∠SDR = 90° ... (i) (Given)

and ∠OSD = ∠ORD = 90° ... (ii)(Angle between tangent and

corresponding radius)Therefore,

∠SOR = 90° ... (iii)⇒ Quadrilateral DROS is a rectangle

[From (i), (ii) and (iii)]But SO = SR (= r)∴ DROS is a square.⇒ r = SO = DR = SD = OR ... (iv)We know that tangents drawn from anexternal point to a circle are equal.∴ BQ = BP = 27 cm

(Tangents from point B)QC = BC – BQ = 38 – 27 = 11 cm

RC = QC = 11 cm(Tangents from point C)

DR = DC – RC = 25 – 11 = 14 cm∴ r = 14 cm [From (iv)]Hence, radius of the incircle is 14 cm.

8. Let AD = x. Weknow that thetangents drawnfrom an externalpoint to a circleare equal.

∴ AD = AF = x

BD = BE and CE = CFNow, BD = AB – AD = 8 – x = BEand CE = BC – BE = 10 – (8 – x)

= 2 + x = CF.AF = AC – CF = 12 – (2 + x)

= 10 – x = ADBut AD = x∴ 10 – x = x ⇒ x = 5 cm, i.e., AD = 5 cm

BE = 8 – x = 8 – 5 = 3 cmand CF = 2 + x = 2 + 5 = 7 cmThus, AD = 5 cm, BE = 3 cm and CF = 7 cm.

CHAPTER TEST

1. (C) ∵ OA ⊥ AT∴ ∠OAT = 90°In ∆OAT,

cos T = ATOT

⇒ cos 30° = AT4

⇒ AT = 2 3 cm.

2. (A) ∠PAO = 90°∠OAB = ∠PAO – ∠PAB

= ∠PAO – 12

(∠PAB + ∠PBA)

(∵ ∠PAB = ∠PBA)

= ∠PAO – 12

(180° – ∠APB) (ASP)

= 90° –12

(180° – 50°) = 25°.

3. In right ∆POQ,

PQ = 2 213 – 5 = 12 cm

PR = PQ = 12 cm

ar( PQOR)

= ar(∆PQO) + ar(∆PRO)

= 1

PQ QO2

× × +

1PR RO

2 × ×


= 1

12 52

× × +

112 5

2 × ×

= 2 × 30 = 60 cm2

4. True, as ∠BPA = 90°, (∵ AB is diameter)

∠PAB = ∠OPA = 60° (∵ OP = OA)

Also OP ⊥ PT.∴ ∠APT = 30°and ∠PTA = 60° – 30° = 30°.

5. Let the given chord be ABand two tangents to thecircle with centre O be APand BP.We need to prove

∠PAB = ∠PBAJoin OA and OB.Proof:In ∆AOB, AO = BO∴∠ABO = ∠BAO ... (i)As, the tangent is perpendicular to the radiuspassing through the point of contact,

∠PAO = ∠PBO = 90° ... (ii)Again, ∠PAO= ∠PBO [Using (ii)]⇒ ∠PAB + ∠BAO = ∠PBA + ∠ABO⇒ ∠PAB + ∠ABO = ∠PBA + ∠ABO

[Using (i)]⇒ ∠PAB = ∠PBA.

6. Hint: AP = AU,BP = BQ,CR = CQ,DR = DS,ET = ESFT = FU.

7. We know that thetangents drawn from anexternal point to a circleare equal in length.

∴ AQ = AR ... (i)BQ = BP ... (ii)

and CP = CR ... (iii)Now,

AQ = AB + BQ= AB + BP [From (ii)]= AB + (BC – PC)= AB + BC – CR [From (iii)]= AB + BC – (AR – AC)= AB + BC + CA – AR= AB + BC + CA – AQ [From (i)]

⇒ AQ + AQ = AB + BC + CA

⇒ AQ = 12

(AB + BC + CA)

Hence proved.8. ∠BCA = 90° ... (i) (Angle in semicircle)

∠PCA = ∠PCB + ∠BCA

⇒ 110° = ∠PCB + 90°⇒ ∠PCB = 110° – 90° = 20° ...(ii)Since PC is a tangent and CB is a chord∴ ∠BAC = ∠PCB = 20° ... (iii) [Using (ii)]Now, applying angle sum property in ∆ABC,we have

∠CBA + ∠BCA + ∠CAB = 180°⇒ ∠CBA + 90° + 20° = 180°

[Using (i) and (ii)]⇒ ∠CBA = 180° – 90° – 20° = 70°.

❑❑

191RTSNOC U C T I O N S

4Chapter

CONSTRUCTIONS

WORKSHEET – 50

1. (D) Since, the angle between two radii of a circle and the angle between corresponding twotangents are supplementary.∴ Required angle = 180° – 35° = 145°.

2. (B) Out of the numbers 8 and 5 in 85

, the greater one is 8, so the required number of

points is 8.3.

AB = 8.5 cm and ACCB

=37

AC = 2.55 cm, CB = 5.95 cm.

4. Steps of construction:

First, we draw a triangle ABC such that AB = 4 cm, AC = 5 cm and ∠BAC = 90°. Further, wewill draw a triangle A′BC′ similar to triangle ABC using the following steps.Step I: Draw a ray BX making an acute angle with BC on the side opposite to vertex A.Step II: Locate five points X1, X2, X3, X4 and X5 on BX so that BX1 = X1X2 = X2X3 = X3X4 = X4X5.Step III: Join X3C and draw X5C′ || X3C to intersect BC extended at C′.Step IV: Draw C′A′ || CA to intersect BA extended at A′.


4cm

5cm

90°

90°

B C C�

A�

A

X

X1

X2

X3

X4

X5

Then, ∆A′BC′ ~ ∆ABC.


Step I: First, draw a circle with radius as 5 cm and centre at O. Then take a point P so thatOP = 11 cm.

Step II: Bisect OP to find mid-point M of OP. Then take M as centre and MP = MO as radius,draw a circle to intersect the previous circle at Q and R.

Step III: Join PQ and PR which are the required tangents.

After measuring PQ and PR, we find PQ = PR = 9.8 cm (approximately).

Justification:

Join OQ and OR.

In ∆OPQ, OP = 11 cm, OQ = 5 cm and PQ = 9.8 cm

∴ OP2 – OQ2 = 112 – 52 = (11 + 5)(11 – 5) = 96

And PQ2 = (9.8)2 = 96.04

Clearly, OP2 – OQ2 ≈ PQ2

⇒ OP2 = OQ2 + PQ2

Also, OP2 = OR2 + PR2

Therefore, ∆POQ and ∆POR are right triangles with ∠PQO = ∠PRO = 90°.


So, tangents are perpendicular to radii passing through their respective points of contact.i.e., PQ ⊥ OQ and PR ⊥ OR.

6. ∠C = 180° – (∠B + ∠A) = 180° – 150° = 30°.Steps of construction:In order to construct a triangle similar to ∆ABC, follow the following steps:Step I: First, construct a ∆ABC in which BC = 7 cm, ∠B = 45° and ∠C = 30°.

Step II: Make an acute angle CBX such that X is on the side opposite to vertex A.


Step III: Locate four points namely X1, X2, X3 and X4 on BX such that BX1 = X1 X2 = X2 X3 =X3 X4.

Step IV: Join X3C and draw a line X4C′ || X3C to intersect BC produced at C′.Step V: Draw a line C′A′ parallel to side CA of ∆ABC to intersect BA produced at A′. Then,∆A′BC′ is the required triangle.

WORKSHEET – 51

1. (B) Since 4 + 7 = 11, therefore, B will be joined to A11.2. (D) The required angle and the angle between the two tangents are supplementary.

∴ Required angle = 180° – 60° = 120°.

3. Here, 5 + 8 = 13AB = 7.6 cmAC : BC = 5 : 8AC = 2.92 cm and BC = 4.68 cm



First, we draw ∆ABC with the givenmeasurements. Then we draw anothertriangle A′BC′ similar to ∆ABC and of

scalar factor 34

using the following steps:

Step I: Draw a ray BX such that ∠CBX isan acute angle.

Step II: Mark X1, X2, X3, X4 on BX suchthat BX1 = X1X2 = X2X3 = X3X4.

Step III: Join X4C and draw X3C′ || X4C.

Also, draw A′C′ || AC.

Thus, ∆A′BC′ ∼ ∆ABC.

5. Justification: In ∆PQR and ∆PQ′R′, ∠P = 45° is common and RQ || R′Q′.∴ ∆PQR ~ ∆PQ′R′We have draw PP1 = P1P2 = P2P3 = P3P4

∴ 3 4

3

P P 1=

PP 3

⇒ 3 4

3

P P 1+ 1 = 1

PP 3+

⇒ 3 4 3

3

P P PP 4=

PP 3+

⇒ 4

3

PP 4=

PP 3

And P3Q || P4Q′

∴ PQPQ 3

′ 4= .

Also, ∆PQR ~ ∆PQ′R′

Hence, PQPQ

′= PR

PR′ =

R QRQ′ ′

.

6. Steps of construction: In order to construct a pair of required tangents, follow the followingsteps:Step I: Draw a circle with radius OA = 3 cm and centre O.


Step II: Take any point P outside the circledrawn in step I and join OP.Step III: Obtain mid-point M of OP obtainedin step II and draw another circle with radiusOM = PM and centre M to intersect thecircle drawn in step I at A and B.Step IV: Join PA and PB.These PA and PB form the required pair oftangents.

WORKSHEET – 52

1. (C) The next step should be the line parallelto B5C should be passed through B4 as the

sides of required triangle are 45

of the

corresponding sides of ∆ABC.2. (C) Two distinct tangents to a circle can be

constructed from P only when P is situatedat a distance more than radius (here 2r)from the centre.

3. False. In the ratio 3 + 2 : 3 – 2 , i.e.,

11 + 6 2 : 7, 11 + 6 2 is not a positiveinteger, while 7 is.

4. Steps of construction: In order to construct a∆ABC and its similar triangle with givenmeasurements, follow the following steps:

Step I: Draw a ∆ABC inwhich BC = 7 cm, ∠B = 45° and∠C = 180° – (45° + 105°) = 30°.

Step II: Make an acute∠CBX such that X is on theopposite side of the vertexA and locate points B1, B2,B3 and B4 on BX such thatBB1 = B1B2 = B2B3 = B3B4.

Step III: Join B3C and drawB4C′ || B3C to intersect BCproduced at C′. Also drawC′A′ || CA to intersect BA produced at A′.Hence, ∆A′BC′ ~ ∆ABC.

5. Steps of construction: In order to draw a pair of tangents to the given circle, follow thefollowing steps:Step I: Draw a radius AO in the given circle with centre O and draw another radius makingan angle AOB of measure 180° – 60° = 120°.


Step II: Make ∠OAP = 90° and ∠BOP = 90° to intersect each other at P.Such obtained AP and BP are the required tangents such that ∠APB = 60°.

6. We are given a ∆PQR with eachside of measure 6 cm.

Steps of construction: In order toconstruct ∆ABC follow thefollowing steps:

Step I: Make an acute angle RQXand locate seven points Q1, Q2,Q3, Q4, Q5, Q6 and Q7 on the rayOX such that QQ1 = Q1Q2 = Q2Q3= Q3Q4 = Q4Q5 = Q5Q6 = Q6Q7.

Step II: Join Q7 R and draw Q6 Cparallel to Q7 R to intersect QR atC.

Step III: Draw CA parallel to RPto intersect BP (B and Q coincide)at A.

Then, ∆ABC is the requiredtriangle.


WORKSHEET– 53

1. (A) Angle between the tangents is less than 180°.

2. (B) Line segment A5B7 divides the linesegment AB in the ratio 5 : 7.

3. False, because the point P lies inside the circle.

4.

Measuring the tangent AP, we get AP = 4.0 cm

5. We are given a circle of radius4 cm and centre O.Steps of construction: In order todraw the required pair of tangents,follow the following steps.Step I: Draw a pair of radius OAand OB inclined at an angle of180° – 120° = 60° to intersect thegiven circle at A and B respectively.Step II: Draw perpendiculars AP andBP which intersect each other at P.Then AP and BP are the requiredtangents.Justification: In quadrilateral,OBPA, applying Angle sum property,we have

∠O + ∠A + ∠B + ∠P = 360°⇒ 60° + 90° + 90° + ∠P = 360°⇒ ∠P = 360° – 240° ⇒ ∠P = 120°.Angle between the tangents is 120°.


6.

In the figure, ∆A′BC′ ~ ∆ABC.

Steps of construction:First, we draw ∆ABC with the givenmeasurements. Then we draw anothertriangle A′BC′ similar to ∆ABC and of

scalar factor 34


Step I: Draw a ray BX such that ∠CBX isan acute angle.

Step II: Mark X1, X2, X3, X4 on BX suchthat BX1 = X1X2 = X2X3 = X3X4.




1. (D) The minimum number of points shouldbe 9 as 9 > 5 out of the numerator and

denominator of 95

. The next step is to be

joined B5 to C.

2. (B) Angle of inclination, here θ, can liebetween 0° and 180°. So, the mostappropriate option is (B), i.e., 0 < θ < 180°.

3. In the adjoining figure,

∆AB′C′ ~ ∆ABC such that

ABAB

′=

B CBC′ ′

=ACAC

′ =

32

.

WORKSHEET – 54



First, we draw ∆ABC with the given measurements. Then we draw another triangle A′BC′

similar to ∆ABC and of scalar factor 34


Step I: Draw a ray BX such that ∠CBX is an acute angle.

Step II: Mark X1, X2, X3, X4 on BX such that BX1 = X1X2 = X2X3 = X3X4.




5.1 5

2 =2 2

Steps of construction: In order to construct an isosceles triangle and another triangle having52

of its corresponding sides, follow the steps given below:

Step I: Construct an isosceles triangle having any length of equal sides by drawing baseQR = 8 cm and altitude PM = 4 cm passing through the mid-point M of side QR.

Step II: Draw a ray QX such that ∠RQX is and acute angle; and divide the ray in five equalparts, namely QQ1, Q1Q2, Q2Q3, Q3Q4 and Q4Q5.

Step III: Join Q2R and draw Q5R′ || QR intersecting QR produced at R′.


Step IV: Draw R′P′ || RP intersecting QP produced at P′.

Hence ∆P′QR′ is formed so that P Q QR P R 5

= = =PQ QR PR 2′ ′ ′ ′

.


6. Steps of construction: In order to draw the required pairs of tangents, follow the followingsteps:Step I: Draw a line segment AB of length 9 cm. Taking A as centre and radius 4 cm; andB as centre and radius 3 cm, draw circles.

Step II: Find the mid-point M of AB. Then, taking M as centre and radius as AM = MB,draw a circle to intersect circles drawn in step I at P, Q and R, S respectively.Step III: Join AR, AS, BP and BQ.Thus, obtained AR, AS and BP, BQ are the required pairs of tangents.


ASSESSMENT SHEET – 71. (C) The required angle say θ and the angle between the tangents are supplementary.

∴ θ + 120° = 180° ⇒ θ = 60°.

2. (C) Since AB is divided in the ratio s : t.Therefore, the minimum number of points on AX would be s + t.

3. AB is the required tangent drawn frompoint A to the circle with centre Osuch that OA = 5 cm and OB = 3 cm.

4. True, because the angle between the tangents must be less than 180°.

5.

We are given a circle with centre O and radius =72

= 3.5 cm. We take points P and Q on

its extended diameter AB. Draw tangents PP1, PP2 from P and QQ1, QQ2 from Q, whichare the required tangents.


6. We are given a ∆PQR withPQ = QR = RP = 5 cm. Todraw ∆AQB ~ ∆PQR andwith given scale factor,draw ∠RQX < 90° and markQ1, Q2, Q3 and Q4 such thatQQ1 = Q1Q2 = Q2Q3 = Q3Q4.Join Q4R and draw Q3B || Q4Rand AB || PR.

Hence ∆AQB ~ ∆PQR with

AQPQ

= QBQR

=ABPR

= 34

.

7. First we draw ∆ABC using the given measurements. Further, we follow the steps givenbelow:

(a) Draw acute angle BCX and mark C1, C2 and C3 on it such that CC1 = C1C2 = C2C3. JoinBC3 and draw B′C2 || BC3 to meet BC at B′.


(b) Draw B′A′ || BA to meet AC at A′.Thus ∆A′B′C ~ ∆ABC.Justification:

In ∆ABC and ∆A′B′C, AB || A′B′ and BC istransversal.∴ ∠ABC = ∠A′B′C′ = 90°Similarly, ∠BAC = ∠B′A′CTherefore, ∆A′B′C ~ ∆ABC.

⇒A B′ ′ΑΒ

=B CBC′

=A CAC

′...(i)

Let us take the ray CX.CC1 = C1C2 = C2C3

⇒ CC1+ C1C2 = 2CC1 and CC1 + C1C2

+ C2C3 = 3CC1

⇒ CC2 = 2CC1 and CC3 = 3CC1

⇒ 2

3

CCCC

= 23

⇒ CBCB

′ = 23

(Using Basic propor-

tionality theorem in ∆BCC3)

From (i) and (ii),

A B′ ′ΑΒ

=B CBC′

=A CAC

′ =

23

.

8. Let the common centreof the two circles be O.The point P is taken onthe outer circle. To drawa pair of tangents fromP to the inner circle, wefollow the instructionsgiven below:(a) Join PO and find itsmid-point M. Taking Mas centre, draw a circlepassing through P andO to intersect the innercircle at A and B.(b) Join PA and PB.PA and PB are therequired tangents.Length of PA: Onmeasuring, the length ofPA is 4.0 cm.Verification: Join AO.

AO = 3 cm.In ∆AOP, ∠OAP = 90°∴ PO2 = AO2 + PA2 (Pythagoras theorem)⇒ 52 = 32 + PA2

⇒ PA2 = 25 – 9⇒ PA2 = 16 ⇒ PA = 4Clearly, PA is 4.0 cm.



1. (C) Q7 to R.

2. (D) 5 + 7 = 12.

3.

AP : PB = 4 : 5.

4. False, because in the ratio 3 – 1 : 3 1+ ,

i.e., 2 – 3 : 1, 2 – 3 is not a positiveinteger, while 1 is.

5. To draw a pair of tangents from P to thecircle with centre O, we follow the steps asgiven:

(a) Join OP and find its mid-point M.

(b) Taking M as centre and radius = MP =MO, draw a circle to intersect the givencircle at A and B.

(c) Join PA and PB.

PA and PB are the required tangents.

On measuring, PA = 6.35 cm and

PB = 6.35 cm. Clearly, PA and PB are ofsame length.

6. We know that the angle between a pair of tangents to a circle and the angle betweentheir corresponding radii are supplementary. Therefore, the angle between these radii= 180° – 90° = 90°.

PA and PB are the required tangents drawn from the external point P to the circle withcentre O and radius 4 cm.

In quadrilateral OAPB formed by tangents PA, PB and radii OA, OB, each internalangle is of 90° and each side is of length 4 cm. Therefore, OAPB is a square. Perimeterof square OAPB = 4 × 4 = 16 cm.


7. First we construct a ∆ABC with thegiven measurements. Then weconstruct a ∆A′BC′ similar to ∆ABC

and scale factor 57

. For it, we follow

the steps given below.(a) Draw a ray BX such that O < ∠CBX< 90° and mark points B1, B2, B3, B4,B5, B6 and B7 on it such that BB1 =B1B2 = B2B3 = B3B4 = B4B5 = B5B6 =B6B7.

(b) Join B7C and draw B5C' || B7C tointersect BC at C′ and hence drawA′C′ || AC to intersect AB at A′.Thus ∆A′BC′ ~ ∆ABC.Justification: In ∆CBB7,

CB7 || C′B5 and 5

7

BBBB

=57

.

So, by Thale's theorem,BCBC

′=

57

...(i)

Similarly, in ∆ABC,A BAB

′=

57

...(ii)


Now, in ∆ABC,

BCBC

′=

A BAB

′=

57

[Using (i) and (ii); and ∠B = 90° (Given)]

∆A′BC′ ~ ∆ABC and A BAB

′=

BCBC

′ =

A CAC′ ′

=57

. Hence justified.

8. First we draw an isosceles triangle ABC with base BC = 7 cm and altitude AD = 4 cm.Altitude passes through the mid-point D of BC. Hence we construct a ∆A′BC′ similar to

∆ABC and of scalar factor 11

2, i.e.,

32

using following the steps given below:

(a) Draw an acute angle CBX opposite to the vertex A with respect to BC.

(b) Mark points X1, X2, X3 on ray BX such that BX1 = X1X2 = X2X3.

(c) Join X2C and draw X3C′ || X2C to meet BC produced at C′.

(d) Draw C′A′ || CA to meet BA produced at A′.

Thus formed ∆A′BC′ is similar to ∆ABC and of scalar factor 32

.


CHAPTER TEST

1. (B) A line segment can't be divided in the ratio 6 1+ : 6 – 1 , i.e., 7 2 6+ : 5 as 7 2 6+ isnot a positive integer while 5 is.

2. (C) Line segment P3Q2 divides PQ in 3 : 2 at M. Therefore, P3M : Q2M = 3 : 2 and soQ2M : P3M = 2 : 3.

3. True, because the irrational ratio 1

3 :3

can be converted into the rational ratio that is 3 : 1.

4. 180° – 60° = 120°Radius = 3 cmRequired tangents arePA and PB.

5. Steps of construction: In order to draw pairs of tangents, follow the steps given below.Step I: Draw a line segment AB = 6 cm and then taking A and B as centres, draw the circlesof radii 3 cm and 2 cm respectively.


Step II: Find M as mid-point of AB and taking it as centre and radius as AM = MB, draw acircle intersecting the circle having the centre as A at P, Q and the circle having the centre asB at R, S.Step III: Join AR, AS, BP and BQ.Thus, AR, AS and BP, BQ are the required pairs of tangents.

6. Steps of construction: In order to construct triangles ABC and AQR, follow the steps givenbelow:Step I: Draw any line XY and take any point D on it.Step II: Draw any ray DZ such that ∠ZDY = 90°. Locate point C on DZ such that CD = 3 cm.Step III: Make an ∠DCB = 30° such that CB intersects XY at B.Step IV: Locate a point A onXB such that AB = 5 cm and byjoining AC, we find ∆ABC.Step V: Make an acute angleYAT and locate T1, T2 and T3on the ray AT such thatAT1 = T1T2 = T2T3.Step VI: Join T2B and drawT3Q || T2B to intersect line AYat Q. Also, draw QR to intersectAC extended at R.Thus, ∆AQR is obtained suchthat∆ABC ~ ∆AQR and

AQ QR AR 3= = =

AB BC AC 2.

211EMOS CILPPA A T I O N S O F NOGIRT O ...

5Chapter

SOME APPLICATIONS OF TRIGONOMETRY

WORKSHEET – 57

1. (A) cos 30° =3l

⇒ l = 2 3 m.

2. (B) sin 30° = 65l

⇒ l = 130 m.

3. Let the height of thetower be h.

tan 30° = Perpendicular

Base

⇒ 13

= 30h

⇒ h = 10 3 m.

4. Let OA be the horizontalground and K be theposition of the kite at aheight h m above theground, then AK = h m. Itis given that OK = 100 m,∠AOK = 60°.In ∆AOK, right angled at A, we have

sin 60° = 100

h ⇒ h = 100 sin 60°

⇒ h = 100 × 3

2 = 50 3 = 50 × 1.732

∴ h = 86.60 m.

5. 4.28 m, 2.14 m

Hint: sin 60° = 3.7l

tan 60° = 3.7x

.

6. Height = 94.64 m, Distance = 109.3 mHint:

tan 45° = QMYM

⇒ YM = QM

But XP = YM∴ XP = QM

tan 60° = 40 + QMQM

.

7. Let BD be the tower ofheight h m and CD bethe pole. In right-angledtriangle ABD,

tan 45° = BDAB

⇒ 1 = ABh

⇒ AB = h

In right-angled triangle ABC,

tan 60° = BCAB

⇒ 3 = BD + CD

AB

⇒ 5h +h

= 3

⇒ h = −

53 1

⇒ h =−

51.732 1

⇒ h = 6.83 m.

8. Let the chimney be AB with base B andanyone is walking from the point C to D.In ∆ABD, ∠B = 90° and ∠D = 45°∴ ∠DAB = 45° ⇒ BD = BA ...(i)In right angled ∆ABC,

tan 30° = ABBC

⇒ 13

= ABCD + BD

⇒13

= AB

50 + AB[Using equation (i)]

⇒ 3 AB = 50 + AB


⇒ AB = 3 150

×3 –1 3 1

++

= ( )+50 3 1

2

⇒ AB = 25 ( )+3 1 = 25 (1.732 + 1)

= 68.30 m.

WORKSHEET – 58

1. (A) ∠ACB = ∠XAC = 45°

sin (∠ACB) = 20x

and tan (∠ACB) = 20y

⇒ x = 20 2 m and y = 20 m.

2. (D) tan 60° =20h

.

⇒ 3 =20h

⇒ h = 20 3 m.

3. Let the length ofshadow of poleAB be BC = x, thenAB = x.Also let θ be theangle of elevation ofSun's altitude.In right-angledtriangle ABC,

tan θ = xx ⇒ θ = 45°

Hence the angle of elevation of the Sun'saltitude is 45°.

4. Let the angle ofelevation be θ. Letthe observer be ABwith his eye at Aand the tower beEC.

∴ CD = AB = 1.5 m

ED = 30 – 1.5 = 28.5 mAnd AD = BC = 28.5 mIn right-angled ∆ADE,

tan θ = = =DE 28.51

AD 28.5 ⇒ θ = 45°.

5. Let the balloon be at the point O, the threadbe OA and the required height be OB.Case I. The cable is inclined at 60°.

⇒ sin 60° = OBOA

⇒ 32

= OB215

⇒ OB = 215 3

2

= 215 × 1.732

2= 186.19 m.

Case II. The cable is inclined at 60° – 15°= 45°

sin 45° = OBOA

⇒ =1 OB2152

⇒ OB = × =215 2 215 222 2

= 215 × 1.414

2 = 152 m (approx.)

So, reduced height = 186.19 m – 152 m= 34.19 m.

6. Let AB is a hill and C and D be two citycentres subject to the angles of elevationof the top A of hill AB at C and D are 30°and 60° respectively, then ∠ACB = 30°,∠ADB = 60°, AC = 9 km.In right-angled ∆ABC,

sin 30° =AB9

⇒ AB = 9 × sin 30° = 9 ×12

= 4.5

In right-angled ∆ABD, we have

sin 60° =ABAD

⇒ AD = AB cosec 60°

⇒ AD = 4.5 ×23

= 9 × 33 × 3

= 3 3 = 3 × 1.732

= 5.196 ≈ 5.20 km.


3. True.

tan C = ABBC

= hx

If AB = 1110

h

and BC = 1110

x

Then, tan C = ABBC

= hx

.

4. Let the height of thetower CD be y metresand the horizontaldistance of point Afrom the building BCis AB = x metres.In right-angled triangleABC,

tan 45° = 20x

⇒ x = 20 m

Also, in right-angled triangle ABD,

tan 60° =+20 yx

⇒ 20 3 = 20 + y

⇒ y = ( )20 3 1− m

Thus, the height of the tower is ( )−20 3 1 m.

5. ( )7 3 1+ m

Hint: tan 45° =AEEC

⇒ EC = 7m

tan 60° = DEEC

.

6. Let the width of the river be AC such thatAC = AB + BC

Let P be the point on the bridge such thatBP = 3 m.

7. Let the ladder and thewall upto which theladder reaches be AC andBC respectively.(a) In ∆ABC,

sin 30° = BCAC

⇒ 12

= BC4

⇒ BC = 2 m.

(b) Also, cos 30° =ABAC

3

2 =

AB4

⇒ AB = 2 3 m.

8. 2 m

Hint: h = height of pedestal

tan 45° = hx

⇒ x = h

tan 60° = 1.46h +h

.

WORKSHEET– 59

1. (B) In ∆ABC,

tan A = BCAB

⇒ tan A = BC3 BC

(... AB = 3 BC)

tan A = 13

= tan 30°

∴ Angle of elevation is 30°.

2. (C) In ∆ABC,

tan 60° = ABBC

⇒ 3 = 30BC

BC = 30

3, ∴ BC = 10 3 m.


In right-angled triangle ABP,

tan 30° = 3

AB⇒ AB = 3 3 m

Also in ∆CBP, tan 45° =3

BC⇒ BC = 3 m

Now, AC = AB + BC= 3 3 + 3 = 3( 3 + 1) mHence, width of the river is 3( 3 + 1) m.

7. Let the tower be BC the flagstaffbe AB and the point on the planebe P.Let BC = hIn right-angled ∆BCP,

tan 30° = PCh

⇒ PC = h cot 30° ...(i)In right-angled ∆ACP,

tan 60° = 5+PC

h

⇒ PC = (5 + h) cot 60° ...(ii)Comparing equations (i) and (ii), we have

h cot 30° = (5 + h) cot 60°

⇒ 3h = (5 + h) 13

⇒ 3h = 5 + h⇒ h = 2.5Hence, the height of the tower is 2.5 m.

ORLet the two planes be at A and B respectively.Also P be the point on the groundIn right-angled triangle APC,

tan 30° = 3125PC

⇒ PC = 3125 3 m

Also in right-angled triangle BPC,

tan 60° = BCPC

⇒ BC = 3125 3 × 3 = 3 × 3125∴ AB = BC − AC = 3 × 3125 – 3125

= 2 × 3125 = 6250 m.Hence, distance between the two planes is6250 m.

8. Let the tower be PQ and the objects be Aand B.∵ ∠XQA = 45°and ∠XQB = 60°∴ ∠QAP = 45°and ∠QBP = 60° (Alternate angles)In right ∆APQ,

∠PAQ + ∠PQA = 90°⇒∠PQA = 90° – 45° = 45° (∵∠PAQ = 45°)∴ AP = PQ = 150 (∵ PQ = 150 m)⇒ AB + BP = 150 ...(i)In right ∆BPQ,

tan 60° = PQBP

⇒ 3 = 150BP

⇒ BP = 150

3...(ii)

Putting BP = 150

3 in equation (i), we get

AB + 150

3= 150 ⇒ AB = 150

11 –

3

⇒ AB = 150 × 3 – 1

3 ×

33

= 50 × ( )3 – 3

= 50 (3 – 1.73) = 50 × 1.27⇒ AB = 63.50 mThus, distance between the two objects is63.50 m.

WORKSHEET – 60

1. (C) tan 45° = OPPQ

= QPa

⇒ 1 = QPa

⇒ QP = a m

∴ ar(∆OPQ) = 12

× QP × OP

= 12

× a × a = 12

a2.


2. (B) tan 60° = TPPO

⇒ 3 = TP40

⇒ TP = 40 3 m.

3. True

tan 30° =ABBC

⇒ 13

= AB81

⇒ AB =81

3 = 81 3

3

⇒ AB =81×1.732

3 = 46.76 m.

4. Let the height of the pole AB = x mLength of the rope AC = 20 m

In ∆ABC,∠ACB = 30º

∴ sin 30° = ABAC

⇒12

= 20x

⇒ x = 10 m∴ Height of the pole = 10 m.

5. Let the tree be PQ, the width of the river beMP and the person moves from M to S.In right ∆PQS,

tan 45° = PQPS

⇒ 1 = PQPS

⇒ PS = PQ⇒ PM + MS = PQ⇒ PM + 40 = PQ ...(i)In right ∆PQM,

tan 60° = PQPM

⇒ = PQ3

PM⇒ PQ = 3 PM ...(ii)From equations (i) and (ii),

PM + 40 = 3 PM

⇒ PM = 3 + 140

×3 – 1 3 + 1

⇒ PM = ( )20 3 + 1 ... (iii)

From equations (ii) and (iii),

PQ = ( ) ( )3 × 20 3 + 1 20 3 + 3=

Hence height of the tree and the width of the

river are ( )20 3 + 3 m and ( )20 3 + 1 m

respectively.

6. 6.34 m

Hint: tan 45° =1 2+

15

y y

y1 + y2 = 15 …(i)

tan 30° =1

15

y

⇒ y1 =15

3…(ii)

∴ y2 = 15 − y1 = 6.34 m.

7. Distance = 17.32 m, Height = 40 m

Hint: tan 30° = 10x

⇒ x = 10 3 m

tan 60° =10 3

y

⇒ y = 30 m.

8. Let the plane moves from P to Q in 30seconds and the points H and K on theground be just below the points P and Qrespectively.

In right ∆AQK,

tan 30° = QKAK

⇒ 13

= 3600 3AK


⇒ AK = 10800 ...(i)In right ∆APH,

tan 60° = PHAH

⇒ 3 = 3600 3

AH

[∵ PH = QK = 3600 3 ]

⇒ AH = 3600 ...(ii)

Now, PQ = HK = AK – AH

⇒ PQ = 10800 – 3600

[Using (i) and (ii)]

⇒ PQ = 7200

∵ Speed = Distance travelled

Time

∴ Speed = PQ 7200=

30 30

= 240 m/s = 3600

240 ×1000

km/h

= 864 km/h.

WORKSHEET – 61

1. (A) tan C = ABBC

⇒ tan C = ABAB

(∵ BC = AB)

⇒ tan C = 1 = tan 45°

⇒ C = 45°.

2. (D) tan C = ABBC

⇒ tan 60° = 15BC

⇒ 3 = 15BC

⇒ BC = 15

3⇒ BC = 5 3 m.

3. False.Let height of the tower is h metres so theangle of elevation is 30°.

tan 30° = BCh

⇒13

= BCh

⇒ BC = h 3 ... (i)

When height = 2h,

tan θ = ABBC

⇒ tan θ = 2

3h

h[From (i)]

⇒ tan θ = 23

≠ tan 60°.

4. Let AB be the ladder leaning against a wallOB such that ∠OAB = 60° and OA = 9.6 m.In ∆OAB right angled at O, we have

cos 60° = OAAB

⇒ AB = OAcos 60°

⇒ AB = 9.60.5

= 19.2 m.

5. Let the point, cloud and reflection of thecloud be at P, Q and Q′ respectively.

Let PM = x, QM = y

We have to find QB, i.e., y + h

In right-angled triangle QPM,

tan α = yx

⇒ x = αtan

y… (i)


Also in right-angled triangle Q′PM,

tan β = + 2y hx

⇒ tantan

y βα

= y + 2h

[From equation (i)]

⇒ y β

− α

tan1

tan= 2h

⇒ y + h = h 2 tan

1tan tan

α+ β − α

=( )tan tan

tan tan

h β + αβ − α

.

Hence proved.

6. Hint:

ED = AC = length of ladder = l (say)

Now, cos α =BC y

l=

, sin α =+ b xl

,

cos β = + y al

, sin β = xl

,

Consider,cos cossin sin

α − ββ − α

=

y y al lx b xl l

+−

+−

=ab

−− =

ab

.

7. 250 m

Hint: tan α = hx

= 57

tan β = +150

hx

= 12

.

8. Let the tower, the flagstaffand the point on the planebe AB, BC and P respec-tively.

Let AB = y and AP = xIn ∆ABP,

tan α = yx

⇒ 1x

= αtan

y...(i)

In ∆ACP,

tan β = +h yx

⇒ 1x

= β

+tanh y

...(ii)


αtany

= β

+tanh y

⇒ y tan β = h tan α + y tan α⇒ y (tan β – tan α) = h tan α

⇒ y = α

β αtan

tan – tanh

. Hence proved.

WORKSHEET– 62

1. (C) sin 30° = ABAC

⇒12

= AB15

⇒ AB = 152

m.

2. (B) tan C = ABBC

= 3h

h

⇒ tan θ = 13

= tan 30°

⇒ θ = 30°.


3. True.

As tan θ = ABBC

⇒ tan θ = ABAB

(∵ AB = BC)

⇒ tan θ = 1 = tan 45°∴ θ = 45°.

4. Let AB be the tower and C be the point onthe ground.

In ∆ABC,

tan 30° = ABBC

⇒13

= AB30

⇒ AB = 30

3 = 10 3 m.

Hence, the height of the tower is 10 3 m.

5. Let the height (PQ) of mountain be h km.In right-angled ∆ABC,

sin 30° =BC1

⇒ BC =12

km

⇒ DP = 12

km …(i)

Also, cos 30° = AB1

⇒ AB =3

2km …(ii)

Similarly, from right-angled ∆APQ,AP = PQ = h …(iii)

And from right-angled ∆CDQ ,

3 = DQCD

⇒ 3 = −−

DPAB

hh

[From (iii)]

⇒ 3 =

123

2

h

h

−

−[From (i) and (ii)]

⇒ h =1

3 1−=

3 12+

= 1.366 km.

6. Let the aero-plane's firstsituation be atA and secondat B. Let thepoint of obser-vation be at O.From right-angled ∆AOD,

tan 45° = ADOD

⇒ OD = 3000 m

Again from right-angled ∆BOC,

tan 30° = 3000

3000 + DC

⇒ DC = 3000 ( )−3 1 m

Now speed of the plane

= Distance

Time=

( )3000 3 1

15

−

[ ]D C AB=∵= 146.42 m/sec

7. Let the height ofthe tower PQ andthe width of thecanal AP be h andx respectively.Let the anotherpoint be B such that

AB = 20 mIn right ∆APQ,

tan 60° = hx

h = 3 x ...(i)In right ∆BPQ,

tan 30° = +20h

x


⇒ h 3 = 20 + x ... (ii)


3 x × 3 = 20 + x ⇒ 2x = 20

⇒ x = 10

Substitute x = 10 in equation (i) to get

h = 10 3

Hence, the height of the tower is 10 3 mand the width of the canal is 10 m.

8. Let O be centre ofthe balloon of radiusr and P the eye ofthe observer. Let PA,PB be tangents fromP to the balloon.Then,

∠APB = α

... ∠APO = ∠BPO =2α

... OL ⊥ PX, ∠OPL = β

... In ∆OAP, sin2α

=OAOP

⇒ OP = r cosec2α

In ∆OPL, sin β = OLOP

⇒ OL = r cosec2α

sin β.


1. (C) Given:

AB : BC = 1 : 13

i.e., AB : BC = 3 : 1

i.e.,ABBC

= 3

1

∴ tan θ = ABBC = 3 = tan 60°

⇒ θ = 60°.

2. (A) sin 30° = BCAB

⇒12

=BC5

⇒ BC =52

m.

3. Wire is AB.CE = BD = 14 m.AE = AC + CE

⇒ 20 = AC + 14⇒ AC = 6 mIn ∆ABC, ∠C = 90°,

sin 30° = 6

AB

⇒12

= 6

AB ⇒ AB = 12 m.

4. False, because the tangent of the angle ofelevation doubles notthe angle of elevation.

5. BC is the multi-storeyedbuilding with the footB and the top C as thepoint of observation.AD is the building withbottom A and the top D. Draw DE || AB(see figure).Given angles are ∠XCD = 30° and∠XCA = 45°. ∠CDE and ∠XCD are alternateinterior angles.∴ ∠CDE = ∠XCD = 30°.Similarly, ∠CAB = ∠XCA = 45°

BE = AD = 8 m

In right triangle ABC,

tan 45° = CE+BE

AB⇒ 1 =

CE+8AB

⇒ AB = CE + 8 ...(i)

Also, in right triangle DCE,

tan 30° = CEDE

⇒ 13

= CEAB

(∵ DE = AB)

⇒ AB = 3 CE ...(ii)


From equations (i) and (ii), we get

( )3 – 1 CE = 8

⇒ CE = 8 3 + 1

×3 – 1 3 + 1

= 8 × (1.73 + 1)

3 – 1 = 10.92 m.

Substituting CE = 10.92 in (i), we getAB = 10.92 + 8 = 18.92 m

Further, BC = BE + CE = 8 + 10.92= 18.92 m

Hence, the height of the multi-storyedbuilding and the distance between the twobuildings is 18.92 metres each.

6. Let AB be the first towerwith bottom A and CDbe the second towerwith bottom C.

BE = 80 m

CD = 160 m

AB = CE

∵ XD || BE and BD is the transversal

∴ ∠DBE = ∠XDB = 30°

In right triangle BDE,

tan 30° = DEBE

= CD – CE

BE

⇒ 13

=160 – AB

80

⇒ 160 – AB = 80

3

⇒ AB = 160 – 80 80 3

160 –33

=

= 480 – 80 3

3

= 480 – 80 × 1.7323

= 113.81

Hence, the height of the first tower is 113.81metres.

7. Let the lower window, upper window andthe balloon be at A, B and C respectively.AB = AD = EF = FG = 2 m ...(i)BG = AF ...(ii)In right triangle BCG,

tan 30° = CGBG

⇒13 =

CGBG

⇒ CG = 13 BG

...(iii)

In right triangle ACF,

⇒ tan 60° = CFAF

⇒ 3 = CG+2

BG[Using (i) and (ii)]

⇒ BG = CG+2

3...(iv)

From equations (iii) and (iv),

CG = 13

× CG+2

3

⇒ 3CG = CG + 2⇒ CG = 1 ...(v)Now, CE = CG + FG + EF

= 1 + 2 + 2 [Using (i) and (v)]= 5

Hence, the height of the balloon is 5 metres.

8. Let the tower be AB, the flagstaff BC andthe point on the plane P.Let AB = y and AP = xIn ∆PAB,

tan α = ABAP

= yx

...(i)

In ∆PAC,

tan β = ACAP

⇒ tan β = +h yx

...(ii)


From equations (i) and (ii),

x = αtan

y =

+βtan

h y

⇒αtan

y = βtan

h +

βtany

⇒ α β

1 1–

tan tany =

βtanh

⇒β αα β

tan – tantan tan

y = βtan

h

⇒ y = α

β αtan

tan – tanh

Hence, the height of the tower AB isα

β αtan

tan – tanh

.


1. (A) cos 45° = 10 2AC

⇒ 12

= 10 2AC

⇒ AC = 20 m.

2. (B) ∠YAM = ∠XYA = 45°; tan 45° = 25

AM⇒ AM = 25 m

Time =255

= 5 seconds.

3. Let the kite be at A and the thread AB.AC = 75 m;

∠ABC = 60°

sin 60° = ACAB

⇒ 32

= 75AB

⇒ AB = ×150 33 3

= 50 3 = 50 × 1.732

⇒ AB = 86.6

⇒ The length of the string to the nearestmetre is 87 metres.

4. False, the angle of elevation will remainunchanged if the height of the tower isincreased by 10% too.

5. Let the trucks be at Aand B; the balloon at Pand the vertical line PQ.∠XPA = 45°∠XPB = 60°∠PAQ and∠PBQ are alternateinterior angles with∠XPA and ∠XPB respectively.∴ ∠PAQ = 45° and ∠PBQ = 60°In right ∆PAQ,

tan 45° = PQAQ

⇒ 1 = PQ

100 + BQ

⇒ BQ = PQ – 100 ...(i)In right ∆PBQ,

tan 60° = PQBQ

⇒ 3 = PQBQ

⇒ PQ = 3 BQ = 3 (PQ – 100)[Using (i)]

⇒ PQ ( )3 – 1 = 100 3

⇒ PQ = 100 3

3 – 1

⇒ PQ = 100 3

3 – 1 ×

++

3 1

3 1

= 50 ( )3 3+

Therefore, height of the balloon is

50 ( )+3 3 metres.

6. Let P be the point on the bridge; A and Bare the two points on the opposite banks ofthe river such that AB is the width of theriver. ∠X1PA = 45°, ∠X2PB = 30°.Draw PC ⊥ AB to meet AB at C.

∠PAB = ∠X1PA

= 45°

∠PBA = ∠X2PB

= 30°.


In right ∆ PAC,

tan 45° = PCAC

⇒ 1 = 30

AC ⇒ AC = 30

In right ∆PBC,

tan 30° = PCCB

⇒ 13

=30CB

⇒ CB = 30 3

Now, AB = AC + CB = 30 + 30 3

= 30 + 30 × 1.732

= 30 + 51.96 = 81.96Hence the width of the river is 81.96 metres.

7. Let the wall be AM.Let the ladder changes its position fromAB to A′B′. Let the length of the ladder bel such that

AB = A′B′ = l

Let BM = y and

A′M = z

In right triangleABM,

sin α = AMAB

⇒ sin α = +

3x

z

l

⇒ 3 sin α = + 3x zl

...(i)

And cos α = BMAB

⇒ cos α = yl

...(ii)

Add equations (i) and (ii) to get

3sin α + cos α = + + 3x y z

l...(iii)

In right triangle A′B′M,

sin β = ′′ ′

A MA B

= zl

⇒ 3 sin β = 3zl

...(iv)

And cos β = ′′ ′

B MA B

=+x yl

...(v)

Add equations (iv) and (v) to get

3sin β + cos β = + + 3x y z

l...(vi)

Divide equation (iii) by equation (vi) to get

α + αβ+ β

3sin cos3sin cos

= 1. Hence proved.

8. Let the observer be at P on the ground.

When the plane is at A vertically above C,∠APC = 60°.When the plane is at B vertically above D,∠BPD = 30°.

AC = BD = 1.2 km = 1200 metres.AB = CD

In right triangle APC,

⇒ tan 60° = ACPC

⇒ PC = AC

tan 60° =

12003

In right triangle BPD,

tan 30° = BDPD

= BD

PC+CD

⇒ 13

= +

12001200

AB3

⇒ 1200

3 + AB = 1200 3

⇒ AB = 1200 3 –1200

3

⇒ AB = 2

1200 ×3

AB = 800 × 3 metres


Speed = =Distance travelled AB

Time 15sec

= 800 3 m

15sec

=

1800 3 km

100015

h60 60

×

×

= 800 3 60 60

1000 15× × ×

× km/h

= 192 3 km/h = 192 × 1.73 km/h

= 332.16 km/hHence, the aeroplane is flying at a speed of332.16 km/hr.

CHAPTER TEST

1. (A) From the adjoin-ing figure, angle ofdepression of P is∠XOP = α and angleof depression of Q is

∠XOQ = 90° – β.

2. (B) tan 30° = 20 3h

⇒ h = 20 m.

3. Let the tower be BC and the length ofshadow be AB.

tan 60° = BCAB

⇒ 3 = 20AB

⇒ AB = 20

3m

⇒ AB = 20 33

m.

4. True, because the vertical tower, length ofthe shadow and the ray of the sun make aright angled isosceles triangle.

5. Let the ships be at A and B; and the towerbe PQ.

∠PAQ = ∠XPA = 30°∠PBQ = ∠XPB = 45°

In right ∆ BPQ,∵ ∠PBQ = 45°, ∴ ∠BPQ = 45°⇒ BQ = PQ = 75 ...(i)In right ∆ PAQ,

tan 30° = PQ

AB + BQ

⇒ AB + 75 = 75 3 [Using (i)]

⇒ AB = ( )75 3 – 1 m.

6. 8 3 m

Hint: A′C = AC

cos 30° = 8

AC

tan 30° = BC8

.

7. Let the window beat P and height ofthe opposite housebe h.In right ∆ APQ,

tan 45° = 60

AQ⇒ AQ = 60 ⇒ BP = 60

In right ∆ BCP,

tan 60° = – 6060

h ⇒ 60 3 = h – 60

⇒ h = 60 + 60 3 = 60 ( )1 3+

Thus, the required height is 60 ( )1 3+ m.


8. Let the cloud be at C, the point ofobservation be at P and the reflection ofthe cloud in the lake be at D. Let Q be anypoint just below the cloud, 60 m above thewater level.

In right ∆CPQ,

tan 30° = CQPQ

⇒13 =

– 60PQ

h

⇒ PQ = ( )3 – 60h ...(i)

In right ∆DPQ,

tan 60° = + 60PQ

h

⇒ PQ = + 60

3h

...(ii)

Using equations (i) and (ii), we have

⇒ ( )3 – 60h = + 60

3h

⇒ 3h – 180 = h + 60⇒ h = 120Hence, height of the cloud is 120 metres.

❑❑

225BORP TILIBA Y

6Chapter

PROBABILITY

WORKSHEET – 661. (C) Sample space: {HH, HT, TH, TT}

Favourable events: {HT, TH}

∴ Required probability = 2 1

=4 2

.

2. (D) Probability of non-happening of anevent = 1 – Probability of happening of

that event

=3

17

− = 47

.

3. (C) Sample space is {1, 2, 3, 4, 5, 6} andfavourable events are {2, 3, 5}

∴ Required probability = 36

= 12

.

4. No, because the number of favourable out-comes of getting ‘6’ and ‘not 6’ are respecti-vely 1 and 5; and so their probabilities are

16

and 56

.

5. Sample space: {1, 2, 3, ......... ,99}∴ n(S) = 99.The numbers divisible by 3 and 5 both arenumbers divisible by 15.So, favourable outcomes are: {15, 30, 45, 60,75, 90}Let E be the event getting a numberdivisible by 3 and 5.∴ n(E) = 6

∴ P(E) = (E) 6 2(S) 99 33

nn

= = .

6. (i) 123

(ii) 546

Hints:(i) Prime numbers are 5 and 7.

(ii) Perfect square numbers are

9, 16, 25, 36, 49.

7. Let A = The event that 5 will not come upeither time.Now sample space is given by

S ={(1, 1), (1, 2), (1, 3), (1, 4), (1, 5),(1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5),(2, 6), (3, 1), (3, 2), (3, 3), (3, 4),(3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4),(4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4),(6, 5), (6, 6)}

Total number of outcomes in sample spacen(S) = 36

∴ A = {(1, 5), (2, 5), (3, 5), (4, 5),(5, 1), (5, 2), (5, 3), (5, 4),(5, 5), (5, 6), (6, 5)}

∴ n ( )A = 11

∴ n(A) = n(S) – n ( )A = 36 – 11 = 25

(i) P(A) = (A)(S)

nn

=2536

(ii) P ( )A = (A)(S)

nn

=1136

.

8. Total number of marbles = 5 + 8 + 4 = 17i.e., n(S) = 17(i) Let E1 be the event ‘a red marble is

taken out’.∴ n(E1) = 5

Now, P(E1) = 1(E ) 5(S) 17

nn

= .

(ii) Let E2 be the event ‘a white marble istaken out’.∴ n(E2) = 8

Now, P(E2) = 2(E ) 8(S) 17

nn

= .

(iii) Let E3 be the event ‘a non-green marbleis taken out’.∴ n(E3) = 17 – 4 = 13

Now, P(E3) = 3(E ) 13=

(S) 17nn

.


9. The sample space isS = {1, 2, 3, 4, 5, 6, 7, 8}

∴ n(S) = 8(i) Let E1 be the event that the arrow will

point at 8, thenn(E1) = 1

∴ P(E1) = 1(E )(S)

n n

= 18

.

(ii) Let E2 be the event that the arrow willpoint at 1, 3, 5 or 7; then

n(E2) = 4

∴ P(E2) = 2(E ) 4 1= =

(S) 8 2nn

.

(iii) Let E3 be the event that the arrow willpoint at 3, 4, 5, 6, 7, or 8; then

n(E3) = 6

∴ P(E3) = 3(E ) 6 3= =

(S) 8 4nn

.

(iv) Let E4 be the event that the arrow willpoint at 1, 2, 3, 4, 5, 6, 7 or 8; then

n(E4) = 8

∴ P(E4) = 4(E )(S)

nn

= 88

= 1.

WORKSHEET – 671. (Α)

Hint: Outcomes in favourable event ofgetting the sum as a perfect square are(1, 3), (2, 2), (3, 1), (3, 6), (4, 5), (5, 4), (6, 3).

2. (D)Hint: |x| ≤ 4 ⇒ – 4 ≤ x ≤ 4

⇒ x = – 4, – 3, – 2, – 1, 0, 1, 2, 3, 4.

3. (B) P(getting a number less than 3)

=2 16 3

= .

4. No, because the theoretical probability of

getting a head on tossing a coin is 12

and

the experimental probability tends to 12

when the number of tosses increases.

OR

n(S) = 100Let E be the event of getting a prime.The primes from 1 to 100 are:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.∴ n(E) = 25

Now, P(E) =(E) 25 1(S) 100 4

nn

= = .

5. Let S be the sample space and E be theevent that the sum of numbers appearingon the dice is a prime.∴ E = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1),

(2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),(5, 2), (5, 6), (6, 1), (6, 5)}

∴ n(E) = 15, n(S) = 36

Now, P(E) =(E)(S)

nn

= 1536

.

6. The sample space isS = {HH, HT, TH, TT}

∴ n(S) = 4(i) The outcomes for at least one head:

{HH, HT, TH}

∴ Probability (at least one head) = 34

.

(ii) The outcomes for at most one head:{HT, TH, TT}

∴ Probability (at most one head) = 34

.

(iii) The outcomes for one head: {HT, TH}

∴ Probability (one head) = 24

= 12

.

OR

Total number of students = 23∴ n(S) = 23Let E be the event that the selected studentis not from A, B and C.∴ n(E) = 23 – 4 – 8 – 5 = 6

Now, P(E) =(E) 6(S) 23

nn

= .

227BORP TILIBA Y

7. All possible outcomes are(1 × 1), (1 × 4), (1 × 9), (2 × 1), (2 × 4), (2 × 9),(3 × 1), (3 × 4), (3 × 9),i.e., 1, 4, 9, 2, 8, 18, 3, 12, 27.∴ n(S) = 9∴ n(favourable outcomes) = 5

Hence, probability (xy is less than 9) = 59

.

8. Let S be the sample space, thenn(S) = 52

(i) Let E1 be the event that the card drawnis neither a hearts nor a king.Number of hearts = 13Number of kings = 4But one king is of hearts.∴ n(E1) = 52 – 13 – 4 + 1 = 36

Now, P(E1) = 1(E )(S)

nn

=3652

=9

13.

(ii) Let E2 be the event that the card drawnis an ace of spades.Since number of ace of spades is 1∴ n(E2) = 1

∴P(E2) = 152

(iii) Let E3 be the event that the card drawn is either a black card or a king.Number of black cards

= Sum of numbersof cards of clubsand spades

= 13 + 13 = 26Number of kings = 4

But 2 kings are black∴ n(E3) = 26 + 4 – 2 = 28

∴ P(E3) =28 752 13

= .

9. Number of outcomes in sample spacen(S) = 62 = 36

(i) Favourable outcomes areE1 = {(1, 1), (1, 2), (1, 4), (1, 6),

(2, 1), (2, 3), (2, 5), (3, 2), (3, 4),(4, 1), (4, 3), (5, 2), (5, 6), (6, 1),(6, 5)}

∴ n(E1) = 15 ∴ P(E1) = 1536

= 5

12 .

(ii) Favourable outcomes areE2 = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5),(6, 6)}

∴ n(E2) = 6 ∴ P(E2) = 636

=16

.

(iii) Favourable outcomes areE3 = (2, 2), (4, 4), (6, 6)

∴ n(E3) = 3 ∴ P(E3) = 3 1

=36 12

.

(iv) Favourable outcomes areE4 = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3),(6, 6), (3, 2), (3, 6), (3, 4), (6, 4), (6, 2)}

∴ n(E4) = 11 ∴ P(E4) = 1136

.

WORKSHEET – 68 1. (C)

Hint: Sample space is {1, 2, 3, 4, 5, 6} andset of favourable numbers is {2, 4, 6}.

2. (A)Hint: The sum of probabilities of having aparticular event and not having the sameevent is one.

3. (D) Total number of coins= 100 + 50 + 20 + 10 = 180

So, total number of elementary events= 180

Favourable number of elementary events= 180 – 10 = 170


= 1718

.

4. False, because the probability of each

outcome will be 12

only when the two

outcomes are equally likely otherwise not.

5. No, because areas of regions 3, 5 and 7 arenot equal.

6. All possible outcomes are given by{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),


(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

(i)536

Hint: Favourable outcomes:(2, 6), (6, 2), (3, 5), (5, 3), (4, 4).

(ii) 0Hint: No, favourable outcome is possible.

(iii) 1Hint: Favourable outcomes are thesame as the outcomes in sample space.

7. Let the number of black balls be x.So, total number of balls = x + 5Probability of drawing a black ball is

P1 = + 5x

x

Also, probability of drawing a red ball is

P2 = 5 + 5x

According to the question, we haveP1 = 2.P2

⇒ 2 × 5

= + 5 + 5x

x x ⇒ x = 10.

Hence, the number of black balls is 10.

8. Number of all cards = 50 – 5 + 1 = 46i.e., n(S) = 46(i) Let E1 be the event that the number

on the card taken out is a prime lessthan 10.Prime numbers from 5 to 9 are 5and 7.∴ n(E1) = 2

∴ P(E1) = 1(E ) 2 1(S) 46 23

nn

= = .

(ii) Let E2 be the event that the number onthe card taken out is a perfect square.The perfect square numbers from5 to 50 are 9, 16, 25, 36 and 49.

∴ n(E2) = 5

∴ P(E2) = 2(E )(S)

nn

=546

.

9. Number of all possible outcomes,n(S) = 52

(i) Number of spades = 13Number of aces = 4

But 1 ace is of spades.∴ Number of favourable outcomes

= 13 + 4 – 1 = 16∴ P(card drawn is a spades or an ace)

= 1652

= 4

13.

(ii) Number of red kings = 2∴ P(card drawn is a red king)

=2 152 26

= .

(iii) Number of kings = 4Number of queens = 4∴ Number of favourable outcomes

= 52 – 4 – 4 = 44∴ P (card drawn is neither a king nor aqueen)

=44 11

=52 13

.

(iv) Number of kings = 4Number of queens = 4

Number of favourable outcomes= 4 + 4 = 8

∴ P(card drawn is either a king or a

queen) =8 2

=52 13

.


1. (D) We know that probability (P) of any

event can be 0 ≤ P ≤ 1. Therefore, 1514

> 1

can’t be probability of an event.

2. (B) Let the number of bolts be x. Then

400x

= 0.035 ⇒ x = 400 × 0.035 ⇒ x = 14.

229BORP TILIBA Y

3. Number of cards = 50Prime numbers from 51 to 100 are: 53, 59,61, 67, 71, 73, 79, 83, 89, 97Therefore, number of all possible outcomes

= 50.And number of favourable outcomes = 10.

∴ Required probability = 10 1

=50 5

.

4. True,Ratio of probabilities

= Ratio of areas of regions a, b and c= Ratio of areas of corresponding sectors= Ratio of corresponding angles= 60° : 120° : 180° = 1 : 2 : 3.

5. A leap year contains 366 days, wherever52 weeks and 2 days. There are 52Thursdays in 52 weeks. Therefore, a leapyear consists 52 Thursdays and 2 days.These two days may be one choice out ofthe seven given below:

(i) Thursday and Friday(ii) Friday and Saturday

(iii) Saturday and Sunday(iv) Sunday and Monday(v) Monday and Tuesday

(vi) Tuesday and Wednesday(vii) Wednesday and ThursdaySo, two days are either Thursday andFriday or Wednesday and Thursday.Clearly, the number of all possibleoutcomes is 7 and the number of favourableoutcomes is 2.

Hence, required probability = 27

.

6. Let number of blue marbles= xNumber of green marbles = yNumber of white marbles = zTherefore, x + y + z = 54

P(selecting a blue marble) =x

x y z+ +

⇒ 13

=54x

⇒ x =54

= 183

P(selecting a green marble) =y

x y z+ +

⇒ 49

=54y

⇒ y =4 × 54

9 = 24

Now, substituting x = 18, y = 24 inx + y + z = 54, we get

18 + 24 + z = 54⇒ z = 54 – 42⇒ z = 12Hence, the jar contains 12 white marbles.

7. Number of blue triangles = 3Number of red triangles = 8 – 3 = 5Number of blue squares = 6Number of red squares = 10 – 6 = 4

Number of all possible outcomes= 8 + 10 = 18

(i) P(a lost piece is a triangle)

=8 4

18 9= .

(ii) P(a lost piece is a square)

=10 518 9

= .

(iii) P(a lost piece is a blue square)

=6 1

18 3= .

(iv) P(a lost piece is a red triangle)

= 518

.

8. Let E be the event of placing at least oneletter in the wrong envelope. Anyoneenvelope can be filled with a letter in 4ways. Second one can be filled by 3 ways.Third one can be filled by 2 ways. And thefourth one can be filled with the remainingone letter in 1 way. Therefore, the fourletters are placed in the four envelopes in4 × 3 × 2 × 1 = 24 ways. Out of these 24


ways, only 1 way is such that all the lettersare placed in the right envelopes and 23ways are such that at least one letter isplaced in the wrong envelope.

∴ P(E) =2324

.


1. (B) P(getting X or Y)= P(getting X) + P(getting Y)

=2 1 3 1

+ = =6 6 6 2

.

2. (C) Given: P(E) = 3 P(E’) ... (i)We have P(E) + P(E’) = 1 ... (ii )(i ) and (ii ) gives P(E) = 3 {1 – P(E)}

i.e., 4 P(E) = 3, i.e., P(E) = 34

.

3. The number of outcomes when a pair ofdice is rolled = 62 = 36.The outcomes such that the sum is divisibleby 3 are:(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2),(4, 5), (5, 1), (5, 4), (6, 3), (6, 6).These are 12 outcomes.The outcomes such that the sum is divisibleby 2 are:(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 1),(3, 3), (3, 5), (4, 2), (4, 4), (4, 6), (5, 1), (5, 3),(5, 5), (6, 2), (6, 4), (6, 6).These are 18 outcomes.The outcomes such that the sum is divisibleby 6 are:(1, 5), (2, 4), (3, 3), (4, 2), (5, 1), (6, 6).These are 6 outcomes.Now, the number of outcomes which aredivisible by 3 or 2 is 12 + 18 – 6 = 24.

Hence, the required probability = 24 2

=36 3

.

4. No.An unbiased die has six equally likelyoutcomes. These are: 1, 2, 3, 4, 5 and 6.Each of them has equal probability.

Therefore, the probability of getting 6 is 16

and that of not 6 is 56

.

5. All possible outcomes are given byS = {1, 2, 3, ....., 1000}∴ n(S) = 1000(i) Let E1 be the event that the first player

wins a prize. Then,E1 = Perfect square numbers

greater than 500 and lessthan 1001.

= 529, 576, 625, 676, 729,784, 841, 900, 961.

∴ n(E1) = 9

Now, P(E1) =( )1E(S)

nn

= 9

1000

(ii) Let E2 be the event that the secondplayer wins a prize, if the first has won.

∴ n(E2) = n(E1) – 1 = 9 – 1 = 8 And number of all possible outcomes

= n(S) – 1 = 1000 – 1 = 999

Now, P(E2) = 2(E )999

n =

8999

.

6. Let E1 be the event ‘the mobile phone isacceptable to Varnika’ and E2 be the event‘the mobile phone is acceptable to thetrader’.∴ n(E1) = Number of good mobile

phones= 42

And n(E2) = Number of good mobilephones + Number of mobilephones having only minordefects

= 42 + 3 = 45Number of all mobile phones is given by

n(S) = 48

(i) P(E1) = 1(E )(S)

nn

= 4248

= 78

(ii) P(E2) = 2(E )(S)

nn

= 4548

= 1516

.

231BORP TILIBA Y

7. When two dice are thrown, the samplespace is given by

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

∴ n(S) = 36(i) Let A1 be the event ‘product of the

numbers on the top of dice is 6’. ThenA1 = {(1, 6), (2, 3), (3, 2), (6, 1)}

n(A1) = 4

∴ P(A1) = 1(A ) 4 1= =

(S) 36 9nn

.

(ii) Let A2 be the event ‘product of thenumbers on the top of dice is 12’.Then,

A2 = {(2, 6), (3, 4), (4, 3), (6, 2)}∴ n(E2) = 4

∴ P(E2) = 2(E )(S)

nn

= 4

36 =

19

.

(iii) Let A3 be the event ‘product of thenumbers on the top of dice is 7’. Then,

A3 = Nil∴ n(E3) = 0

∴ P(E3) = 3(E )(S)

nn

= (0)36

= 0.

8. On tossing a coin 3 times, all possibleoutcomes are given by the sample space asS = {HHH, HHT, HTH, HTT, THH, THT,TTH, TTT}∴ n(S) = 8

(i) Let E1 be the event of getting all heads,then

E1 = {HHH}∴ n(E1) = 1

Now, P(E1) = 1(E )(S)

nn

= 18

.

(ii) Let E2 be the event of getting at least2 heads, thenE2 = {HHH, HHT, HTH, THH}∴n(E2) = 4

Now, P(E2) = 2(E )(S)

nn

= 48

= 12

.

CHAPTER TEST

1. (B) Number of faces having B or C = 2 + 1 = 3

Number of all faces = 6

P(getting B or C) =3 1

=6 2

.

2. (C) P(drawing a green ball)= 3 × P(drawing a red ball)

⇒5 +

nn

= 3 × 5

5 n+ ⇒ n = 15.

3. (C) The even prime numbers from 1 to 6 is2 only

∴ P(getting an even prime number) = 16

.

4. Case I. 2 dice are thrown.Number of all outcomes in the samplespace, n(S) = 62 = 36Favourable numbers, n(E1) = 1

∴ P(E1) = 1

36

Case II. 1 die is thrown.Number of all outcomes, n(S) = 6Favourable numbers, n(E2) = 1

∴ P(E2) = 16

So, the student throwing one die has thebetter chance because he has moreprobability.

ORThe sample space isS = {HHH, HHT, HTH, HTT, THH, THT,

TTH, TTT}∴ n(S) = 8The outcomes having at least two heads areE = {HHH, HHT, HTH, THH}∴ n(E) = 4

∴ P(E) =(E) 4 1(S) 8 2

nn

= = .


5. False, because there are equal probabilities

of getting the head or tail, that is 12

.

ORTotal number of outcomes, n(S) = 36Favourable outcomes,E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}∴ n(E) = 5

∴ P(E) =536

.

6. There are 52 cards in the pack. Therefore,the number of outcomes in sample space isgiven by

n(S) = 52Number of hearts cards = 13

Number of queens = 4Number of queens of hearts = 1

So, number of favourable outcomes is givenby

n(E) = 52 – (13 + 4 – 1) = 36

Now, the required probability will be givenby

P(E) =(E) 36 9

= =(S) 52 13

nn

.

7. Area of the rectangular region= Length × Breadth= 3 × 2 = 6 m2

Area of circular region= π × Radius2

=22 1

×7 4

=1114

m2

Now, required probability

=Area of circular region

Area of rectangular region

=

11146

= 11

14 6× =

1184

.

8. (i) 12

(ii) 14

Hint: A, B, C and D can be arranged in 24ways. A before B can be arranged in thefollowing ways:CADB, DACB, CABD, DABC, ABCD,ABDC, CDAB, DCAB, ACDB, ADCB,ACBD, ADBC.Out of these A just before B occurs in6 ways.

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

Therefore,2x

=3y

⇒ 3x – 2y = 0 ... (i)and 2(x + y) = 3y + 8⇒ 2x – y = 8 ... (ii)Solving equations (i) and (ii ), we get

x = 16, y = 24Number of non-red balls = number of whiteballs = 16P(choosing ball is not red)

= 1616 24+

= 1640

= 25

.

❑❑

233OOC MOEGET EDR I AN T R Y

7Chapter

COORDINATE GEOMETRY

= { }1– 5(5 – 5) – 4(5 – 7) 4(7 – 5)

2+

= 1

(0 8 8)2

+ + = 8 sq. units.

7. Let A(1, 7), B(4, 2), C(−1, −1), D(− 4, 4)be the vertices.

∴ AB = ( ) ( )2 24 1 2 7− + − = 34

BC = ( ) ( )2 21 4 1 2− − + − − = 34

CD = ( ) ( )− + + +2 24 1 4 1 = 34

AD = ( ) ( )2 24 1 4 7− − + − = 34

Also,

AC = ( ) ( )2 21 1 1 7− − + − − = 68

BD = ( ) ( )2 24 4 2 4+ + − = 68

So, AB = BC = CD = AD and diagonals,i.e., AC = BD.⇒ ABCD is a square.

8. Draw DE ⊥ AB and join BD.Since, diagonals of a parallelogram bisecteach other.

∴ Mid-point of AC= Mid-point of BD

⇒1

2a+

=2 – 4

2{∵ Comparing only x-coordinate}

⇒ 1 + a = – 2∴ a = – 3.

WORKSHEET –721. (C)

Hint: Condition of collinearity must besatisfied,i.e., k(− 2 − 4) + 6(4 − 3) + (−3) (3 + 2) = 0

⇒ − 6k + 6 − 15 = 0 ⇒ k = 3

2−

.

2. (B)Hint: Let the ratio is k : 1∴ Using section formula

0 =5

1k

k− +

+ ⇒ k = 5.

3. 2 units

Hint: Distance = 2

22 8(2 2)

5 5 + + −

= 105

= 2 units.

4. 2Hint: Use AO = OB and apply distanceformula.

5. P(– 7, 0)Hint: Any point on x-axis is P(x, 0)

ORSince, P is equidistant from A and B,∴ PA = PB ⇒ PA2 = PB2

⇒ (x – 3)2 + (y – 6)2 = (x + 3)2 + (y – 4)2

⇒ x2 – 6x + 9 + y2 + 36 – 12y= x2 + 6x + 9 + y2 – 8y + 16

⇒ – 6x – 6x – 12y + 8y + 9 + 36 – 9 – 16 = 0⇒ – 12x – 4y + 20 = 0⇒ 3x + y – 5 = 0. Hence proved.

6. LetA(– 5, 7) ≡ A(x1, y1); B(– 4, 5) ≡ B(x2, y2)and C(4, 5) ≡ C(x3, y3)Now, area of ∆ABC

= { }1 2 3 2 3 1 3 1 21

( – ) ( – ) ( – )2

x y y x y y x y y+ +


Now,

ar(∆ABD) = { }11(3 3) 2(– 3 2) – 4(– 2 – 3)

2+ + +

= +1(6 – 2 20)

2 = 12 ...(i)

Also,

ar(∆ABD) = 12

× base × altitude

= 12

× AB × DE

= 12

× 26 × DE ...(ii)

= 2 2AB = 1 + 5 26∵

Comparing equations (i) and (ii), we have

12

× 26 × DE = 12 ⇒ DE = 2426

× 2626

∴ DE = 12 26

13Hence, a = – 3 and required height

= 12 26

13 units.

WORKSHEET – 731. (B) Let coordinates of A be (x, y).

As O will be mid-point of AB,

∴ 2 =1

2x +

⇒ x = 3

and − 3 =4

2y +

⇒ y = − 10.

2. (C) Any point on y-axis be (0, y)

∴ ( ) ( )226 5 y+ − = ( ) ( )2 20 4 3 y+ + −

⇒ y = 9∴ Point is (0, 9).

3. (A)Hint: Let the ratio is k : 1.Now use section formula.

4. Since diagonals of parallelogram bisect eachother.∴ Mid-point of AC = mid-point of BD

i.e.,6 9 1 4

,2 2+ +

=

8 5,

2 2p+

⇒ 152

=8

2p+

⇒ p = 7.

OR

Given vertices are:A(– 3, 0), B(5, – 2) and C(– 8, 5). We know that

centroid G is 1 2 31 2 3 ,3 3

y y yx x x + ++ +

∴ Centroid = 3 5 8 0 2 5

,3 3

− + − − +

= (−2, 1).

5. True, because distance between (6, 4) and(7, – 2); and distance between (6, 4) and(5, – 2) are equal.

6. (1, 3)Hint: Use section formula.

7. Area of triangle

= ( ) ( ) ( )1 2 3 2 3 1 3 1 212

x y y x y y x y y − + − + −

= ( ) ( ) ( )12 0 4 1 4 – 3 +2 3 – 0

2 + − −

= 1

8 7 62

+ + =212

sq. units.

8. As the given points are collinear, the areaof the triangle formed by these points mustbe zero.

Let (2, 1) ≡ (x1, y1); (p, –1) ≡ (x2, y2); and(– 1, 3) ≡ (x3, y3)

Now, area of the triangle = 0

⇒12

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

⇒ 12

[2(–1 – 3) + p(3 – 1) – 1(1 + 1)] = 0

⇒ 12

(– 8 + 2p – 2) = 0

⇒ 2p = 10 ⇒ p = 5.

9. Hint: Proceed as done in solved example 5.


WORKSHEET – 74

1. (D)

Area = { }+ + +14(– 6 5) 1(– 5 – 5)– 4(5 6)

2

= 1

(– 4 – 10 – 44) – 292

=

= 29 sq. units.

2. (B) Condition of collinearity must be satisfied∴ − 5(p + 2) + 1(−2 − 1) + 4(1 − p) = 0⇒ −5p − 10 − 3 + 4 − 4p = 0⇒ p = −1.

3. 2 : 3Hint: Use section formula.

4. TrueHint: Any point P on x-axis will be of typeP(x, 0)∴ Let A(7, 6), B(−3, 4)∴ Use PA = PB.

5. k = – 8Hint: Using section formula

Coordinates of A

= 1 × 6 + 2 × 3 6 × 1 + 2 × 3

,3 3

−

= (4, 0)

∴ As it lies on 2x + y + k = 0⇒ 2 × 4 + 0 + k = 0∴ k = – 8.

6. Let O be the centre and P be the point onthe circumference such that O ≡ (2a, a – 7)and P ≡ (1, – 9).

Radius = OP = 10 2

2

= 5 2 units

i.e., ( ) ( )+ +2 22 – 1 – 7 9a a = 5 2

Squaring both sides, we get

(2a – 1)2 + (a + 2)2 = ( )25 2

⇒ 4a2 – 4a + 1 + a2 + 4a + 4 = 50⇒ 5a2 = 45 ⇒ a = ± 3Thus, a = ± 3.

7. Let the required ratio be k : 1.

Using x = 1 2 2 1

1 2

++

m x m xm m

and y = 1 2 2 1

1 2

++

m y m ym m

, we have

12

= – 7 3

1k

k+

+ and 6 =

++

9 51

kk

⇒ k + 1 = –14k + 6 and 6k + 6 = 9k + 5

⇒ k = 13 and k =

13


8. Let D = Mid-point of BC

= 7 5 3 7

,2 2+ +

= (6, 5)

∴ Median AD = ( )2 26 3 0− +

= 3 units.

9. Hint: Show that all sides are equal.

WORKSHEET – 75

1. (A) Centroid is 1 2 3 1 2 3,3 3

x x x y y y+ + + +

= 3 5 8 0 2 5

,3 3

− + − − +

= (− 2, 1).

2. (C) ar(∆ABC)

= { }12(1 2)– 2(– 2 – 3) 3(3 – 1)

2+ +

= + +1(6 10 6)

2 = 11 sq. units.


3. (D) AB = 2 2(– 2 – 2) (3 – 4)+ = +16 1

= 17 .

4. Let ratio is k : 1.∴ Using section formula

2 = 3 2

1k

k−

+; y =

7 21

kk

++

⇒ 4 = k ; y = 7 2

1k

k+

+

⇒ k = 4 ; y = ( )7 4 24 1

++

= 6∴ Ratio is 4 : 1; y = 6.

5. True,

∵ Pythagoras Theorem is satisfied.AB2 = (– 2)2 + (1 – 3)2

= 4 + 4 = 8AC2 = (– 1)2 + (4 – 3)2 = 2BC2 = (1)2 + (3)2 = 10

∴ AB2 + AC2 = BC2

⇒ ∠A = 90°.6. k = 0, 6

Hint: PQ = 58 ⇒ PQ2 = 58

⇒ (k − 3)2 + (2 + 5)2 = 58⇒ (k − 3)2 = 9⇒ k − 3 = ± 3 ⇒ k = 0 or 6.

7. Let the coordinates of R be (x, y)

Then, x = × + ×

+4 2 3 1

4 3 and y =

× + ×+

4 3 3 24 3

⇒ x = 117

and y = 187

Therefore, the coordinates of R are

11 18,

7 7 .

8. 25 sq. units

Hint: Join SQ.

Find ar(∆PQS)

and ar(∆RQS)

∴ Required area = ar(∆PQS) + ar(∆RQS).

9. Each side of a square and rhombus areequal, but the diagonals of a square areequal and that of a rhombus may or maynot equal.

Side PQ = + +2 2(3 – 2) (4 1)

= +1 25 = 26

Side QR = 2 2(– 2– 3) (3 – 4)+

= +25 1 = 26

Side RS = 2 2(– 3 2) (– 2 – 3)+ +

= +1 25 = 26

Side SP = 2 2(2 3) (–1 2)+ + +

= 25 1+ = 26

Diagonal PR = 2 2(– 2 – 2) (3 1)+ +

= 16 16+ = 4 2

Diagonal QS = 2 2(– 3 – 3) (– 2 – 4)+

= 36 36+ = 6 2

Clearly, the sides are equal but thediagonals are not equal. Hence, PQRS is arhombus but not a square.

WORKSHEET – 76

1. (C) As given points are collinear.So, a(b − 1) + 0 (1 − 0) + 1 (0 − b) = 0⇒ ab − a − b = 0⇒ a + b = ab


⇒1 1a b

+ = 1.

2. (B)

∵ A, B, C are collinear so condition ofcollinearity must be satisfied⇒ x1(y2 – y3) + x2(y3 – y1) + x3(y1– y2) = 0

⇒ – 1(3 – 3) + (– 5) (3 – p) + 0(p – 3) = 0⇒ 0 – 15 + 5p = 0⇒ 5p = 15

p = 3.3. (A)

Distance = ( ) ( )2 2cos sin sin + cosθ − θ + θ θ

= ( )2 22 sin cosθ + θ = 2 .

4. (A) Let A ≡ (x, – 1) and B ≡ (3, 2).AB = 5

⇒ + +2 2(3 – ) (2 1)x = 5

⇒ 9 – 6x + x2 + 9 = 25 (On squaring)⇒ x2 – 6x – 7 = 0⇒ (x – 7) (x + 1) = 0 ⇒ x = 7 or –1.

5. False.

As mid-point of AC = 6 9 1 4

,2 2+ +

= 15 5

,2 2

and mid-point of BD = 8 2 3

,2 2

p+ +

∴ Mid-point of AC = Mid-point of BD

⇒152

= 8

2p+

⇒ p = 7.

6. ar(∆ABD)

= { }1– 5(– 5 – 5)– 4(5 – 7) 4(7 5)

2+ +

= 1

(50 8 48)2

+ + = 53 sq. units.

ar(∆BCD)

= { }1– 4(– 6 – 5)– 1(5 5) 4(– 5 6)

2+ + +

= 1

(44 – 10 4)2

+ = 19 sq. units.

Now, ar(quadrilateral ABCD)= ar(∆ABD) + ar(∆BCD)= 53 sq. units + 19 sq. units= 72 sq. units.

7. Let the points P, Q and R divide AB into fourequal parts AP = PQ = QR = RB as shownbelow in the adjoining figure.

Clearly, Q is the mid-point of AB

∴ Q ≡ – 2 2 2 8

,2 2+ +

i.e., Q ≡ (0, 5)P is the mid-point of AQ

∴ P ≡ – 2 0 2 5

,2 2+ +

i.e., P ≡ 7

–1,2

and R is the mid-point of QB.

∴ R ≡ 0 2 5 8,

2 2+ +

i.e., R ≡ 131,

2

.

Hence, the required points are P7

–1,2

,

Q(0, 5) and R13

1,2

.


8. Let P(2, – 3) and Q(10, y) be the givenpoints. Then, PQ = 10

⇒ ( ) ( ){ }2210 2 – –3y− + = 10

⇒ ( )264 3y+ + = 10

⇒ 264 9 6y y+ + + = 10

Squaring both sides, we gety2 + 6y + 73 = 100

⇒ y2 + 6y – 27 = 0⇒ y2 + 9y – 3y – 27 = 0⇒ y(y + 9) – 3(y + 9) = 0⇒ (y + 9) (y – 3) = 0⇒ y = – 9 or 3.

OR

91,

2 − Hint:

R divides AB internally.i.e. ratio 3 : 1.Now, use section formula.

9. Let P (x, y), A (7, 1), B (3, 5)As P is equidistant from A and B... PA = PB ⇒ PA2 = PB2

⇒ (7 – x)2 + (1 – y)2 = (3 – x)2 + (5 – y)2

⇒ 49 + x2 – 14x + 1 + y2 – 2y= 9 + x2 – 6x + 25 + y2 – 10y

⇒ – 8x + 8y = – 16 ⇒ x – y = 2.

OR

Let the required ratio be k : 1 and the pointof division be (l, m) using section formula,we have

l = +

+2 1

1kk

and m = ++

7 31

kk

So, the point of division is 2 1 7 3

,1 1

k kk k

+ + + +

.

This point lies on the line 3x + y – 9 = 0.

∴ + +

+ + +

2 1 7 33 – 9

1 1k kk k

= 0

⇒ 6k + 3 + 7k + 3 – 9k – 9 = 0

⇒ 4k = 3 ⇒ k = 34


WORKSHEET–77

1. (A)Hint: Use condition of collinearity.

2. (D)Hint: Any point on x-axis be (x, 0).Let ratio be k : 1

3. (C)Hint: Origin is (0, 0).

4. FalseHint:A, B, C will not form a triangle.∵ ar(∆ABC) = 0

5. Diagonals AC and BD cut each other at themid-point P.

∴ 132

x+= 2 ; 12

2y+

= − 5

⇒ x1 = 1 ; y1 = −12.

Similarly, 2 12

x −= 2 ; 2 0

2y +

= − 5

⇒ x2 = 5 ; y2 = −10.

Hence, two other vertices of the parallelo-gram are (1, – 12) and (5, – 10).

6. Area of quadrilateral ABCD

= ar(∆ABD) + ar(∆BCD)


= { }11(– 3 – 21) 7(21– 1) 7(1 3)

2+ + +

+ { }17(2 – 21) 12(21 3) 7(– 3 – 2)

2+ + +

= 1

(– 24 140 28)2

+ + +1

(– 133 288 – 35)2

+

= 72 + 60 = 132 sq. units.

7. 1 3 7 5 3,

2 2 ± ±

Hint:As AB = BC = AC∴ AB = BC⇒ AB2 = BC2

⇒ (x − 3)2 + (y − 4)2 = 26 ...(i)Similarly,

AC = BC ⇒ (x + 2)2 + (y − 3)2 = 26 ...(ii)Solve (i) and (ii).

8. PA = PB⇒ PA2 = PB2

⇒ (x − 3)2 + (y − 6)2 = (x + 3)2 + (y − 4)2

⇒ x2 + 9 − 6x + y2 + 36 − 12y= x2 + 9 + 6x + y2 + 16 − 8y

⇒ 3x + y = 5.

9. Coordinates of P are +

– 6 68 – 4

,2 2

,

i.e., (2, 0).

Coordinates of Q are – 4 – 10 6 – 8,

2 2

i.e., (– 7, – 1).

Coordinates of R are – 6 – 88 – 10

,2 2

,

i.e., (– 1, – 7).

Now, ar(∆ABC)

= { }+ +18(6 8)– 4(– 8 6)– 10(– 6 – 6)

2

= + +1(112 8 120)

2= 120 sq. units.

and ar(∆PQR)

= { }12(–1 7) – 7(–7 – 0) – 1(0 1)

2+ +

= +1(12 49 – 1)

2 = 30 sq. units.

So,( PQR)( ABC)

arar

= 30

120 =

14

⇒ ar(∆PQR) = 14

ar(∆ABC).

Hence proved.

WORKSHEET –781. (A)

Hint: Use formula of centroid of a triangle.

2. (C)

Using condition of collinearity,k[3k – 1] + 3k[1 – 2k] + 3[ – k] = 0

⇒ 3k2 – k + 3k – 6k2 – 3k = 0⇒ – 3k2 – k = 0⇒ – k(3k + 1) = 0

⇒ k = 13

− or 0.

3. (D) Using mid-point formula:

1 = 2 2

2a −

; 2a +1 = 4 3

2b+

⇒ a = 2 ; b = 2.


4. Let the ratio is k : 1∴ Using section formula, the point is

8 3 9 1,

1 1k k

pk k

+ − + +

As it lies on x − y − 2 = 0⇒ 8k + 3 − 9k + 1 − 2k − 2 = 0⇒ − 3k + 2 = 0

⇒ k = 23

∴ Ratio is 2 : 3.

5. Let the coordinates of vertices of the triangleABC are A(x1, y1), B(x2, y2) and C(x3, y3).

Since, the point (1, 2) is the mid-point of AB.

∴+1 2

2x x

= 1, +1 2

2y y

= 2

i.e., x1 + x2 = 2, y1 + y2 = 4 ...(i)Similarly, x2 + x3 = 0, y2 + y3 = –2 ...(ii)and x3 + x1 = 4, y3 + y1 = –2 ...(iii)∴ x1 + x2 + x3 = 3, y1 + y2 + y3 = 0

...(iv)Solving results (i), (ii), (iii) and (iv), we get x1 = 3, y1 = 2, x2 = – 1, y2 = 2, x3 = 1, y3 = – 4.

Hence the required vertices are A(3, 2),B(– 1, 2) and C(1, – 4).

6. (7, 2) or (1, 0)

Hint: ar(∆ PAB) = 10

⇒ ar(∆ PAB) = + 10 or ar(∆ PAB) = −10

Let P(x, y) ∴ Use area of triangle

= ( ) ( ) ( )1 2 3 2 3 1 3 1 212

x y y x y y x y y − + − + − .

7. The given points would be collinear, if thearea of triangle formed by them as verticesis zero.

i.e., ( ){}

1– – ( – – )

2( – – ) 0

a c a a b b a b b c

c b c c a

+ + +

+ + =

Here, LHS

= ( ){ }1– ( – ) ( – )

2a c b b a c c b a+ +

= ( )1– – –

2ac ab ab bc bc ac+ + = 0

= 0 = RHS. Hence proved.

8. 24 sq. units

Hint: Area of rhombus = 21

d1 × d2

where d1, d2 = length of diagonals.

9. ar(∆ABC) = 5

⇒ ( ) ( ) ( ){ }16 – 1 – 2 1 – 2 3 2 – 6

2k k k+ = 5

⇒ ( )15 – 2 4 6 –18

2k k k+ + = 5

⇒ ( )115 – 20

2k = 5 ⇒ 15

– 102

k = 5

⇒ 15– 10

2k = ± 5 ⇒

152

k = 10 ± 5

⇒152

k = 15 or 5 ⇒ k = 2 or 23

.

WORKSHEET – 79

1. (B) A(0, 0), B(3, 3 ), C(3, λ)As AB = BC = AC

∴ AB = AC ⇒ AB2 = AC2

⇒ 12 = 9 + λ2

⇒ λ2 = 12 − 9

⇒ λ2 = 3 ∴ λ = ± 3

∴ Required value of λ = − 3 .

2. (D) Mid-point of (6, 8) and (2, 4) is P(4, 6).∴ If A(1, 2), then

AP = ( ) ( )2 24 1 6 2− + −

= 9 16+

= 5 units.


3. 2 or – 4

Hint: Use Pythagoras Theorem.

4.

∴PAPQ =

23

⇒ PA : AQ = 2 : 1

∴ Using section formula.

x = 23

; y = 13

.

∴ Coordinates of A are 2 1,

3 3

.

5. 11 52 ,

2 2

−

; 11 52 ,

2 2

+

Hint:Let AB = AC = 3.Use distance formula.

6. Let the points of trisection be P(x1, y1) andQ(x2, y2) such that P is the mid-point ofA(3, – 2), Q(x2, y2) and Q is the mid-pointof P(x1, y1), B(– 3, – 4).

i.e., AP = PQ = QB

⇒ AP : PB = 1 : 2

Using section formula,

∴ x1 = – 3 2×3

1 2++

, y1 = – 4 2 (–2)

1 2×+

+

⇒ x1 = 1, y1 =8

–3

Again AQ : QB = 2 : 1

∴ x2 =2×(– 3)+ 3

2 1+, y2 =

2×(– 4) – 22 1+

⇒ x2 = – 1, y2 =10

–3

.

Hence, the required points are P

81, –

3

and Q

10–1, –

3.

7. A(1, 10), B(– 7, – 6), C(9, 2)Hint: x1 + x2 = 2 × (–3) = – 6

x2 + x3 = 2Adding, x3 + x1 = 10

2(x1 + x2 + x3) = 6x1 + x2 + x3 = 3

∴ x1 = 1, x2 = – 7, x3 = 9

Using same method,y1 + y2 + y3 = 6

and y1 = 10, y2 = – 6, y3 = 2.

8. Area of ∆DBC

= { }15 – (– 2) + ( 3)(– 2 – 3 )+ 4(3 – 5)

2x x x −

= 1(28 – 14) (14 – 7)

2x x=

Area of ∆ABC

= { }16 5– (– 2) + ( 3)(– 2 – 3)+4(3 – 5)

2 −

=1 49

(42 +15 – 8)2 2

=

From question

( DBC)( ABC)

arar

∆∆

=12

⇒14 – 7

492

x=

12

⇒14 – 7

492

x=

12

or – 14 +7

492

x=

12

⇒ 8x – 4 = 7 or – 8x + 4 = 7

⇒ x = 118

or – 38

.


WORKSHEET – 801. (B) As the distance of a point from x-axis is

equal to its y-coordinate, i.e., 3.

2. (A) A(a + b, a − b), B(2a + b, 2a − b),C(a − b, a + b), D(x, y)Mid-point of AC = Mid-point of BD⇒ x = −b⇒ y = b.

3. – 51– , 0 ; , 2

3 3

Hint:

Use AP : PB = 1 : 2and AQ : QB = 2 : 1

4. Let the third vertex be C(x, y) of the given∆ABC. Using, AC = AB and BC = AB,we have x2 + y2 = 12 and

(x – 3)2 + 2( – 3 )y = 12

Solving these, we obtain

x = 0, y = 2 3 or x = 3, y = – 3 .

Hence, the required vertex is ( 0, 2 3 ) or(3, – 3 ).

5. Let ABCD be the given square such thatA(3, 4) and C(1, – 1). Let D(a, b) be theunknown vertex.

Using AD2 = CD2, we have

(a – 3)2 + (b – 4)2 = (a – 1)2 + (b + 1)2

⇒ 4a + 10b – 23 = 0

⇒ b = 23 – 4

10a

...(i)

Using AD2 + CD2 = AC2, we have(a – 3)2 + (b – 4)2 + (a – 1)2 + (b + 1)2

= (3 – 1)2 + (4 + 1)2

⇒ a2 + b2 – 6a – 8b + 9 + 16 + a2 – 2a + b2

+ 2b + 1 + 1 = 4 + 25⇒ a2 + b2 – 4a – 3b = 1 ...(ii)Using equations (i) and (ii), we get

a2 +

223 – 4 23 – 4– 4 – 3

10 10a a

a = 1

⇒ 116a2 – 464a – 261 = 0⇒ 4a2 – 16a – 9 = 0

(Dividing by 29)

⇒ a = 16 256 4 4 9

2 4

± + × ××

= 92

or 1

–2

Substitute a = 92

and a = 1

–2

successively

to get b = 12

and b = 52

.

Hence, the required vertices are 9 1,

2 2

and

1 5– ,

2 2.

OR

(4, 5), (2, 3), (6, 9)

Hint:

x1 + x2 = 6 y1 + y2 = 8∴ x2 + x3 = 8 y2 + y3 = 12

x1 + x3 = 10 y1 + y3 = 14

Adding,2(x1 + x2 + x3) = 24 2(y1 + y2 + y3) = 34


⇒ x1 + x2 + x3 = 12 ⇒ y1 + y2 + y3 = 17∴ x1 = 4 ∴ y1 = 5

x2 = 2 y2 = 3x3 = 6; y3 = 9.

6. Let O(x, y) be the circumcentre passingthrough A(3, 0), B(– 1, – 6) and C(4, – 1).Then OA = OB = OCTaking OA = OB⇒ OA2 = OB2 (Squaring)

⇒ (x – 3)2 + (y – 0)2 = (x + 1)2 + (y + 6)2

⇒ x2 – 6x + 9 + y2 = x2 + 2x + 1 + y2 + 12y+ 36

⇒ 2x + 3y + 7 = 0 ...(i)and OA = OC ⇒ OA2 = OC2 (Squaring)⇒ (x – 3)2 + y2 = (x – 4)2 + (y + 1)2

⇒ x2 – 6x + 9 + y2 = x2 – 8x + 16 + y2 + 2y+ 1

⇒ x – y – 4 = 0 ...(ii)Solving equations (i) and (ii), we get

x = 1, y = – 3.Thus, the coordinates of the centre are (1, – 3)

Now, radius = OA = ( ) ( )+ +2 23 –1 0 3

= +4 9 = 13 units.

OR

The area of the triangle formed by the givenpoints must be zero.

i.e., ( ){12 – 6 2 – ( –1) (6 – 2 – 2+2 )

2k k k k k k+

( )( )}– 4 + 2 – 2 – 2k k k = 0

⇒ k(4k – 6) – (k – 1) × 4 – (4 + k) (2 – 4k) = 0

⇒ 4k2 – 6k – 4k + 4 – 8 + 16k – 2k + 4k2 = 0

⇒ 8k2 + 4k – 4 = 0 ⇒ k2 + 12

k – 12

= 0

⇒ k2 + 12

k + 1

16–

116

– 12

= 0

⇒

21+

4k –

234

= 0

⇒ + + + 1 3 1 3

–4 4 4 4

k k = 0

⇒ k = – 1 or 12

.

7. SP = +2 2 2( – ) (2 – 0)at a at

= +2 2 2 2 2( – 1) 4a t a t

= a + +4 2 2– 2 1 4t t t = a + +4 22 1t t

= a(t2 + 1)

SQ =22

2 2

– 2– – 0

aaa

t t +

= + +2 2 2

24 2 2

2 4–

a a aa

t t t

= a + +4 21 2

1t t

= a + 2

11

t

Now,1 1

SP SQ+ = ( )2

2

1 11+1 +1a t at

+

=

+

2

2 21 1

+1 1+t

a t t

= + +

2

2

111

ta t

=1a

which is independent of t.

8. As P(x, y) is mid-point of AB,

x = +32

k and y =

+4 62

i.e., x = +32 2

k and y = 5


The value of x and y will satisfy x + y – 10 = 0

∴ +32 2

k + 5 – 10 = 0 ⇒ 2k = 5 –

32

⇒ k = 7.

WORKSHEET – 81

1. (A) Let the coordinates of the third vertexC be (x, y).

∴ +3 – 23

x=

53

and + +2 1

3y

= 13

i.e. x = 4 and y = – 2

∴ C ≡ (4, – 2).

2. (B) As BP = ( )32

2a×

= 3a

and OP = 12

OA = a

∴ Coordinates of B are (a, 3 a).

3. (A) Let the required point be (h, k).

Then

h = 3×(– 4) + 2 × 6

3 + 2 and k =

× + ×+

3 5 2 33 2

i.e., h = 0 and k = 215

.

So, the required point is 21

0,5

.

4. (A)Hint: Use AB = BC = AC

Take AB2= BC2

and BC2 = AC2.

5. True.Let O(0, 0), A(5, 5) and B(– 5, 5) be the threepoints.

∴ OA = 5 2 = OB

and AB2 = 100 = OA2 + OB2.

ORTrue,

AB = 2 2(– 4 6) (6 – 10)+ + = 20 = 2 5

BC = 2 2(3 4) (– 8 – 6)+ + = 245 = 7 5

AC = 2 2(3 6) (– 8 –10) = 405 = 9 5+ +

∴ AB + BC = AC.

Also, AB 2 2= AB = AC

AC 9 9⇒ .

6. As A, P, B will be collinear,

∴ .( ) ( 0) 0 (0 )a y b x b y− + − + − = 0

⇒ ay − ab + xb = 0 ⇒ ay + xb = ab

⇒ yb

+ xa = 1 ⇒

xa +

yb

= 1.

7. Let the coordinates of P be (x, y).

⇒APAB

= 37

⇒ ABAP

= 73

⇒ABAP

–1 = 73

–1 ⇒ PBAP

= 43

⇒ APPB

= 34

.

Using section formula, we have

x = ( )3×2+4 – 2

3+4 and y =

( )3×(– 4)+4× – 2

3+4

i.e., x = –27

and y = – 207

Thus, the coordinates of P are 2 20

– , –7 7

.

8. Let the coordinates of P be (x, y).PA = PB

⇒ PA2 = PB2 (Squaring)⇒ (x – 3)2 + (y – 4)2 = (x – 5)2 + (y + 2)2

⇒ x2 – 6x + 9 + y2 – 8y + 16= x2 – 10x + 25 + y2 + 4y + 4

⇒ 4x – 12y – 4 = 0⇒ x – 3y – 1 = 0 ...(i)


Area of ∆PAB = 10

⇒ ( ) ( ) ( ){ }+ + +14 2 3 –2 – 5 – 4

2x y y = 10

6x – 6 – 3y + 5y – 20 = ± 206x + 2y – 26 = ± 20

⇒ 3x + y – 3 = 0 ...(ii)or 3x + y – 23 = 0 ...(iii)Now, we have to solve equations (i) and(ii) as well as equations (i) and (iii).Solving equations (i) and (ii), we get

x = 1, y = 0Solving equations (i) and (iii), we get

x = 7, y = 2Hence, the coordinates of P are (1, 0) or(7, 2).

OR

∵ P is mid-point of AB

∴ P ≡ +

1 3 5 – 7

,2 2

,

i.e., P ≡ (2, – 1)

∵ Q is mid-point of BC

∴ Q ≡ 3 0 – 7 4

,2 2+ +

, i.e., Q ≡

3 3

, –2 2

∵ R is mid-point of CA

∴ R ≡ + +

0 1 4 5

,2 2

, i.e., R ≡

1 9,

2 2

Now, ar(∆PQR)

= 1 3 9 3 9

2 – – 12 2 2 2 2

+ + + +

1 3–1

2 2

= + + 1 33 1

–122 4 4

= –7 74 4

= ...(i)

ar(∆ABC)

= { }11(– 7 – 4) 3(4 – 5) 0(5 7)

2+ + +

= 1

(–11– 3 0) – 7 72

+ = = ...(ii)

Dividing equation (i) by equation (ii), we

have ∆∆

( PQR)( ABC)

arar

=

747

⇒ ar(∆PQR) = 14

ar(∆ABC).

9. Since, A is onthe x-axis, so itscoordinates willbe of the form(x, 0). Similarly,the coordinatesof B will be ofthe form (0, y).Since P is the mid-point of AB.

∴ – 2 =0

2x +

and 3 = +02

y

∴ x = – 4 and y = 6∴ Coordinates of A are (– 4, 0) and coordi-nates of B are (0, 6).Now,

PO = 2 2(– 2) (3) 4 9 13+ = + =

PA = 2 2(– 4 2) (0 – 3) 4 9+ + = + = 13

Clearly, PA = PB = PO⇒ P is equidistant from A, B and theorigin O.

WORKSHEET– 82

1. (A) (– 5, 1), (1, p) and (4, – 2) are collinear.⇒ – 5(p + 2) + 1 (–2 –1) + 4(1 – p) = 0⇒ – 5p – 10 – 3 + 4 – 4p = 0 ⇒ 9p = – 9⇒ p = –1.

2. (C)

Area = { }11(4 – 6)– 2(6 – 3) 0

2+

= 1(– 2 – 6)

2= 4 sq. units.

3. See Worksheet −−−−− 78, Sol. 4.4. False, because P does not lie on the line

segment AB.


5. See Worksheet – 80, Sol. 5.6. As P is equidistant from A and B,

PA = PB ⇒ PA2 = PB2 (Squaring)⇒ (a + b – x)2 + (b – a – y)2

= (a – b – x)2 + (a + b – y)2

⇒ (a + b – x)2 – (a – b – x)2

= (a + b – y)2 – (b – a – y)2

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

⇒ 2(a – x) × 2b = 2(b – y) × 2a⇒ ab – bx = ab – ay ⇒ bx = ay

Hence proved.OR

Let the third vertexbe C(x, y) of thegiven ∆ABC suchthat A(0, 0) and

B (3, 3).

Using AC = ABi.e. AC2 = AB2 (Squaring)⇒ x2 + y2 = 9 + 3⇒ x2 + y2 = 12 ...(i)Also, using BC = ACi.e., BC2 = AC2 (Squaring)

⇒ (x – 3)2 + 2( – 3)y = x2 + y2

⇒ x2 – 6x + 9 + y2 – 2 3y + 3 = x2 + y2

⇒ 3x + 3 y – 6 = 0 ...(ii)Solving equations (i) and (ii), we obtainx = 0, y = 2 3 or x = 3, y = – 3

Hence, the third vertex is (0, 2 3) or

(3, – 3).

7.ADAB

= AEAC

= 14

⇒ABAD

= ACAE

= 41

⇒ ABAD

– 1 = ACAE

– 1 = 41

–1

(Subtracting 1 throughout)

⇒AB – AD

AD=

AC – AEAE

= 4 – 1

1

⇒ BDAD

= CEAE

= 31

⇒ AD : BD = AE : EC = 1 : 3Let the coordinates of D and E be (x1, y1)and (x2, y2) respectively.Let us use section formulae.

x1 = 1 × 1+3 × 4

1+3, y1 =

1 × 5+3 × 61+3

and x2 = 1 × 7 +3 × 4

1+3, y2 =

1 × 2+3 × 61+3

i.e., x1 = 134

, y1 = 234

and x2 = 194

, y2 = 5

So, the coordinates of D are 13 23

,4 4

and

of E are 19, 5

4

.

ar(∆ADE)

= + + 1 23 13 19 23

4 – 5 (5 – 6) 6 –2 4 4 4 4

= × + ×

1 3 13 19 14 –

2 4 4 4 4

= 1532

sq. units.

Again, ar(∆ABC)

= { }14(5 – 2) 1(2 – 6) 7(6 – 5)

2+ +

=1

(12 – 4 7)2

+ = 152

sq. units

∴ ar(∆ADE) : ar(∆ABC) =

1532152

= 1 : 16.


8. P(at2, 2at), Q 22

,aa

tt−

, S(a, 0)

∴ SP = 2 2 2( ) (2 )at a at− +

= 2 4 2 2 2 2 22 4a t a a t a t+ − +

= 2 4 2 2 22a t a a t+ +

= 2 2( )at a+ = at2 + a

SQ = 22

22 aa

att

− − +

= 22 2

24 2 2

2 4aa aa

t t t+ − +

= 2

2a

at

+ = 2

at

+ a

∴ 1

SP+ 1

SQ= 2

1at a+

+ 2

2t

a at+

= 2

2

1

( 1)

t

a t

++

= 1a

which is independent of t.


1. (B) x = 2×7 +3×3

2+3 =

235

y = 2×9+3×5

2+3 =

335

Hence, the coordinates of P are 23 33,

5 5

.

2. (B) Let the centroid be P(x, y).

x = 2 + 5 + 3

3; y =

1+ 2+43

;

i.e., x = 103

; y = 73

Thus P ≡ 10 7

,3 3

.

3. D is mid-point of BC

∴ D ≡ 1 – 3 5 – 1

,2 2

;

i.e., D ≡ ( – 1, 2)

AD = 2 2(5 1) (1– 2)+ +

= 36 1+

= 37 units.

4. False, because area of the triangle formedby joining the given points is not zero as,

area = 12

4(6 – 3) + 7(3 – 5) + 6(5 – 6)

= 12

12 – 14 – 6 = 4 ≠ 0.

5. As P(x, y) is the mid-point of A(3, 4) andB(k, 6),

x = 3

2k+

and y = 4 6

2+

(Using mid-point formula)

i.e. x = 3

2k+

and y = 5.

Substituting these values of x and y inx + y – 10 = 0, we get

32

k++ 5 – 10 = 0 ⇒

32

k+ = 5

⇒ 3 + k = 10 ⇒ k = 7.

6. Using the distance formula

Distance = 2 22 1 2 1( – ) ( – )x x y y+ ,

we get

AB = { }2 2– 4 – (– 5) (– 2 – 6)+

= 1 64+ = 65


BC = { } { }2 27 – (– 4) 5 – (– 2)+

= 121 49+ = 170

and CA = ( ) ( )2 2– 5 – 7 6 – 5+

= 144 1+ = 145

Since the lengths of all the sides of ∆ABCare different, therefore, ∆ABC is a scalenetriangle.

7. Let the mid-points of sides AB and AC beD and E respectively. Such that D(2, – 1)and E(0, – 1).

AD = 12

AB and AE = 12

AC ...(i)

According to the converse of BasicProportionality Theorem, DE || BC, andtherefore, ∆ADE ~ ∆ABC.

Now,( ABC)( ADE)

arar

= 2 2

2 2AB (2AD)

=AD AD

[Using (i)]

⇒ ar(∆ABC) = 4 × ar(∆ADE) ...(ii)

Now, ar(∆ADE)

=12

x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)

= 12

1(– 1 + 1) + 2(– 1 + 4) + 0(– 4 +1)

= 12

0 + 6 + 0 = 3 square units

Substituting this value of ar(∆ADE) inequation (ii), we havear(∆ABC) = 4 × 3 = 12 square units.

8. Let the given line divides the given linesegment at P(h, k) in the ratio λ : 1.Using section formula, we have

h = × 2 1 8

1λ + ×

λ + and k =

1 1 (– 9)1

λ× + ×λ +

i.e., h = 2 8

1λ +λ +

and k = – 9

1λλ +

Since, P(h, k) lies on the line2x + 3y – 5 = 0, therefore, x = h and y = kmust satisfy it.

Therefore, 2h + 3k – 5 = 0

⇒2 8 – 9

2 3 – 51 1

λ + λ+ λ + λ +

= 0

⇒ 4λ + 16 + 3λ – 27 – 5λ – 5 = 0⇒ 2λ = 16 ⇒ λ = 8 ⇒ λ : 1 = 8 : 1

∴ h = 2 8 8

8 1× +

+ and k =

8 – 98 1+

i.e., h = 83

and k = – 19

∴ P(h, k) = P8 1

, –3 9

.

Hence, the given line divides the given linesegment in the ratio 8 : 1 at the point

8 1, –

3 9

.


1. (C) Let the coordinates of P and Q be (x, 0)and (0, y) respectively.

∴0

2x +

= 3 and 0

2y+

= – 7

i.e., x = 6 and y = – 14Here, P ≡ (6, 0) and Q ≡ (0, – 14).

2. (C) Area of triangle formed by A, B and theorigin = 0


⇒12

1(b – 0) + a(0 – 2) + 0(2 – b) = 0

⇒ ± (b – 2a) = 0 ⇒ 2a = b.

3. Let the required point be P(h, k).Then PO = PA = PB∴ h2 + k2 = (h – 2x)2 + k2

and h2 + k2 = h2 + (k – 2y)2

⇒ h2 = h2 + 4x2 – 4xhand k2 = k2 – 4yk + 4y2

⇒ h = x and k = y∴ P is (x, y).

4. False, because AB ≠ CD and BC ≠ AD as

AB = 2 22 1+ = 5 ;

BC = 2 2(–1) (–10)+ = 101;

CD = ( )2 2– 8 11+ = 185;

AD = ( )2 2– 7 2+ = 53 .

5. Radius = r = Diameter

2

= 10 2

2 = 5 2 units

Coordinates of centre O are (2a, a – 7).Coordinates of a point P on the circumferenceare (11, – 9)∴ r = OP

⇒ 5 2 = 2 2(11– 2 ) (– 9– 7)a a+ +

⇒ ( )25 2 = (11 – 2a)2 + (– a – 2)2

(On squaring)⇒ 121 + 4a2 – 44a + a2 + 4a + 4 = 25 × 2⇒ 5a2 – 40a + 75 = 0

(Dividing throughout by 5)

⇒ (a – 5) (a – 3) = 0 ⇒ a = 5 or 3

Hence, a = 5, 3.

6. Let the required ratio be λ : 1.Here, we will use section formula as givenbelow.

x = 2 1mx nxm n

++

and y = 2 1my nym n

++

In this question,

– 3 = – 2 – 5

1λ

λ + and k =

3 – 41

λλ +

⇒ – 3λ – 3 = – 2λ – 5 and k =3 – 4

1λλ +

⇒ λ = 2 and i.e., k =3 2 – 4

2 1×

+

⇒ λ : 1 = 2 : 1 and k =23

Hence, the ratio is 2 : 1 and k =23

.

7. First, we find the length of each side ofquadrilateral ABCD.

AB = 2 2(1– 5) (5 – 6)+ = 2 2(– 4) (–1)+

= 16 1+ = 17

BC = 2 2(2 –1) (1– 5)+ = ( )221 – 4+

= 1 16+ = 17

CD = 2 2(6 – 2) (2 – 1)+ = 2 24 1+

= 16 1+ = 17

AD = 2 2(6 – 5) (2 – 6)+ = 2 21 (– 4)+

= 1 16+ = 17

Clearly, AB = BC = CD = ADAll the sides of quadrilateral ABCD areequal.Therefore, ABCD is a rhombus. It may be asquare if diagonals are equal. To confirm it,we have to find out the lengths of diagonalAC and BD.

AC = 2 2(2 – 5) (1– 6)+ = 2 2(–3) (–5)+

= 9 25+ = 34


BD = 2 2(6 – 1) (2 – 5)+ = 2 25 (– 3)+

= 25 9+ = 34

Clearly, AC = BD.Hence, quadrilateral ABCD is a square.

8. To find area of quadrilateral ABCD, wedivide it into two parts by either diagonal(see graph).Area of a triangle ABC

= 12

x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)

= 12

– 5 (– 5 + 6) – 4(– 6 – 7) – 1(7 + 5)

= 12

– 5 + 52 – 12 = 12

× 35

= 352

sq. units

ar(∆ACD) =12

– 5(– 6 – 5) – 1 (5 – 7) + 4(7 + 6)

= 12

55 + 2 + 52 =12

× 109

= 109

2 sq. units

Now, ar(quadrilateral ABCD)= ar(∆ABC) + ar(∆ACD)

= 352

+ 1092

= 1442

= 72 square units.

Alternative Method:ar(quadrilateral ABCD)

= 12

[(x1– x3) (y2 – y4) + (x2 – x4) (y3 – y1)]

= 12

[(– 5 + 1) (– 5 – 5) + (– 4 – 4) (– 6 – 7)]

= 12

[40 + 104] = 72 sq. units.

CHAPTER TEST

1. (D) Let the point of division be (x, y)

x = 1 × 3 + 2 × 7

1+2, y =

1 × 4 + 2(– 6)1+ 2

⇒ x = 173

, y = – 83

⇒

17 8, –

3 3 lies in the IVth quadrant.

2. (A) Mid-point of hypotenuse AB isequidistant from the vertices A, B and O.Therefore, the required point is

0 2 2 0,

2 2x y+ +

, i.e., (x, y).

3. Let the required point be P(0, y) such that

PA = PB

⇒ +2 2(0 – 1) ( – 5)y = +2 2(0 – 4) ( – 6)y

⇒ 1 + y2 – 10y + 25 = 16 + y2 – 12y + 36

(Squaring)

⇒ y = 13

Hence the coordinates of the point P be(0, 13).


4. False, because Q lies outside the circle asOQ > radius of circle.

5.

9a – 2 = 3 8 1 (3 1)

3 1a a× + × +

+

and – b = 3 5 1(– 3)

3 1× +

+

⇒ 36a – 8 = 24a + 3a + 1and – 3b – b = 15 – 3⇒ 9a = 9 and 4b = – 12Thus, a = 1 and b = – 3.

6. We know that area of a triangle is fourtimes the area of a triangle formed byjoining the mid-points of it.

D

1 5– ,

2 2, E(7, 3), F

7 7,

2 2

ar(∆DEF)

= 1 1 7 7 5 7 5

– 3 – 7 – – 32 2 2 2 2 2 2

+ +

= 1 1 77 –

2 4 4 +

= 114

sq. units.

Since, D, E and F are the mid-points ofsides of ∆ABC,∴ ar(∆ABC) = 4 × ar(∆DEF)

= 4 × 114

= 11 sq. units.


8. Let the coordinates of the vertices beA(x1, y1), B(x2, y2) and C(x3, y3)Observe the adjoining figure.

3 = +1 2

2x x

and 4 = +1 2

2y y

∴ x1 + x2 = 6 ...(i)and y1 + y2 = 8 ...(ii)Similarly,

x2 + x3 = 16 ...(iii)y2 + y3 = 18 ...(iv)x3 + x1 = 12 ...(v)y3 + y1 = 14 ...(vi)

Add equations (i), (iii) and (v) to getx1 + x2 + x3 = 17 ...(vii)

Add equations (ii), (iv) and (vi) to gety1 + y2 + y3 = 20 ...(viii)

Subtract equation (iii) from equation (vii)to get

x1 = 1Similarly, x2 = 5 and x3 = 11Subtract equation (iv) from equation (viii)to get

y1 = 2Similarly, y2 = 6 and y3 = 12Hence, the coordinates of the vertices ofthe ∆ABC are (1, 2), (5, 6) and (11, 12).

❑❑


8Chapter

AREAS RELATED TO CIRCLES

WORKSHEET– 85

1. (A) A =2

360rπ θ

°=

8 8 135360

π × × × °°

= 24π cm2.

2. (B) Perimeter = AB + BC + CD + length of

arc AED= 20 + 14 + 20 + π × (7)= 76 cm.

3. Perimeter = Outer arc length + Innerarc length + 2 × (14).

= 30

(21 7) 28180

π × °+ +

°

= 22 28

287 6

× + = 44

283

+

= 44 84

3+

=128

3 =

242

3 cm.

4. No.The correct statement is: "Area of a segmentof a circle = Area of the correspondingsector – Area of the corresponding triangle."

5. 154 m2

Hint: Find area of shaded part.

6. (i) 778

cm2 (ii) 498

cm2

Hint: Area of quadrant =2

4rπ

7. 686.18 cm2, 20.32 cm2

Hint: Area of minor segment

= 2 sin360 2

rπθ θ − °

∴ Area of major segment= Area of circle − Area of minor segment.

8. (i) Length of wire used= 2πr + 5 × (2r)

= 22 35 35

2 107 2 2

× × + ×

= 110 + 175= 285 mm.

(ii) Area of each sector

= 2

360rπ θ

°

=

22 35 3536

7 2 2360

× × × °

°

= 3854

mm2.

WORKSHEET– 86

1. (B) πR12 + πR2

2 = πR2

⇒ π(R12 + R2

2) = πR2

⇒ R12 + R2

2 = R2.

2. (C)Hint: Area of shaded part

= 2 24521 7

360π × ° − °

= 154 cm2.

3. 228.57 cm2

Hint: Shaded portion= Area of quadrant – Area of square ABCD Radius OB = 20 2×

= Diagonal of square.

4. No, because radius of the largest circle must

be 2b cm such that the area will be

2

4bπ cm2.

253AERA S ETALER D T O ELCRIC S

5. Area of rectangle ABCD = AB × BC= 50 × 28= 1400 cm2

Radius of each semicircle (r) = BC 282 2

=

= 14 cm.

Area of each semicircle = 12

πr2

= 12

×22

14 147

× × = 308 cm2.

Now, area of remaining paper= Area of rectangle ABCD – 2 × area

of either semicircle= 1400 – 2 × 308= 784 cm2

Further, observing theadjoining figure, wehave length of curveAPD or curve BQC.

πr = 22

147

×

= 44 cm.Now, perimeter of remaining paper

= AB + CD + curve APD + curve BQC= 50 + 50 + 44 + 44 = 188 cm.

6. Radius = 6 cm∴ Diameter PS

= 2 × 6 = 12 cmAlso, PQ = QR = RS

= 123

= 4 cm.

Perimeter = length of PQ + length of QS

+ length of PS

= π [2 + 4 + 6] = 12 ×227

= 37.71 cm.

Area = ( ) ( ) ( )2 2 26 4 22 2 2π π π− +

= 2π

(36 − 16 + 4) = 22

7 2× × 24

= 5282 7×

= 2647

= 37.71 cm2.

7. 98.4 cm2

Hint: Inner radius = r

∴ r = S∆

;

where ∆ = area of ∆ABCS = semiperimeter of ∆ABC.

∴ Shaded part = ar(∆ABC) − ar(circle)

8. (i) Perimeter of sector = 2r + Arc length

= 2 × 5.7 +180

rπ θ°

∴ 27.2 = 11.4 + 180

rπ θ°

⇒180

rπ θ°

= 27.2 – 11.4

= 15.8 m∴ Arc length = 1580 cm.(ii) 450300 cm2

Hint: Area of sector = 12

l r.

WORKSHEET–871. (A)

Hint: 2πr = πr 2

⇒ r = 2 units.

2. (A)Hint: 2πr = 2πr1 + 2πr2⇒ r = r1 + r2

= 19 + 9 = 28 cm.

3. False, because radii are different so theirareas will also be different if theircorresponding arc lengths are equal.

4. 22 cm

Hint: Arc length = 180

rπ θ°

.

5. Shaded part = Area of square − 4 × Areaof circle

= (14)2 − 4(πr 2)

= 196 − 4 ×22 7 77 2 2

× ×

= 196 − 154= 42 cm2.


OR

Length of arc AEB = π × 2.82

= 1.4π cm.

Length of arc BFC = π × 1.42

= 0.7π cm.

Length of arc ADC = π2.8 1.4

2+

= 2.1π cm

Now, perimeter of shaded region= Sum of lengths of the arcs AEB, BFC

and ADC= 1.4π + 0.7π + 2.1π

= 4.2π = 4.2 × 227

= 13.2 cm.

6. 1543

cm2

Hint: r = 14 cmAngle swept in 1 min = 6°Angle swept in 5 min = 30°

∴ Area = 2

360rπ θ

°.

7. 88.44 cm2

Hint: Use 2

21sin

360 2r

rπ θ

− θ°

sin120° = sin60° = 32

.

8. 66.5 cm2

Hint: See solved example 7.

WORKSHEET– 88

1. (C) Circumference = 2πr

= 22

2 427

× ×

= 264 cm.

∴ No. of revolutions = 792 100

264×

= 300.

2. (B) 2πr = 22

⇒ r = 222π

= 11π

∴ Area of quadrant = 214

rπ

= 1 11 11

× × ×4

ππ π

= 1214π

cm2.

3. Area of sector = 2

360rπ θ

°

=3.14 × 4 × 4 × 30

360°

°= 4.19 cm2

Area of major sector = ar(circle) – 4.19= 3.14 × (16) – 4.19= 46.05 cm2

4. True.Hint: ∠AOB = 60°

ar(∆OAB) = 3

4× (side)2.

5. 151

3cm2

Hint: Use, area of shaded part

= 2 240 R360

rπ × ° − °

R = Outer radiusr = Inner radius.

6. (i) The horse can graze in the shape of aquadrant of a circle with radius 5 m

∴ Required area =14

π × (radius)2

=14

× 3.14 × 5 × 5

= 19.625 m2.(ii) Radius of quadrant of circle

= Length of the rope = 10 m

Area of the sector =14

× π × 102

=14

× 3.14 × 100

= 78.50 m2

So, the increase in the grazing area= (78.50 – 19.625) m2

= 58.875 m2.


7. Let us mark the four unshaded parts asI, II, III, IV in figure.

∴ Area of I + area of III = area of ABCD −area of two semicircles of radius 5 cm each.

= 100 − 3.14 × 25= 21.5 cm2

Similarly, area of II + area of IV = 21.5 cm2

Area of shaded part= ar(ABCD) − ar(I + II + III + IV)= 100 − 2 × 21.5= 57 cm2.

8. Radius of the circle having ABC as quadrant= AB = AC = 14 cm

Area of this quadrant ABC = 14

× π × 142

= 49π cm2

Area of isosceles ∆ABC = 12

× AC × AB

=12

× 14 × 14 = 98 cm2

BC = 2 2AB AC+

= 2 214 14+ = 14 2 cm.

Area of semicircle having diameter as BC

= 12

π × 2

14 22

= 12

π × ( )27 2 = 49π cm2

Now, area of shaded region = Area of ∆ABC+ Area of the semicircle with BC asdiameter – Area of quadrant ABC

= 98 + 49π – 49π = 98 cm2.

WORKSHEET – 89

1. (A) Angle = 360°× 35 210

60= ° .

2. (C) Perimeter of square= Circumference of circle

= 2πr = 2 × 227

× 21 = 132 cm

∴ Side of the square = 132

4 = 33 cm.

3. Let O be the centre of circle.

In ∆OBD, cos 60° =ODOB

and sin 60° = BDOB

⇒ OD32

=12

and 32

=BD32

⇒ OD = 16 and BD = 16 3

⇒ BC = 2BD = 32 3

... Area of design= Area of circle – Area of ∆ABC

= ( )22 3× 32 32 3

4

π − ×

cm2

= 322 × 3 1.73

3.14 –4

×

cm2

= 1024 × 12.56 – 5.19

4

cm2

= 256 × 7.37 = 1886.72 cm2.

4. 161.31 cm2

Hint: Required area = 12

πr2 – ar(∆PRQ).

5.

(i) The distance around the track along theinner edge = 106 + 106 + (π × 30 + π × 30)

(∵ Inner radius = 30 m)


= 212 + 227

× 60

= 212 + 1320

7 =

28047

m = 4400

7m.

(ii) The area of track

= (106 × 80 − 106 × 60) + 2 × 2 240 302π −

= 2120 + 700 × 227

= 2120 + 2200 = 4320 m2.

6. AB = 56 mArea of lawn ABCD = AB2 = 562 = 56 × 56

= 3136 m2 ... (i)

Draw OM ⊥ AB.

AC = 2AB = 56 2 m

AO =AC2

= 56 2

2 = 28 2 m

⇒ Radius (r) of circular flower bed

= 28 2 m

OM = BC 56

=2 2

= 28 m

Area of each circular flower bed= Area of sector AOBP – Area of ∆AOB

=14

πr2 – 12

× AB × OM

= 14

× 227

× 28 2 × 28 2 –12× 56 × 28

= 1232 – 784= 448 m2 ...(ii)

Using equations (i) and (ii), we get therequired area

= 3136 + 2 × 448= 4032 m2.

7. Inner area of ring I = π × (Radius)2

= π × 232

2

= 256π cm2

Outer area of ring I = π × 234

2

= 289π cm2

So, area of ring I = Outer area – Innerarea

= 289π – 256π

= 33π cm2

Inner area of ring II = π × 219

2

= 90.25π cm2

Outer area of ring II = π221

2

= 110.25π cm2

So, area of ring II = 110.25π – 90.25π= 20π cm2

Therefore, total area of these two rings= (33π + 20π) cm2

= 53π cm2.

8. Sum of areas of 4 quadrants

= 4 × 21Radius

4 π ×

= π × 12 = π cm2

Area of the circle in the middle= π × (Radius)2

= π × 22

2

= π cm2

Area of the square = (Side)2

= 42 = 16 cm2.Now, area of the remaining portion

= 16 – π – π = 16 – 2π

= 16 – 2 × 227

= 9.71 cm2.



1. (A) πR2 = π216

2

+ π212

2

⇒ R2 = 64 + 36 = 100⇒ R = 10 m.

2. (C) Area = π(21)2 × 120360

°°

=22 1

× 21 × 21 ×7 3

= 462 cm2.

3. Let central angle be θ.

Area = πr2 ×360

θ°

⇒ 54π = π × 362 ×360

θ°

⇒ θ =54 ×36036×36×

π °π

= 15°

Now, length of arc

(l) =180

rπ θ°

=

22× 36 × 15

7180

°

°

= 22× 3

7= 66

7= 9.43 cm.

Alternative Method:

Area of sector =12

lr.

⇒ 54π = 12

× l × 36

∴ l =×54 × 22 2

7 × 36=

667

= 9.43 cm.

4. False, as the corresponding radii areunequal so the area is different.

5. Join SP and QR.ar(rectangle ABCD) = AB × CD = 26 × 12

= 312 m2

Semicircles SPM and RQN are congruent.Radius(r) of the semicircle SPM

=1

(12 4 4)2

− − = 2 m

∴ ar(semicircle SPM) + ar(semicircle RQN)

= 12

πr2 + 12

πr2 = πr2 = 227

× 4 = 887

m2

SP = RQ = 2r = 4 mPQ = SR = AB – 3 – 3 – r – r

= 26 – 3 – 3 – 2 – 2 = 16 mNow, ar(rectangle SPQR) = SP × PQ

= 4 × 16 = 64 m2

Area of the shaded region

= 312 – 88

647

−

= 312 – 12.57 – 64= 235.43 m2.

6. ACB is the minor segment of a circle withcentre O and radius OA = OB = r = 14 mJoin AB.In ∆OAB, OA = OB (Radii of same circle)⇒ ∠2 = ∠1 ... (i)

(Angles opposite to equal sides)∴ ∠1 + ∠2 + 60° = 180°

(Angle sum property for a triangle)⇒ ∠1 + ∠1 + 60° = 180° [Using (i)]⇒ ∠1 = ∠2 = 60° [Using (i)]⇒ ∆AOB is equilateral

∴ ar(∆AOB) =3

4r2 =

34

× 142

= 49 23 m


ar(sector AOBC) = πr2 × 60360

°°

= 22 114 14

7 6× × ×

= 2308 m

3Now, ar(segment ACB)

= ar(sector AOBC)– ar(∆AOB)

= 230849 3 m

3 −

.

7. Sides of ∆ABC are AB = 15 m, BC = 16 mand CA = 17 m. Areas grazed by threeanimals tied at A, B and C are respectivelyAPU, BQR and CST.The ungrazed area is shaded, i.e., PQRSTU(see figure) which to be required.In ∆ABC,

θ1 + θ2 + θ3 = 180° ... (i)

(Angle sum property of a triangle)

ar(sector APU) + ar(sector BQR)+ ar(sector CST)

= πr2 × 1

360θ

°+ πr2 × 2

360θ

°+ πr2 × 3

360θ

°

= 2

360rπ

° (θ1 + θ2 + θ3)

= 222 7

7 360×

° × 180° [Using (i)]

= 77 m2

For ∆ABC, semiperimeter

s =15 16 17

2+ +

= 24 m

∴ ar(∆ABC) = 24(24 15)(24 16)(24 17)− − −

= 24 9 8 7× × × = 24 221 m

Now, area of shaded region

= ( )24 21 77− m2

Hence, area of ungrazed region

= ( )24 21 77− m2.

8. ar(∆ABC) =12

× Base × Height

(∵ ∠A = 90°)

=12

× 10 × 10 = 50 cm2.

ar(sector APR) = π(7)2 × 90360

°°

=227

× 7 × 7 × 14

= 38.5 cm2.ar(PBCR) = ar(∆ABC) – ar(sector APR)

= 50 – 38.5 = 11.5 cm2.Cost of the silver plating on PBCR

= ar(PBCR) × Rate= 11.5 × 20= ` 230.00.


1. (B) πr2 = 1.54 ⇒ r2 =1.5422

× 7 ⇒ r = 0.7 m

Number of revolutions = 17622

2 0.77

× × = 40.

2. (A) This rhombus must be a square withdiagonals as the diameters of the circle.

πr2 = 1256

⇒ r2 = 1256 7

22×

⇒ r = 20 cm (approx.)

∴ Diameters = d1 = d2 = 40 cm

∴Area of the rhombus = 12

× d1 × d2

= 12

× 40 × 40

= 800 cm2.


3. Minutes elapsed by minute hand= 6.40 – 6.05 = 35

∵ Angle covered by minute hand in60 minutes = 360°∴ Angle covered by minute hand in

35 minutes = 36060

°°

× 35 = 210°

∴ Area = πr2 ×210360

°°

= 222 7 275× 5 × =

7 12 6

=5

456

cm2.

4. True, because when areas are equal, theradii are equal and so circumferences areequal.

5. In ∆AOB, AB2 = 202 = 400

And AO2 + BO2 = ( ) ( )2 210 2 10 2+

= 400

i.e., AO2 + BO2 = AB2 ⇒ ∠AOB = 90°

∴ ar(∆AOB) = 12

× AO × BO

= 12

× 10 2 10 2×

= 100 cm2

ar(sector AOB) = π × (AO)2 × 90360

°°

= 50π cm2

ar(rectangle ABCD) = AB × BC= 20 × 10 = 200 cm2

Now, area of the shaded region= ar(rectangle ABCD)

– ar(sector AOB) + ar (∆AOB)= 200 – 50π + 100= 300 – 50π = 50(6 – π) cm2.

6. r = 10 cmLet AB is the chordwhich subtend anangle of 60° at centreof circle.

∴ Area of corresponding segment= Area of APB

A =2

360°rπ θ

– ar(∆AOB)

As ∆AOB is equilateral

∴ A =2 ×60π(10)

360°–

3× 10 × 10

4

=157

25 33

− cm2.

7. Area of square ABCD = BC2

= 142 = 196 cm2

EFG, GHI, IJK and KLE are four semicircles

of radius 14 3 3

4− −

= 2 cm each.

EGIK is a square of side as the diameter ofeither circle, that is, 4 cm.

Sum of areas of the four semicircles= 4 × area of one semicircle

= 4 × 12

π × 22 = 8π cm2

Area of square EGIK = (Side)2 = 42 = 16 cm2

Now, area of shaded region= Area of square ABCD

– Sum of areas of four semi-circles – Area of square EGIK

= 196 – 8π – 16= (180 – 8π) cm2.

8. In right ∆ABC,AC2 = AB2 + BC2

(Pythagoras theorem)


⇒ AC2 = 142 + 142

⇒ AC = 14 2 cm

ar(∆ABC) =12

× Base × Height

=12

× 14 × 14

= 98 cm2

Radius of quadrant ABCP = AB = BC = 14 cmar(quadrant ABCP)

=14

× Area of corresponding

circle

=14

× π(AB)2 = 14

×227

× 14 × 14

= 154 cm2

ar(semicircle ACQ) = 12

× π × 2AC

2

=12

×227

× 14 22

×14 2

2[∵ AC = 14 2 ]

= 154 cm2

Now, area of shaded portion= ar(semicircle ACQ)

– ar(quadrant ABCP)+ ar(∆ABC)

= 154 – 154 + 98 = 98 cm2.Hence, area of shaded portion is 98 cm2.

CHAPTER TEST

1. (A) Two vertices of thetriangle should coincidewith the two extremitiesof the diameter of thesemicircle and the thirdone lies on the curve.

ar(∆ABC) =12

× BC × AO

= 12

× 2r × r

= r2 sq. units.

2. (B)Hint: Diameter of circle = 16 cm∴ Diagonal of a square = 2 × side.

2 × side = 16side = 8 2

∴ Area of square = 2(8 2)

= 128 cm2.3. Let radius of given circle and side of given

square be r and a respectively.Perimeter of the circle = perimeter of the

square

⇒ 2πr = 4a ⇒ r =2aπ

Required ratio =2

2r

aπ

=2a

π ×2

2

4aπ

=4π

= 4

227

= 1411

i.e., 14 : 11.

4. True.

Let radius of smaller circle = r = 52

Let radius of larger circle = R = 5.22

∴ Ratio of areas = 2

2Rr

ππ

=

2

2

52

2

52

= 2 : 1.

5. Length of arc = Length of the wire = 20 cm

Angle = 60° = 60180

° × π°

radian =3π

radian

Radius of the curve ABC

= Length of arc

Angle

⇒ r = 20

3π =

60π cm.


Area of sector OABC = 32

π

π× πr2

= 16

× π × 60 60×π × π

= 600π

cm2.

6. Let the angle subtended by arc at the centrebe θ.

Area of sector = 54π

⇒360

θ°

× π × (Radius)2 = 54π

⇒360

θ°

× π × (36)2 = 54π

⇒ θ = 360° × 54

36 36× = 15°

Now, length of the arc = Radius × Angle inradian

= 36 ×15

180°°

× π

= 3π cm.

7. In ∆ABC,BC = 48 cm, AC = 50 cm, AB = 14 cmHere, AB2 + BC2 = 142 + 482

= 196 + 2304= 2500 = (50)2

= AC2

So, ∆ABC is a right-angled triangle with∠ABC = 90°

ar(∆ABC) =12

× AB × BC

=12

× 14 × 48

(A1) = 336 cm2 ... (i)

Now area of sectors drawn at each cornerof triangle ABC

(A2) =22 2

31 2+360 360 360

rr r π θπ θ π θ+

° ° °

= 2

360rπ

°(θ1 + θ2 + θ3)

= × 5 ×5360

π°

× 180°

(Using Angle sum property)

= 25

2π

= 25 × 3.14

2= 25 × 1.57 = 39.25 cm2

∴ Required area = A1 – A2= 336 – 39.25 = 296.75 cm2.

8. See Assessment Sheet – 16, Sol. 7.

❑❑


9Chapter

SURFACE AREAS AND VOLUMES

WORKSHEET– 931. (C) Edge of cube = 4 cm

∴ Length of resulting cuboid = 4 + 4 = 8 cmBreadth of resulting cuboid = 4 cmHeight of resulting cuboid = 4 cm

∴ Surface area = 2(lb + bh + hl)= 160 cm2.

2. (B) Volume of big sphere

V1 =43

π (3)3

Volume of 1 small sphere

V2 =43

π (0.3)3

∴ Number of balls obtained

= 1

2

VV

= 270.027

= 1000.

3. 3 : 1.Hint: Volume of cylinder = V1 = πr2h

Volume of cone = V2 =13

πr2h

∴ V1 : V2 = 1 :13

= 3 : 1.

4. 176 mm2

Hint:

Surface area of capsule= Lateral surface area of cylinder+ 2 × Lateral surface area of hemisphere

5. Volume of water filled in the cone

=13

π r2h =13

π × 4 × 4 × 12

= 64π cm3

Let the number of lead shots be x.

So, volume of x lead shots = 14

× capacity

of the cone

⇒ x × 43

π × (0.5)3 =14

× 64π

⇒ x =16 3

4 0.125×

×=

480.5

= 96

Hence, the required number of shots is 96.

6. Volume of earth dug out = πr2h

= 227

× 7 × 7 × 20 = 154 × 20 = 3080 m3

Area of platform = 22 × 14 − π r2

= 22 × 14 − 227

× 72

= 154 m2

Now, height of the platform

= Volume of earth dug out

Area of platform

= 3080154

= 20 m.

7. Radius of the pipe = r = 142

= 7 cm =7

100m

Rate of flowing of water= 15 km/h

=15 1000

3600×

m/s =256

m/s

Let the required time be t.Water flowing through the pipe in 1 second

= πr2 × Rate of water flowing

= 322 7 7 25× × × m

7 100 100 6= 377

m1200

∴ Water flowing into the tank in t seconds

=77

1200t m3

263FRUS AERACA S NA D ULOV EM SE

This volume of the water is the same asthe volume of the water in the tank at theheight of 21 cm.

∴77

1200× t = 50 × 44 ×

21100

⇒ t = 50 44 21 1200

77 100× × ×

×

= 7200 seconds⇒ t = 2 hrs.

8. r = 8 cm; R = 20 cm; h = 16 cm

∴ Slant height = l = ( )22 + R –h r

= 2 216 +12

= 400 = 20 cm.

∴ Total surface area of container= πl (R + r) + πr2

= π [l (R + r) + r2]= 3.14 [20 × 28 + 64]= 1959.36 cm2

∴ Cost of metal sheet used

= 1959.36 ×15

100= ` 293.90.

WORKSHEET– 94

1. (A) Volume of sphere = 43

π(3)3

Let length of wire = l.

Radius of wire (r) =42

= 2 mm = 2

10cm.

∴ Volume of wire = πr2l = π22

10

× l

∴ Volume of sphere = Volume of wire

⇒43

× 27 = 4100

l

⇒ l = 900 cm.

2. (B)

Hint: V = 1 2 1 2A A A A3h + +

= 5(20 + 8) = 140 cm3.

3. Let the base radii be 3x and 5x respectivelyand let the same height be h.

V1 = 21(3 ) ×

3x hπ ; V2 = 21

(5 ) ×3

x hπ

∴ 1

2

VV

=2

2925

xx

, i.e., V1 : V2 = 9 : 25.

4. 53.625 cm2

Hint:H = height of cone

= 7.75 – 1.75= 6 cm

l = slant height of cone

= 2 2Hr +

∴ T.S.A. = C.S.A. of cone + C.S.A. of hemisphere

= πr l + 2π r 2

= π r [l + 2r].

5. T.S.A. = C.S.A. + 2 × (C.S.A. of hemisphere)= 2πrh + 2(2πr 2) = 2πrh + 4π r 2

= 2πr [h + 2r] = 2 × 227 × 3.5 [10 + 7]

= 2 × 22 × 0.5 × 17= 374 cm2.

6. 4158 cm3

Hint:Volume of solid= Volume of cone + Volume of hemisphere

= 13 πr2h + 2

3πr3 =

2

3rπ (h + 2r).


7. Volume of material of hollow spherical shell= External volume – Internal volume

= 3 34 4(5) – (3)

3 3π π

=3 34

(5 – 3 )3

π = 34 × 98cm

3π

Let the radius of the cylinder be r.

Volume of the cylinder = πr2 × 38cm

3

As volume of the material will remainsame.

∴ πr2 ×83

=4 98

3×

π

∴ r2 = 49 ⇒ r = 7 ⇒ 2r = 14Hence, the diameter of the cylinder is14 cm.

8. Radius of bottom = r1 = 202

= 10 cm

Radius of top = r2 = 402

= 20 cm

Height = h = 12 cm

Volume of the bucket = 3

πh(r1

2 + r22 + r1 r2)

= 3.14 12

3×

(102 + 202 + 10 × 20)

= 12.56 × 700 = 8792 cm3

Slant height of the bucket

l = ( )221 2–+h r r

= ( )2212 20 – 10+

= 244 = 2 61 = 2 × 7.81

= 15.62 cm

Surface area of the bucket

= π (r1 + r2) l + πr12

= 3.14 × (10 + 20) × 15.62 + 3.14 × 100

= 1785.40 cm2

= 17.85 dm2

Now,Cost = Surface area × Rate

= 17.85 × 1.20 = 21.42Hence, the volume is 8792 cm3 and cost is` 21.42.

WORKSHEET – 95

1. (A) Let the increase in level be h.

Increase in volume of water = Volume ofthe sphere

⇒ π × 82 × h = 34× × 6

3π

⇒ h = 4.5 cm.

2. (D) Let radius of the cone = r

213

rπ x = 343

xπ

⇒ r = 2x cm.

3. Let the required number of cubes be n.Volume of n cubes = Volume of the ball

⇒ n × 13 =4 22 21 21 21

× × × ×3 7 2 2 2

⇒ n = 4851.

4. T.S.A. of remaining solid= C.S.A. of cylinder + Area of circular base + C.S.A. of cone.= 2πrh + πr2 + πr l= πr [2h + r + l];

= 227 × 0.7 [4.8 + 0.7 + 2.5]

( ) ( )2 22.4 0.7 2.5 cml ∴ = + = = 17.6 ≈ 18 cm2.

5. Volume of 1 conical depression

V1 = 213

r hπ

= ( )21 22× × 0.5 × 1.4

3 7

= 22 × 0.25 × 0.2

3= 0.37


∴ Volume of 4 conical depressions

V2 = 4 × V1

⇒ V2 = 1.48 cm3

Volume of wood in stand= 15 × 10 × 3.5 – 1.48= 523.52 cm3.

6. Let n cones are required.Volume of ice cream in n cones

= Volume of ice cream in 4 cylinders

⇒ 2 31 7 2 7

× × × 12 + ×3 2 3 2

n π π

= 4 × π ×221

2

× 38

⇒21 7 7

× × × × 2 6 +3 2 2

π n

= 4 × π ×21× 21

4× 38

⇒× 49 × 19

12n

= 21 × 21 × 38

⇒ n = 3 × 3 × 2 × 12

⇒ n = 216.

7. 157.5 cm.

Hint: Volume of water flows out of pipe in1 hour (V1) = π (1)2 × 70 × 3600 cm3

Let level of water rise in tank = h

Volume of tank (V2) = π (40)2 × h

∴ V1 = V2

⇒ h = 70 3600

40 40×

×

= 157.5 cm.

8. 4

7827

cm2

Hint: Area of sheet required = Surface areaof cylinder + Surface area of frustum

= 2π r1h + π(r1 + r2) l;where r1 = radius of smaller base

and r2 = radius of larger base.

WORKSHEET – 961. (A) ∆ABC ~ ∆ADE

2

1

rr

= 2hh

⇒ r2 = 2r1

Volume of upper part

V1 = 13

π r12h

Volume of lower part

V2 = 13

πh (r12 + 4r1

2 + 2r12) = 2

173

πr h

∴ V1 : V2 = 1 : 7.

2. (C) Let required height = H.

πr2H + 213

πr h = 3 × 213

πr h

⇒ H +13

h = h ⇒ H = 23h

.

3. Volume of whole solid= Volume of cone + Volume of cylinder

+ Volume of hemisphere.

= 13 πr2H + πr2h + 2

3 πr3

= 13 π r2 [H + 3h + 2r]

= ( )1 22 7 72.8 19.5 7

3 7 2 2× × × × + +

= 22 7

29.33 4

× ××

= 376.016 cm3.

4. False, because the volume of the ball

having radius as2a

= 43

π3

2a

= 34

24aπ .

5. Height of the cylinder = h = 20 cm.Radius of the cylinder or of the hemisphere

= r =10.5

2= 5.25 cm


Total surface area= Curved surface area of the cylinder

+ 2 × curved surface area ofeither hemisphere

= 2πrh + 2 × 2πr2 = 2πr (h + 2r)

= 2 × 227

× 5.25 (20 + 10.5)

= 33 × 30.5 = 1006.5 cm2.

6. Let the radius and slant height of theconical heap be r and l respectively.Volume of sand filled in the bucket

= π × (18)2 × 32Volume of sand in conical form

= 21× × 24

3π r

As volume of sand will remain same.

∴ 21× 24

3πr = π × (18)2 × 32

⇒ r2 =3 × 18 × 18 × 32

24 = 1296

⇒ r = 36 cm

Further, l = ( ) ( )22Radius Height+

= 2 236 24+ = 1872

= 12 13 cm

Thus, radius = 36 cm and slant height

= 12 13 cm.

7. Rate of water flowing = 56 m/sce

∴In 1 second water flows = 56 m

Internal diameter = 15 m

∴ Volume of water flows out from pipe

in 1 second = πr2h

= 222 1 5

× ×7 10 6

= 311m

420.

Volume of water in tank = πr2h

= 222 10

× × 27 2

= 31100m

7

∴ Time taken to fill the tank

=

1100711420

= 100 × 60 seconds

= 1 hr 40 min.

8. h = 16 cm, r1 = 8 cm, r2 = 20 cmCapacity of container (frustum)

= ( )2 21 2 1 2+ +

3h

r r r rπ

=3.14 × 16

3 (82 + 202 + 8 × 20)

=50.24

3 (64 + 400 + 160)

=50.24

3 × 624 = 50.24 × 208

= 10449.92 cm3 = 10449.92

1000 l

= 10.44992 lCost of milk

= Capacity in litres × Rate per litre= 10.44992 × 20= ` 208.9984 ` 209.

If l be the slant height of the frustum, then

l = ( )222 1–+h r r = 2 216 12+

= 400 = 20 cm

Total surface area of the frustum= π (r1 + r2) l + πr1

2

= 3.14 [(8 + 20) × 20 + 82]= 3.14 × 624 = 1959.36 cm2

Cost of metal sheet used

= 1959 . 36

× 8100

= 156.7488 ≈ ` 156.75

Thus, cost of milk is ` 209 and cost of metalsheet is ` 156.75.


WORKSHEET– 97

1. (A) Surface area of cone= Surface area of hemisphere

⇒ πr 2 2+r h = 2πr2

⇒ r2 + h2 = 4r2

⇒ 3r2 = h2

⇒2

2rh

=13

⇒ r : h = 1 : 3 .

2. (D) N × 34 4.2

×3 2

π = 66 × 42 × 21

⇒ N = 66 × 42 × 21× 3 × 7

4 × 22 × 2.1× 2.1× 2.1 = 1500.

3. Let H = 120 cm = height of coneh = 180 cm = height of cylinder

r = 60 cm = radius of cone, cylinderand hemisphere

∴ Volume of water left in cylinder = Volume of cylinder – Volume of cone

– Volume of hemisphere

= 2 2 31 2H

3 3r h r rπ − π − π

= 2 H 23 3

r h r π − −

= 227

× 60 × 60 [180 – 40 – 40]

= 227

× 60 × 60 × 100 cm3 = 1.131 m3.

4. False, because 34R

3π = 8 × 34

3πr ⇒ r =

R2

.

5. r = 14 m

h = height of cone

= 13.5 – 3 = 10.5 m

H = height of cylinder

= 3 m

∴ l = 2 214 + 10.5 = 196 + 110.25

= 306.25 = 17.5 m

Area to be painted = C.S.A. of cone + C.S.A.

of cylinder.

= πr l + 2π rH = πr [l + 2H]

= 227

× 14 × 23.5 = 1034 m2

∴ Cost = 1034 × 2 = ` 2068.

6.

r = radius of cone =402

= 20 cm

h = height of cone = 24 cm

∴ Volume of cone (V1) = 13 π r2h cm3

Also, volume of water flows out of pipe in1 min

V2 = πR2H

= 2

351000 cm

20 π ×

Let conical vessel fills in 't' min.

∴ V1 = t × V2


⇒ t = 1

2

VV

= ( )

2

120 24

35

100020

2π ×

π × 400 × 8 × 400

25 ×1000= 51.2 min.

= 51 min 12 sec.

7. r1 = radius of small cylinder = 8 cmh1 = height of small cylinder = 60 cmr2 = radius of big cylinder = 12 cmh2 = height of big cylinder = 220 cm

Volume of pole= Volume of small cylinder

+ Volume of big cylinder.

= πr12h1 + πr2

2h2

= 3.14 × [64 × 60 + 144 × 200]

= 3.14 × [3840 + 31680]

= 111532.8 cm3

∴ Mass = 111532.8 × 8 g

= 892.26 kg.

8. 340π m2

Hint:

Quantity of canvas required

= C.S.A. of cone + C.S.A. of frustum.

WORKSHEET – 98

1. (A)

2. (C)

3. Number of bags = Volumeof circular drum

Volumeof each bag

= 23.14×(4.2) × 3.5

2.1≈ 92.

4. L = 15 cm, B = 10 cm and H = 5 cmVolume of block = L × B × H

= 15 × 10 × 5= 750 cm3.

The hole is only possible throughout thesurface having area 15 cm × 10 cm.Volume of circular hole

= π r2h = 222 7

× × 57 2

= 22 7

54× ×

=154 5

4×

= 192.5 cm3

∴ Volume of remaining solid= 750 − 192.5= 557.5 cm3

5. 6.521 kg (approx.)

Hint: Volume of metal in pipe = πh (R2 – r2)where, R = outer radius and r = inner radius.Also, mass = volume × density.

6. Canal is in the form of cuboid.So, width = 30 dm = 3 m

height = 12 dm = 1.2 mlength = distance travelled by water

in 30 min.= 5 km = 5000 m

∴ Volume of water in canal = Volume of cuboid

= 3 × 1.2 × 5000= 18000 m3

Let the required area for irrigation bex square metre. Then

8×

100x = 18000

{... Area × height = volume}⇒ x = 225000 m2.


7. Let BO = r

∴ OD = r

AC = 2 215 20+

= 225 400+

= 625 = 25 cm.

Area of ∆ABC = 1

× BC × AB2

=1

15 202

× × = 150 cm2 ...(i)

Again area of ∆ABC

= 1

× AC × BO2

= 1

× 25 ×2

r

...(ii)From (i) and (ii), we obtain

r = 12 cm.From ∆AOB,

AO = 2 220 12− = 16 cm.

Now, volume of double cone= Volume of cone BADOB + Volume

of cone BCDOB

= ( ) ( )2 21 1× 12 × 16 + × 12 × 9

3 3π π

= ( )1 22× × 144 16 + 9

3 7= 3771.43 cm3

Surface area of double cone

= ×12 × 20 + ×12 ×15π π

= ( )22× 12 20 + 15

7 = 1320 cm2.

8. Let height and radius of the original coneABC be AO = 3h and BO = 3r respectively.As planes PQR and STU trisect the coneABC,AQ = QT = TO = h

In ∆ABO, PQ BO,

∴PQBO

= AQAO

⇒ PQ3r

= 3hh

⇒ PQ = r

Similarly, ST = 2r

Now, VAPR = 213

πr h ...(i)

VPSUR = 3

πh[r2 + (2r)2 + r × 2r]

= 273

πr h ...(ii)

VSBCU = 3

πh[(2r)2 + (3r)2 + 2r × 3r]

= 2193

r hπ ...(iii)

Using equations (i), (ii) and (iii), we getVAPR : VPSUR : VSBCU := 1 : 7 : 19

Hence proved.OR

1.05 l; 1961.14 cm2

Hint: Capacity = Volume

= ( )2 21 2 1 23

hr r r r

π+ +

Surface area = ( )21 1 2r r r lπ + π + , r1 < r2


1. (C) π × 4x × 7x = 792

⇒ x2 = 792 7

22 4 7×

× × ⇒ x = 3

⇒ Radius = 4x = 12 cm.

2. (B) ∆AMN ~ ∆AOC

⇒h

h h′

+ ′= 2

1

rr

⇒ h′r1 = hr2 + h′r2⇒ h′(r1 – r2) = hr2

⇒ h′ = 2

2–hr

r r′

⇒ h′ + h = 1

1 2–hr

r r.

3. h = 12 m

π 21r = 9 and π 2

2r = 4


⇒ 21r =

9π and 2

2r = 4π

Volume = 3

πh× ( 2

1r + 22r + rl r2)

= × 123

π 9 4 9 4+ + ×

π π π π

= 4π × 1π

(9 + 4 + 6) = 76 m3 .

4. False, volume of the ball with radius 2a

=34

3 2 π

a= 31

6πa .

5. Edge of the cube = lRadius of the hemisphere

= r = 2l

After cutting out thehemispherical depres-sion, the surface area ofthe remaining solid

= Surface area of the cube + Innercurved surface area of hemisphere– Base area of the hemisphere

= 6l2 + 2πr2 – πr2 = 6l2 + πr2

= 6l2 + π2

2l

= 21

(24 )4

l + π sq. units.

6. Let the width of the embankment be x.Let radius of well (r)

= 72

m

Volume of the earth

dug out = πr2h

= π × 27

2

× 22.5

Volume of the embankment= π (r + x)2H – πr2H= πH (r + x + r) (r + x – r)= π × 1.5 × (7 + x) x

Since, the embankment is formed by usingthe earth dug out. Therefore, their volumesshould be equal.

i.e.,π × 1.5 (7 + x) x = π × 27

2

× 22.5

⇒ 7x + x2 = 15 × 7 × 7

2 × 2⇒ 4x2 + 28x – 735 = 0This is a quadratic equation in x. Now, wehave to solve it.

∴ ( )22 7+x = 735 + 49 = 282

⇒ 2x + 7 = ± 28⇒ x = 10.5, – 17.5Since the width cannot be negative.Therefore, x = 10.5Hence, width of the embankment is 10.5 m.

7. Radius of the pipe = r = 142

= 7 cm

Rate of flowing the water = 15 km/hr= 1500000 cm/hr

Volume of water flowing per hour from thepipe = πr2h

= 2227 1500000

7× ×

= 22 × 7 × 1500000 cm3

Volume of water required in the tanks= Length × Width × Height= 50 m × 44 m × 21 cm= 5000 × 4400 × 21 cm3

Required time

= Volume of water required

Volume of water flowing per hour

= 5000 × 4400 × 2122 × 7 × 1500000

= 2 hours.

8. In the adjoining figure,in ∆ABE and ∆ACD,BE CD, CD ⊥ AC andBE ⊥ AC⇒ ∆ABE ~ ∆ACD

⇒ABAC

= BECD

⇒2hh

= Rr

⇒ r = R2

⇒ r = 5 cm ... (i)

(∵ R = 10 cm as given)


V1 = Volume of upper part = 213

r hπ

= 1× 25

3hπ ...(ii)

[Using (i)]V2 = Volume of lower part

= 2 21( R R)

3π + +h r r

= ( )125 + 100 + 50

3πh

= 1

×1753

πh ... (iii)

∴ Required ratio 1

2

VV

=

1× 25

31

×1753

π

π

h

h

[Using(ii) and (iii)]

= 17

Hence, the ratio of the volumes of the twoparts is 1 : 7.


1. (B) C.S.A. = Inner C.S.A. + Outer C.S.A.

= 2π 21r + 2π 2

2r = 2π ( )2 21 2+r r .

2. (A) n ×43

π × 36

2

= ( )2112 24

3π × ×

⇒ n = 12 ×12 × 8 × 34 × 3 × 3 × 3

n = 32.

3. πr2h = 448 π ⇒ r2 × 7 = 448

⇒ r = 64 ⇒ r = 8 cm

Curved surface area = 2πrh = 2 ×227

× 8 × 7

= 352 cm2.

4. False, sides become a, a and 2a and so surfacearea will become2(a × a + a × 2a + 2a × a) = 2(a2 + 2a2 + 2a2)

= 10a2.

5. Let the edge of the cube be a and the radiusof the sphere be r.Surface area of the sphere

= Surface area of the cube

⇒ 4πr2 = 6a2 ⇒ r2 = 26

4πa ⇒ r =

32π

a

Volume of the sphere (V1) = 343

rπ

=

3234 3

3 2a π π

Volume of the cube (V2) = a3

∴ Ratio in their volumes 1

2

VV

=

3234 3

3 2a π π

: a3

=

–3 –32 2 2

–32

2 × 2 × ×

3 × 3

π πa3 : a3

=

12 ×

13

π a3 : a3 = 6π

: 1.

∴ V1 : V2 = 1 :6π

.

6. h = 7 cm; r = 52

mm = 0.25 cm

Volume of the barrel = πr2h

= 22

× 0.25 × 0.25 × 77

= 1.375 cm3

Volume of ink in the bottle

= 15

litre = 311000 cm

5× = 200 cm3

Number of barrels filled by the ink of bottle

= 200

1.375=

2000001375

= 1600

11

∵ Number of words written by 1 barrel = 330


∴ Number of words written by 1600

11 barrels

= 330 × 1600

11 = 48000.

7. The given hollowcone of base radius r2is cut at a distance h1from the vertex of it.The upper part is alsoa cone of height h1,base radius r1 and theslant height l1. Theremaining part is a frustum of height h2, baseradii r1, r2 and slant height l2. Thus, the heightand slant height of the whole cone will be h1+ h2 and l1 + l2 respectively (see figure).

∆ABC ~ ∆ADE (AA criterion of similarity)

⇒ADAB

= DEBC

= AEAC

⇒ 1 2

1

+h hh

= 2

1

rr

= 1 2

1

+l ll

... (i)

⇒ 2

11 + h

h= 1 + 2

1

ll

⇒ 1

2

hh

= 1

2

ll

... (ii)

Curved surface of the frustum = 89

× curved

surface of the whole cone

⇒ π (r1 + r2) l2 = 89

πr2 (l1 + l2)

⇒ 9r1l2 + 9r2l2 = 8r2l1 + 8r2l2

⇒ 8r2l1 – 9r1l2 = r2 l2

⇒ 1

2

8 ll

– 1

2

9 rr

= 1

(Dividing throughout by r2l2)

⇒ 8 1

2

hh

– 9 1

1 2

+

hh h

= 1

[Using (i) and (ii)]

⇒21 1 2 1 2

2 1 2

8 + 8 – 9( + )

h h h h hh h h

= 1

⇒ 8 21h – h1 h2 = h1 h2 + 2

2h

⇒ 8 21h – 2

2h = 2h1h2

⇒ 8 1 2

2 1–

h hh h

= 2

(Dividing throughout by h1h2)

⇒ Put 1

2

hh

= x to get

⇒ 8x – 1x

= 2

⇒ 8x2 – 2x – 1 = 0

⇒ (2x – 1) (4x + 1) = 0

⇒ x = 12

or x = 1

–4

Since ratio of heights can't be negative,

so x ≠ 1–

4

∴ x = 12

⇒ 1

2

hh

= 12

⇒ h1 : h2 = 1 : 2

Hence the plane divides the altitude of thecone in the ratio 1 : 2.

8. Radius of cylindrical vessel = R = 212

cm and

height of it = H = 38 cmVolume of each cylindrical vessel = πR2HVolume of ice cream in 4 vessels = 4πR2H

Radius of cone = r = 72

cm

Height of cone = h = 12 cmVolume of ice cream in the

each conical shape = 213

πr h

Radius of hemisphere = r = 72

cm

Volume of ice cream in each hemispherical

shape = 323

πr


∴ Volume of ice cream in one cone

= 2 31 23 3

π + πr h r

= 21( 2 )

3r h rπ +

Number of required cones

=

Volume of ice cream in 4cylindrical vessels

Volume of ice cream in 1 conewith hemispherical top

= ( )

2

2

4 R H1

23

r h r

π

π + =

( )2

212R H

2+r h r

= ( )

21 2112 × × × 38

2 27 7× × 12 + 72 2

=12 × 3 × 3 × 38

19

= 216.

CHAPTER TEST

1. (D) 1

2

VV

= 6427

⇒

31

32

4343

r

r

π

π=

3

343

⇒ 1

2

rr = 4

3

1

2

SS

= 2

12

2

44

ππ

rr

= 2

1

2

rr

=24

3

=169

∴ S1 : S2 = 16 : 9.

2. (A) 49 × 33 × 24 = 34 22× ×

3 7r

⇒ r = 3 9261 ⇒ r = 21 cm.

3. r1 = 442

= 22 cm,

r2 = 242

= 12 cm, h = 35 cm

Capacity = 3

πh 2 21 2 1 2( )r r r r+ +

= ( )22 35× × 484 +144 + 264

7 3

= 32706.67cm3 = 32706.67

1000l

= 32.7 l.

4. True, because capacity

= 2 32–

3π πr h r

= 2

(3 – 2 )3

πrh r .

5. Radius of each cone = r = 62

= 3 cm

Let the heights of the cone be h1 and h2respectively.∴ h1 + h2 = 21 cm ...(i)

Given: 1

2

VV

= 21

∴

21

22

1313

r h

r h

π

π= 2

1 ⇒ h1 = 2h2

Substitute h1 = 2h2 in equation (i) to geth2 = 7 cm ∴ h1 = 14 cm

Now, V1= 21

13

r hπ = 21 22× × 3 × 14

3 7= 132 cm3

and V2 = 22

13

πr h = 21 22× ×3 ×7

3 7= 66 cm3

Volume of remaining portion= Volume of the cylinder – (V1 + V2)= πr2 × 21 – (132 + 66)

= ( )2223 21 – 132 66

7× × +

= 594 – 198 = 396 cm3.

6. Let the edge of the new formed cube be a.Note that volume of a cube = (Edge)3

Sum of the volumes of three cubes= Volume of the new formed cube

⇒ 33 + 43 + 53 = a3

⇒ a3 = 27 + 64 + 125

⇒ a = ( )13216

⇒ a = 6 cmHence, the required edge is 6 cm.



8. Let height of the building be h.∴ Radius of dome or cylinder

= r =23

h × 12

= 13

h

Capacity of the building = 67 31m

21

⇒ πr2 (h – r) + 323

πr = 1408

21

⇒ πr2h – πr3 + 323

πr = 1408

21

⇒13

πr2 (3h – r) = 1408

21

⇒13

×227

× 219

h 13 –

3

h h =1408

21

⇒ h3 = 1408 × 21× 9 × 3

22 × 8 × 21 ⇒ h = 3 216

⇒ h = 6 m

Hence, height of the building is 6 m. ❑❑

275R SIT C E P A EP A CR P

Practice Paper–1

SECTION-A

1. (C) D = b2 – 4ac = (– 4)2 – 4 ( ) ( )2 – 2

= 16 + 8 = 24.

2. (B) a = – 5, d = 12

∴ an = a + (n – 1)d

⇒ an = – 5 + (n – 1) 12

= – 5 + 2n –

12

= – 11

2 2n+ = ( )1

– 112

n .

3. (C) M is the nearest point from P because

PMO is a straight line.

PO2 = OQ2 + PQ2

(∵ ∠Q = 90°)

⇒ (PM + 6)2 = 62 + 82

⇒ PM + 6 = 10⇒ PM = 4 cm.

4. (C) ∵ tan P = ABBP

= 2

2 3 =

13

= tan 30°

∴ P = 30°.

5. (D)As we know AB + DC= AD + BC⇒ 18 + x = 16 + 10⇒ 18 + x = 26⇒ x = 8 cm.

6. (C)∵ Arc length =180π θ

°r

⇒ 3π =(6)

180π θ

°⇒

3 1806

π × °π ×

= θ

∴ θ = 90°.

7. (D)∵∠OPQ = ∠OPT – ∠QPT= 90° – 60°= 30°.

8. (C) N × 43

× π ×36

2

= π ×24

2

× 45

⇒ N × 36 × π = π × 4 × 45

⇒ N =18036

= 5.

9. (A) In a single throw of die, the evennumber may be 2, 4 or 6.

∴ P(getting an even number) = 36

= 12

.

10. (D) Let the required angle be θ.We know that angle between tangentsand the angle between correspondingradii are supplementary

∴ θ + 35° = 180°⇒ θ = 145°.

SECTION-B

11. V1 = Volume of cone = 213

πr h

= ( )212.1 8.4

3× π × ×

V2 = Volume of sphere = 34R

3π

As V1 = V2

∴ 34R

3π = ( )21

× 2.1 ×8.43

π

⇒ R3 = (2.1)3

⇒ R = 2.1 cm.

12. As ∠APB = 80°⇒ ∠APO = 40°

(... OP bisect ∠APB)Also ∠PAO = 90°∴ In ∆OAP,

∠A + ∠P + ∠O = 180°⇒ ∠O = 50°⇒ ∠POA = 50°.

PRACTICE PAPERS


13. We have kx2 – 5x + k = 0

∴ Comparing it with ax2 + bx + c = 0,

we get a = k, b = – 5 and c = k.For real and equal roots,

b2 – 4ac = 0 ⇒ 25 – 4k2 = 0

⇒ 4k2 = 25 ⇒ k = ±52

.

14. Let the common difference of the twoA.P.’s be d. Then, their nth terms are:

an = 3 + (n – 1).dand a′n = 8 + (n – 1).d∴ an – a′n = 3 + (n – 1).d – 8 – (n – 1).d

= – 5 for all n∈N∴ a2 – a′2 = – 5.

15. The coordinates of centroid are:

1 2 3 1 2 3+ + + +,

3 3x x x y y y

= 0 + 8 + 6 8 + 12 + 0

,3 3

= 14 20,

3 3

.

16. Since A(x, y), B(1, 2), C(7, 0) are collinear.So ar(∆ABC) = 0

⇒ 12

x(2 – 0) + 1(0 – y) + 7(y – 2) = 0

⇒ 2x – y + 7y – 14 = 0⇒ x + 3y – 7 = 0.

17. Total number of cards = 1000The cards each bearing a perfect squarenumber greater than 500 are: 529, 576,625, 676, 729, 784, 841, 900 and 961.∴ Number of such cards = 9(i) P(the first player wins a prize)

= 9

1000(ii) After winning the first player:

Number of total cards = 999and number of card each bearing aperfect square number greater than500 = 8

∴ P(the second player wins a prize, if

the first has won) =8

999.

18. Let OAB be the given sector.

Then perimeter of OAB = 22 cm

⇒ OA + OB + l = 22 cm

⇒ 6 + 6 + l = 22 cm

⇒ l = 10 cm

∴ Area of sector =12

lr

=12

× 10 × 6 = 30 cm2.

OR

ND = CD

2=

122

= 6 cm

MB = AB2

= 162

= 8 cm

In ∆ODN, ON = 2 2OD – ND

∴ ON = 2 210 – 6 = 64 = 8 cm

In ∆ OBM, OM = 2 2OB – MB

∴ OM = 2 210 – 8 = 36 = 6 cm

∴ Required distance,

i.e., MN =ON – OM = 8 – 6 = 2 cm.

SECTION-C

19. tan θ = 158

= ABBC

⇒ AB = 15k and BC = 8k

⇒ AC = 2 2225 64+k k

= 17k.

∴ 17k = 90

⇒ k = 9017

∴ AB = 15k = 15 × 90

17 = 79.41 m.

⇒ Height of kite above the ground is79.41 m.


20. Given equation is 4x2 + 4bx – (a2 – b2) = 0Comparing with Ax2 + Bx + C = 0, we getA = 4, B = 4b, C = – (a2 – b2)∴ D = B2 – 4AC⇒ D = 16b2 + 16(a2 – b2)

= 16a2 > 0∴ Two distinct real roots are given by

x = – B ± D

2 A =

2– 4 168

±b a

= – 4 4

8±b a

∴ x = –2

a b or –2

a b+ .

21. Let first term of an A.P. = aand common difference = d

∴ am = 1n

⇒ a + (m – 1)d = 1n ...(i)

and an = 1m

⇒ a + (n – 1)d = 1m ...(ii)

Subtract equation (ii) from (i),

(m – n)d = 1n –

1m

⇒ (m – n)d = –m n

mn

⇒ d = 1

mn

Using d in (i), we get

a + (m –1) 1

mn=

1n

⇒ a + 1n

– 1

mn= 1

n

⇒ a = 1

mn

∴ Smn = 2

mn {2a + (mn – 1)d}

= 2

mn ( )– 12 +

mnmn mn

Smn = 1

( 1)2

+mn .

Hence proved.22. We have

SP = ( ) ( )2 22 – 2 – 0+at a at

= a ( )22 2– 1 4t t+

= 4 2 21 – 2 4a t t t+ +

= 4 22 1a t t+ +

= ( )22 1a t + = a ( )2 1+t

and SQ = 2 2

22

– – 0a a

att

+

= ( )22 2 2

4 2

1 – 4a t at t

+

⇒ SQ = 2at ( )22 21 – 4+t t

= 2at ( )221 + t = 2

at

(1 + t2)

∴1 1

SP SQ+ = ( ) ( )

2

2 2

1

1 1+

+ +t

a t a t

= ( )2

2

1

1

+

+t

a t =

1a

;

which is independent of t.Hence proved.

23. As we know that opposite sides of thecircumscribing quadrilateral sub-tendsupplementary angles at the centre of thecircle.


∴ ∠DOC + ∠AOB = 180°⇒ 95° + ∠AOB = 180°⇒ ∠AOB = 180° – 95° = 85°Similarity, ∠AOD + ∠BOC = 180°⇒ ∠AOD + 92° = 180°⇒ ∠AOD = 180° – 92° = 88°.

24. Let A ≡ (3, – 4) and B ≡ (1, 2).

As AP = PQ = QB.So AP : PB = 1 : 2∴ Using section formula,

p = .1 (1) + 2(3)

1 + 2 =

73

again Q is mid-point of PB.∴ Using mid-point formula,

q = – 2 + 2

2 = 0.

Thus, p = 73

; q = 0.

25.

∴ A’BC’ is required triangle.

OR

AP and AQ are required tangents.

26. OA = 7 cm = R (say)

∴ Area of outer circle

= πR2 = π (7)2

= 227

× 49 = 154 cm2

Also OD = OA = 7 cm

∴ r = 72

cm

∴ Area of smaller circle= πr2

= 22 7 77 2 2

× × = 772

= 38.5 cm2.∴ Required area= 154 – 38.5

= 115.5 cm2.

ORThe sectors made by drawing arcs areequal in area.


Sum of areas of the three sectors

= 2603 × × × 5

360° π°

.

⇒ Area of shaded region

= 13 × × × 25

6π

= 252

π cm2

Area of equilateral triangle ABC

= 2310

4× = 225 3 cm

Now, area of unshaded region= Area of ∆ABC – Area of shaded

region

= 25

25 3 –2

π

= 25 × 1.732 – 25

× 3.142

= 43.3 – 39.25

= 4.05 cm2.

27. Since each designoccupy an area equal toa segment of a circle ofradius 28 cm and havingcentral angle 60°.

∴ Area of 6 designs

= 6 × 2 2 2– sin cos cm360 2 2

r rθ θ θ × π °

= 6 × 60 22×

360 7°

°× 28 × 28 – 28 × 28

× sin 30°.cos 30°

cm2

= 6 × 28 × 281 22 1 3

× – ×6 7 2 2

= 2464 – 1999.2 = 464.8 cm2

∴ Cost of making the designs

= 464.8 × 3.50

= ` 1626.80.

28. Hint: Total events = 36Even sum: 2; (1, 1)

4; (2, 2), (1, 3), (3, 1)6; (1, 5), (5, 1), (2, 4), (4, 2), (3, 3)8; (2, 6), (6, 2), (3, 5), (5, 3), (4, 4)10; (4, 6), (6, 4), (5, 5)12; (6, 6)

∴Number of favourable events = 18

∴ Required probability = 18 136 2

= .

OR

Total number of balls = 10 + 5 + 7 = 22(i) Probability of drawing a red ball

= Number of red balls

Total number of balls=

1022

= 5

11

(ii) Probability of drawing a green ball

= Number of green ballsTotal number of balls

= 722

(iii) Probability of drawing a blue ball

= Number of blue ballsTotal number of balls

= 522

∴ Probability of drawing not a blue ball

= 1 – Probability ofdrawing a blue ball

= 1 – 522

=1722

.

SECTION-D

29. Height of the cylinder h = 2.4 cm

and radius r = 1.42

= 0.7 cm

Also height of the conical cavity= 2.4 cm

and radius = 0.7 cm∴ Slant height of the cavity l

= 2 2+l h

= 2 2(0.7) +(2.4)

= 6.25 = 2.5 cm.


Now, T.S.A. of remai-ning solid = C.S.A. ofcylinder + C.S.A. ofcone + Area of baseof cylinder.⇒ 2πrh + πrl + πr2

⇒ πr (2h + l + r)

=227

× 0.7

(4.8 + 2.5 + 0.7)= 17.6 cm2 ≅ 18 cm2.

30. Let V = Volume of poolLet x = Number of hours required by

second pipe to fill the pool∴ x + 5 = Number of hours taken by first

pipe to fill the poolx – 4 = Number of hours taken by third

pipe to fill the pool.

The parts of poolfilling in one hourby first, second andthird pipe would be

V+ 5x

, Vx

and V– 4x

respectively.∴ According to question,

V V+

+ 5x x= V

– 4x

⇒ 1 1+

+ 5x x=

1– 4x

⇒ (2x + 5)(x – 4) = x2 + 5x⇒ x2 – 8x – 20 = 0⇒ x2 – 10x + 2x – 20 = 0⇒ (x – 10)(x + 2) = 0⇒ x = 10 or x = – 2∴ x = 10.(Value with negative sign is rejected)Hence times required by first, second andthird pipe to fill the pool individually are15 hours, 10 hours and 6 hoursrespectively.

ORLet Nisha’s present age be x years, so atpresent Asha’s age be (x2 + 2) years whenNisha grows to her mother’s present age,i.e., (x2 + 2) years then Asha’s age wouldbe (10x – 1) years.That means Nisha grows (x2 + 2x – x) yearand hence Asha also grows the sameyears. That means Asha’s age would be{(x2 + 2) + (x2 + 2 – x)} years.Now equating Asha’s both of abovementioned ages, we get:

(x2 + 2) + (x2 + 2 – x) = 10x – 1⇒ 2x2 – x + 4 = 10x – 1⇒ 2x2 – 11x + 5 = 0⇒ 2x2 – 10x – x + 5 = 0⇒ 2x(x – 5) – 1 (x – 5) = 0

⇒ (x – 5) (2x – 1) = 0 ∴ x = 5, 12

But x = 12

does not satisfy all the given

conditions.

Therefore, Nisha’s present age = 5 yearsand Asha’s present age = x2 + 2 = 27 years.

31. Join OA, OC, OB;and also join OE,OF; E and F beingpoint of contacts.Now, radius ofcircle= OD = OE= OF = 4 cmand BC = CD + DB = 14 cmAlso CD = CF = 6 cmand BD = BE = 8 cm.

Let AF = AE = x⇒ AC = 6x; AB = 8 + x

∴ a =14, b = 6 + x, c = 8 + x Now, using Heron’s formula

ar(∆ABC) = ( – )( – )( – )s s a s b s c

where, s =+ +

2a b c

=14 + 6 + + 8 +

2x x

= 14 + x


∴ar(∆ABC) = (14 + ) × × 8 × 6x x

= 48 (14 + )x x ... (i)

Also,ar(∆ABC) = ar(∆OBC) + ar(∆OAB)

+ ar(∆OAC)

=1 1

× BC × OD + AB × OE2 2

1+ × AC × OF

2

=1

× 4 × (BC + AB + AC)2

= 2(14 + 6 + x + 8 + x)

= 56 + 4x ... (ii)

∴ From (i) and (ii),

48 (14 + )x x = 56 + 4x

4 3 (14 + )x x = 56 + 4x

3 (14 + )x x = 14 + x

Squaring both sides, we get⇒ 3x (14 + x) = (14 + x)2

⇒ 3x = 14 + x⇒ 2x = 14⇒ x = 7∴ AB = 8 + x = 15 cmAnd AC = 6 + x = 13 cm.

OR

We have, radius (OP) = 5 cm = OE andOT = 13 cm.OP ⊥ PT so in right-angled ∆OPT,

OP2 + PT2 = OT2 (Pythagoras Theorem)

⇒ 52 + PT2 = 132

⇒ PT = 169 – 25 = 12 cm.

Let AE = x so PA = x.

⇒ TA = PT – PA = 12 – x

As OE ⊥ AB, in right-angled ∆AET,

AE2 + ET2 = AT2

⇒ x2 + (8)2 = (12 – x)2

⇒ x2 + 64 = 144 – 24x + x2

⇒ 24x = 80 ∴ x = 8024

= 103

cm

Similarly, BE = 103

cm

Therefore, AB = AE + BE = 203

cm.

32. r1 = Radius of top = 28 cmr2 = Radius of bottom = 7 cm

V = Capacity

= 21560 cm3

h = Height of bucket

∴ V= 2 21 2 1 2

1( + + )

3h r r r rπ

⇒ 21560 =1 22

×3 7

h (784 + 49 + 196)

⇒ 20580 = h × 1029

⇒ h = 20 cm. Now, whole surface area (S)

= πr22 + π(r1 + r2)l

Here, l = 2 2

1 2+ ( – )h r r

= 400 + 441 = 841 = 29 cm

∴ S = 227

× (7)2 + 227

(28 + 7) × 29

= 227

× (49 + 35 × 29)

= 3344 cm2.

33. Cash prizes are:

` 320, ` 280, ` 240, ` 200, ` 160, ` 120, ` 80.


Hint: S7 = 1400; d = – 40; n = 7

∴ 1400 = 72

[2a – 6 × 40]

(∵ Sn = 2n

{2a + (n – 1)d})

⇒ 200 = a – 120

⇒ a = 320.

34. In Figure, PCdenotes the multi-storeyed buildingand AB denotesthe 8 m tallbuilding. We areinterested to deter-mine the height of the multi-storeyedbuilding, i.e., PC and the distance betweenthe two buildings, i.e., AC.Look at the figure carefully. Observe thatPB is a transversal to the parallel lines PQand BD. Therefore, ∠QPB and ∠PBD arealternate angles, and so are equal. So∠PBD = 30°. Similarly, ∠PAC = 45°.In right-angled ∆PBD, we have

PDBD

= tan 30° = 13

or BD = PD 3

In right-angled ∆PAC, we have

PCAC

= tan 45° = 1

i.e., PC = ACAlso, PC = PD + DC.Therefore, PD + DC = AC.Since, AC = BD and DC = AB = 8 m,we get

PD + 8 = BD = PD 3This gives

PD = ( )

( )( )8 3 + 18

=3 – 1 3 + 1 3 – 1

= ( ) m4 3 + 1 .

So, the height of the multi-storeyed

building = ( ){ }4 3 + 1 + 8 m = 4 ( )3 + 3 m

and the distance between the two buildings

is also 4 ( )3 + 3 m.

Practice Paper –2

SECTION-A

1. (A) D > 0

⇒ 62 – 4 × 1 × k > 0

⇒ k < 9.

2. (D) 5th term from the end= 201 + (5 – 1) × (– 2)= 201 – 8= 193.

3. (B) 7

Hint: To draw a triangle similar to a given

triangle with a scalar factor pq , p > 0, q > 0,

we should locate the number of pointswhich is greater of p and q.

4. (C) Sum of opposite sides of aquadrilateral having a circle inscribed itare equal.

∴ PQ + RS = PS + QR.

5. (B) Join AC and PC.PC bisects ∠APB⇒ ∠APC = 45°⇒ ∠ACP = 45°In ∆APC,

∠APC = ∠ACP⇒ AC = AP = 4 cm⇒ Length of each tangent = 4 cm.


6. (A)43

π R3 = 49 × 33 × 24

⇒ 34 22R

3 7× × = 49 × 33 × 24

⇒ R3 = 49 × 33 × 24 × 3 × 7

4 × 22

= 49 × 7 × 3 = 73 × 33 = (21)3

∴ R = 21 units.

7. (B) Average

=3 5 5 7 7 7 9 9 9 9

10+ + + + + + + + +

= 7

P(selected number is 7) = 3

10.

8. (B) In quadrilateral POQT,

∠P + ∠O + ∠Q + ∠T = 360°

⇒ 90° + 110° + 90° + ∠T= 360°

⇒ ∠PTQ = 70°.

9. (D) r =4.22

= 2.1 cm;

h = 4.2 cm

Volume= 213

r hπ

h = ×1 22× 2.1× 2.1 × 4.2

3 7

= 19.404 cm3 ≈ 19.4 cm3.

10. (A) In right ∆ABC,

sin 30° = 6

AC

⇒ 12

= 6AC

⇒ AC = 12 m.

SECTION-B

11. r1 = 82

= 4 m , r2 = 42

= 2 m; h = 6 m

Volume of frustum

= π + +2 21 2 1 2

1( )

3h r r r r

= 1 22

× × 6(16 + 4 + 8)3 7

= 447

× 28 = 176 m3.

Thus, the capacity of the reservoir is 176 m3

or 176 kl.

12. Let E be the event of getting the sum as aperfect square.∴ E = {(1, 3), (2, 2), (3, 1), (3, 6), (4, 5),

(5, 4), (6, 3)}∴ n(E) = 7,

Now, P(E) = (E)(S)

nn

= 736

. [∵ n(S) = 36]

13. Since –4 is a root of x2 + px – 4 = 0⇒ (– 4)2 + p(– 4) – 4 = 0⇒ 16 – 4p – 4 = 0⇒ 4p = 12 ⇒ p = 3.Also as the equation x2 + px + k = 0 hasequal roots.∴ D = 0 ⇒ b2 – 4ac = 0⇒ p2 – 4k = 0⇒ 9 – 4k = 0

(∵ p = 3)

⇒ k = 94

.

ORFalse, because consider the equation2x2 + 3x – 1 = 0; its coefficients are all rational

but its roots are x = – 3 9 + 8

2 2±

× =

– 3 174

±

both of them are irrational.

14. The numbers are: 3, 9, 15, 21,.......,99.∴ It is an A.P.⇒ Let a = 3, d = 6

an = 99⇒ a + (n – 1)d = 99⇒ 3 + (n – 1) × 6 = 99 ⇒ n = 17


∴ Required sum = Sn =2n

(a + an)

S17 =172

(3 + 99) = 867.

15. True.Let PQ be the tangent at A to the circum-circle of isosceles ∆ABC.Given: AB = AC.∴ ∠ABC = ∠ACB

Construction: Draw AD; perpendicularbisector of BC as AB = AC.∴ Perpendicular bisector of BC will passthrough A as well as centre of circle O.

∠1 + ∠2 = 90° and ∠3 + ∠4 = 90°⇒ ∠2 = ∠3⇒ ∠CAQ = ∠ACB.So ∠ACB = ∠QACTherefore, PQ || BC

(∵ Alternate interior angle pair is equal).

16. Here, r1 = 6 cm,r2 = 12 cmand h = 8 cm

Slant height (l)

= +2 22 1( – )r r h

= +2 2(12 – 6) 8

= 36 64+ = 10 cm.

17. Let the required ratio be λ : 1.

We will use section formula:

x = 1 2 2 1

1 2

m x m xm m

++

Here, x = 34

, x1 = 12

, x2 = 2, m1 = λ, m2 = 1

∴ 34

=

1× 2 + 1 ×

2+1

λ

λ

⇒ 8λ + 2 = 3λ + 3

⇒ 5λ = 1 ⇒ λ = 15

∴ λ : 1 = 15

: 1 = 1 : 5.

Thus, required ratio is 1 : 5.

18. Area of the triangle

= 12

{x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)}

= 12

{9(7 + 4) + 4(– 4 – 2) + 7(2 – 7)}

= 12

(99 – 24 – 35)

= 12

× 40 = 20 sq. units.

SECTION-C

19. Let the point on the y-axis be P(0, y) andon the x-axis be Q(x, 0).For point P:∵ PA = PB⇒ PA2 = PB2

⇒ (0 – 6)2 + (y – 5)2 = (0 + 4)2 + (y – 3)2

⇒ 36 + y2 – 10y + 25 = 16 + y2 – 6y + 9⇒ 4y = 36 ⇒ y = 9.For point Q:

QA = QB⇒ QA2 = QB2

⇒ (x – 6)2 + (0 – 5)2 = (x + 4)2 + (0 – 3)2

⇒ x2 – 12x + 36 + 25 = x2 + 8x + 16 + 9⇒ 20x = 36

⇒ x = 95

.

∴ Required points are (0, 9) and 9, 0

5

.


OR

Let in ∆ABC;D(1, 2), E(0, – 1)and F(2, – 1) bethe mid-pointsof sides BC, ACand AB respec-tively.Let A(x1, y1), B(x2 , y2), C(x3, y3) be verticesof ∆ABC.Now, using mid-point formula

1 2+2

x x= 2 ⇒ x1 + x2 = 4 ...(i)

+2 3

2x x

= 1 ⇒ x2 + x3 = 2 ...(ii)

+3 1

2x x

= 0 ⇒ x3 + x1 = 0 ...(iii)

Similarly,y1 + y2 = – 2 ...(iv)y2 + y3 = 4 ...(v)y1 + y3 = – 2 ...(vi)

Adding equations (i), (ii) and (iii), we getx1 + x2 + x3 = 3 ...(vii)And from equations (iv), (v) and (vi), we gety1 + y2 + y3 = 0 ...(viii)By solving equation (vii) with equations (i),(ii) and (iii) respectively pairwise, we have

x3 = –1, x1 = 1, x2 = 3Similarly, we can find that

y3 = 2, y1 = – 4, y2 = 2Thus, A = (1, – 4), B = (3, 2), C = (– 1, 2)

∴ Coordinates of the vertices are(1, – 4); (3, 2) and (– 1, 2).

20. Let ABCD bethe square inwhich A(–1, 2)and C(3, 2) aregiven vertices.Let coordinatesof B are (x, y).As AB = BC ⇒ AB2 = BC2

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

⇒ (x + 1)2 = (x – 3)2

⇒ x2 + 2x + 1 = x2 + 9 – 6x

⇒ 8x = 8 ⇒ x = 1Also ∠ B = 90°.So, AB2 + BC2 = AC2

⇒ (x + 1)2 + (y – 2)2 + (x – 3)2 + (y – 2)2

= (4)2 + 02

Put x = 1,⇒ 4 + (y2 + 4 – 4y) + 4 + (y2 + 4 – 4y) = 16⇒ 2(y2

+ 4 – 4y) = 8⇒ y2 + 4 – 4y = 4⇒ y2 – 4y = 0 ⇒ y(y – 4) = 0⇒ y = 0 or y = 4∴ Coordinates of B and D may be taken as (1, 0) and (1, 4).

21. a = 5, l = 45, Sn = 400

Now, Sn = 2n

{a + l} ⇒ 400 = 2n {5 + 45}

⇒ 50n = 800 ⇒ n = 16Now an = 45⇒ a + (n – 1).d = 45⇒ 5 + (16 – 1).d = 45 ⇒ 15d = 40

⇒ d = =40 815 3

Hence, number of terms is 16 and common

difference is 83

.

ORConsider:

– 3 – 2 5 – 3+

+ + +a b a b a ba b a b a b

+ + ............ 11 terms

=3 2

– + –+ + + +

5 3 .............–+ +

a ab ba b a b a b a b

a ba b a b

+ +

=1 3 5 .........11terms

A.P.a

a b+ + +

+ –

ba b+

1 2 3 ......... 11 termsA.P.

+ + + +

= a

a b+{ }11 .2 1 (11 – 1) 2

2 × +

–b

a b+

( )11 111

2+


= ( ) ( )121 – 66a b

a b a b+ +

=121 – 66a b

a b+ = 11(11 – 6 )a b

a b+ .

22. Let AB be the tower of height 22 m andCD be an observer of height 1.5 m.In ∆ACE, CE = DB = 20.5 m,

AE = AB – BE (i.e., CD)= 22 – 1.5 = 20.5 m

and ∠ACE = θ

So, tan θ =AECE

⇒ tan θ =20.520.5

= 1

⇒ tan θ = tan 45° ∴ θ = 45°.

23. Join AB.As OA = r; OP = diameter = 2rAs AP is tangent,⇒ OA ⊥ AP⇒ ∠OAP = 90°∴ In right-angled ∆OAP,

sin (∠1) = OA 1

= =OP 2 2

rr

⇒ ∠1 = 30°Similarly, ∠2 = 30° ⇒ ∠APB = 60°As AP = PB ⇒ ∠PAB = ∠PBAUsing Angle sum property in ∆ APB,

∠PAB = ∠PBA = 60° = ∠APB⇒ ∆APB is equilateral. Hence proved.

24. The ratio between sides is 3 : 2Let ∆ ABC; AB = AC and AM ⊥ BC;

AM = 4 cm∴ A′BC′ is required triangle such that

∆ A′BC′ ~ ∆ ABC.


25. As AB = 28 cm;BC = 21 cm;∠B = 90°

⇒ AC2 = AB2 + BC2

⇒ AC2 = 282 + 212

= 784 + 441= 1225

⇒ AC = 1225 = 35 cm∴ Area of semicircle at AC as diameter

A1 = π

21 352 2

= 35 × 351 22

× ×2 7 2 × 2

= 55 × 35

4 = 481.25 cm2

Area of ∆ABC

A2 = 1

× BC × AB2

= 1

× 21 × 282

= 294 cm2

Area of quadrant PCB

A3 = 21(21)

4π

= × × ×1 2221 21

4 7= 346.50 cm2.

∴ Required area= Area of ∆ ABC + Area of semicircle – Area of quadrant= A2 + A1 – A3

= 294 + 481.25 – 346.50= 428.75 cm2.

26. The given equation is

3x2 + 5 5 x – 10 = 0The discriminant = b2 – 4ac

= (5 5 )2 – 4 × 3 × (– 10)= 125 + 120 = 245 > 0

So, the given equation has two distinctreal roots.

Now, 3x2 + 5 5 x – 10 = 0

⇒ 3x2 + 6 5 x – 5 x – 10 = 0

⇒ 3x(x + 2 5 ) – 5 (x + 2 5 ) = 0

⇒ (x + 2 5 ) (3x – 5 ) = 0

Therefore, x + 2 5 = 0 or 3x – 5 = 0

∴ x = – 2 5 or 3x = 5 ⇒ x = 5

3

∴ The two real roots are – 2 5 and 5

3.

27. The total surface area of the cube = 6 × (edge)2

= 6 × 5 × 5 cm2 = 150 cm2.Note that the part of the cube where thehemisphere is attached is not included in thesurface area.So, the surface area of the block

= T.S.A. of cube – base area of hemisphere + C.S.A. of hemisphere= 150 – πr2 + 2πr2 = (150 + πr2) cm2

= + × × 2 222 4.2 4.2

150 cm cm7 2 2

= (150 + 13.86) cm2 = 163.86 cm2.

OR

∆AOB ~ ∆AO’B’

⇒AO'AO =

O'B'OB

⇒6

12=

8r

⇒ r = 4 cm

Volume of the whole cone (V )

=13

π × 82 × 12

= 256π cm3

Volume of the upper part (small cone)

V1 = 13

π × r2 × 6

= 13

π × 42 × 6

= 32π cm3


Now,volume of upper partvolume of lower part

= 1

1

VV – V

= 32

256 – 32π

π π= 1

7.


28. Total number of coins = 100 + 50 + 20 + 10= 180

(i) Total number of 50 paise coins = 100∴ Probability of getting a 50 paise coin

=100 5

=180 9

.

(ii) Total number of 5 rupee coins = 10∴ Total number of coins other than 5 rupee

= 180 – 10 = 170∴ Probability of getting a coin other than

of 5 rupee = 170180

= 1718

.

SECTION-D

29. First instalment = ` 1000Second instalment = ` 1100

Third instalment = ` 1200...........................................................................................................................................................................

30th instalment = ? (to be calculated)Here, 1000, 1100, 1200,........ forms an APwith a = 1000

d = 100n = 30

Using an = a + (n – 1)d,a30 = a + 29d

= 1000 + 29 × 100= 1000 + 2900= 3900

∴ 30th instalment will be of ` 3900

Now, using Sn = 2n

(a + an)

Also S30 = 302

(1000 + 3900) = 15 (4900)

= ` 73500∴ Total amount left to be paid is 118000 – 73500 = ` 44500.

30. Let unit’s digit = xand ten’s digit = y

Then, original number = 10y + xNow, xy = 15 ... (i) (Given)According to question,

(10y + x) + 18 = 10x + y⇒ – 9x + 9y = – 18⇒ y – x = – 2⇒ y = x – 2Using it in equation (i), we get

x(x – 2) = 15⇒ x2 – 2x – 15 = 0⇒ x2 – 5x + 3x – 15 = 0

x(x – 5) + 3(x – 5) = 0(x + 3)(x – 5) = 0

⇒ x = – 3 or x = 5As negative digits are not acceptable here,∴ x = 5∴ Unit’s digit = 5And ten’s digit = 3Hence, number is 35.

31. Let C(O, r) be a circlewith centre at O andradius r. The circletouches the sides AB,BC, CD and DA of aquadrilateral ABCDat the points P, Q, Rand S respectively.

To show: ∠AOB + ∠COD = 180° ∠AOD + ∠BOC = 180°

Construction: Join OP, OQ, OR and OS.Proof: Since two tangents drawn from anexternal point to a circle subtend equalangles at the centre∴ ∠1 = ∠2; ∠3 = ∠4;

∠5 = ∠6; ∠7 = ∠8Now∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6

+ ∠7 + ∠8 = 360°⇒ 2 (∠2 + ∠3 + ∠6 + ∠7) = 360°and 2 (∠1 + ∠8 + ∠4 + ∠5) = 360°⇒ (∠2 + ∠3) + (∠6 + ∠7) = 180°


and (∠1 + ∠8) + (∠4 + ∠5) = 180°⇒ ∠AOB + ∠COD = 180°and ∠AOD + ∠BOC = 180°

Hence proved.OR

We are given a circlewith centre O, anexternal point T andtwo tangents TP andTQ to the circle,where P, Q are the points of contact (seefigure) . We need to prove that

∠PTQ = 2∠OPQLet ∠PTQ = θAs we know that tangent segmentsdrawn from an external point to the circleare equal. So, TP = TQ

⇒ TPQ is an isosceles triangle.

Therefore, ∠TPQ = ∠TQP = 12

(180° – θ)

= 90° – 12

θ

Also, OP ⊥ PT,i.e., ∠OPT = 90°So, ∠OPQ = ∠OPT – ∠TPQ

= 90° – 1

90 –2

° θ

= 12

θ = 12

∠PTQ

This gives ∠PTQ = 2 ∠OPQHence proved.

32. Speed of water = 3 km/h

Let r1 = radius of pipe (cylindrical)

=202

= 10 cm = 0.1 m.

And r2 = radius of tank = 102

= 5 m

Also, H = height of tank = 2 m.

∴ Volume of cylindrical tank V1= πr22 H

⇒ V1 = × ×2 322 1100(5) 2 = m

7 7

Now let time required to fill the tank = t hours∴ Volume of water delivered by pipe int hours

V2 = t × π (0.1)2(3000)

= × ×2230

7t

⇒ V2 = 6607

t

Consider: V1 = V2

⇒ 11007

= 6607

t

⇒ t = 1100 5

h660 3

=

= ×560

3min= 100 min

or 1 h 40 min.

33. 2πr1 = 18 ⇒ πr1 = 92πr2 = 6 ⇒ πr2 = 3

∴ π(r1 + r2) = πr1 + πr2

= 9 + 3 = 12Curved surface area

= π(r1 + r2) l = 12 × 4 = 48 cm2.

34. Let AB = DE = h = Height of each poleC = Position of point on street

BE = 30 m∠ACB = 45°, ∠DCE = 30°

∴ In ∆ABC, tan 45° =ABBC


⇒ 1 =BCh

⇒ BC = h ... (i)

In ∆CED, tan 30° =DEEC

13

=ECh

⇒ EC = 3h ... (ii)∴ On adding (i) and (ii),

⇒ BC + CE = + 3h h

⇒ 30 = ( )+1 3h

(... BC + CE = BE = 30 m)

⇒ h = −

×+ −

1 3301 3 1 3

⇒ h = ( )− ×30 3 1 30 0.73

=2 2

= 15 × 0.73⇒ h = 10.95 m∴ BC = 10.95 mand EC = 30 – 10.95 = 19.05 m.

OR

Let C = Position of cloudAB = 60 m, ∠CAD = 30°

F = Position of reflection of cloud in thelake

∠ DAF = 60°Let CE = h = Height of cloud above the lake∴ CE= EF = hNow in right-angled ∆CDA,

tan 30° =CDAD

⇒13

=− 60

ADh

⇒ AD = − ×( 60) 3h ... (i)

Now in right-angled ∆FDA,

tan 60° =+ 60

ADh

⇒ 3 =+ 60

ADh

⇒ AD =+ 60

3h

... (ii)

∴ Equations (i) and (ii) give

−( 60) 3h =+ 60

3h

⇒ 3h – 180 = h + 60⇒ 3h – h = 60 + 180⇒ 2h = 240⇒ h = 120 m.

Practice Paper–3

SECTION-A

1. (C) 64x2 – 1 = 0 ⇒ (8x)2 = 1

⇒ 8x = ± 1 ⇒ x = ±18

Out of which positive real root is x = 18

.

2. (A) Given A.P. becomes: 5, 8, 11, ....., 113.an = a + (n – 1) d

⇒ 113 = 5 + (n – 1)3

⇒ n – 1 = 113 5

3−

= 36 ⇒ n = 37.


3. (A) In ∆OAD,OA = 5 cm, OD = 3 cm

∠ODA= 90°∴ AD = 2 2OA – OD

= 2 25 – 3= 4 cm

∴ AB = 2AD = 2 × 4 cm = 8 cm.

4. (C)R2 + 82 = 102

⇒ R2 = 100 – 64= 36

∴ R = 6 cm.

5. (D) ∠PBA + ∠PAB + ∠APB = 180° (ASP)⇒ ∠PBA + ∠PBA + 52° = 180°

(∵ PA = PB)

⇒ ∠PBA = 180 52

2° − °

= 64°.

6. (B) ar(APQB) = ar(sector AOB) – ar(sector POQ)

= π × (21)2 ×60360

°°

– π × (14)2 ×60360

°°

= π 60×

360[212 – 142]

=22 1

× × 35 × 77 6

=385

3 cm2.

7. (D) A non-leap year having 365 dayshas 52 weeks and one day out of 7 days.That one day could be Sunday, Monday,Tuesday, Wednesday, Thursday, Friday orSaturday.∴ Total probabilities = 7. Favourable outcomes = 1

∴ Required probability

(i.e., having 53 sundays) = 17

.

8. (C) Let radius = 4x; slant height = 7xConsider, π × 4x × 7x = 792⇒ x2 = 9⇒ x = 3∴ Radius = 4x = 12 cm.

9. (C) From figure,AB = AP + BP

= AQ + BS= AQ + (BC – CS)= AQ + (BC – CR)= 5 cm + (7 cm – 3 cm)= 9 cm. ∴ x = 9 cm.

10. (D) Let height of pole be x mIn ∆ABC,

sin 30° = ABAC

⇒ 12

= ABAC

⇒ 12

= 20x ⇒ x = 10 m.

SECTION-B

11.

Using distance formula,

2 2(10 2) ( 3)y− + + = 10

⇒ 64 + (y + 3)2 = 100⇒ (y + 3)2 = 100 – 64 = 36⇒ y + 3 = ± 6 ⇒ y = 3 or – 9.

12. Let number of blue balls = x

∴ Total balls = x + 6

∴ P(drawing a blue ball) = 6

xx +

and P(drawing a red ball) = 6

6x +According to questions,

6x

x + = 6

36x

+

⇒ x = 18.OR

No, the sample space has 8 outcomes. Outof them only one outcome say, TTT,represents no heads, so the probability of

no heads is 18

.


13. For real roots D ≥ 0 ⇒ k2 – 4 ≥ 0

⇒ (k – 2)(k + 2) ≥ 0⇒ k ≤ –2 or k ≥ 2.∴ Required value of k is 2.

14. No.Each side of the outer square

= diameter of the circle= d

So, area of square = d2

Each diagonal of theinner square = d

i.e., 2 × side = d

∴ Side = 2

d ⇒ Area =

2

2d

.

Hence area of inner square

=12

× area of outer square.

15. an = 2n + 1 ⇒ a1 = 3

∴ Sn =2n

(a1 + an) ⇒2n

(3 + 2n + 1)

= n(n + 2).

16. Let third vertex be C(x, y).

We have A(2, 3), B(– 2, 1) and G(1,23

)

Using Centroid formula,

1 = 2 – 2 +

3x

; 23

= 3 +1+

3y

⇒ x = 3, y = – 2⇒ Vertex C is (3, – 2).

17. As XP = XQ ... (i)AP = AR ... (ii)BQ = BR ... (iii)

From equation (i),⇒ XA + AP = XB + BQ

{... XP = XA + AP; XQ = XB + BQ}⇒ XA + AR = XB + BR [Using (ii ) and (iii)]

Hence proved.

18. As A =12

lr

⇒ 20π =12

× 5π × r

⇒ r = 8 cm.

SECTION-C

19. The possible outcomes of the experimentare listed in the given table the first numberin each ordered pair is the numberappearing on the blue die and the secondnumber is that on the grey die.

So, the number of possible outcomes= 6 × 6 = 36.

(i) The outcomes favourable to the event‘the sum of the two numbers is 8’denoted by E, are: (2, 6), (3, 5), (4, 4),(5, 3), (6, 2) (see figure).i.e., the number of outcomesfavourable to E = 5.

Hence, P(E) =536

(ii) As you can see from figure, there isno outcome favourable to the eventF, ‘the sum of two numbers is 13’.

So, P(F) =036

= 0

(iii) As you can see from figure, all theoutcomes are favourable to the eventG, ‘sum of two numbers ≤ 12’.

So, P(G) =3636

= 1.


20. Comparing given equation withax2 + bx + c = 0, we geta = p2, b = p2 – q2, c = – q2

∴ D = b2 – 4ac⇒ D = (p2 – q2)2 – 4(p2)(– q2)

= p4 + q4 + 2p2q2

= (p2 + q2)2 > 0∴ Given equation has two real roots given by:

x =– D

2b

a±

= 2 2 2 2

2

–( – ) ( + )

2

p q p q

p

±

⇒ x =2

2

q

p or –1.

OR23 2 5 2x x− − = 0

We need to divide –5x into two parts suchthat sum of them is –5x and product ofthem is 23 2 ×(– 2)x = – 6x2. Such partsare – 6x and x.∴ 23 2 6 2x x x− + − = 0or 3 2 ( – 2) 1 ( 2)x x x+ − = 0

or ( 2)(3 2 1)x x− + = 0

i.e., x – 2 = 0 or 3 2x + 1 = 0

i.e., x = 2 or x = 1

3 2− = –

26

Hence, the required roots are 2 and 2

6− .

21. As a3 + a7 = 6 and a3.a7 = 8

⇒ (a + 2d) + (a + 6d) = 6 ... (i)and (a + 2d)(a + 6d) = 8 ... (ii)From (i), 2a + 8d = 6⇒ a + 4d = 3 ⇒ a = 3 – 4d∴ Using it in (ii), we get

(3 – 4d + 2d)(3 – 4d + 6d) = 8⇒ (3 – 2d)(3 + 2d) = 8⇒ 9 – 4d2 = 8 ⇒ 4d2 = 1

⇒ d2 = 14

⇒ d = ± 12

Case I. If d = 12

, then a = 3 – 4 12

= 1.

∴ S16 = 162

{2 × 1 + (16 – 1) ×12

}

= 8 { }152

2+ =

8 192

×= 76.

Case II. If d = 1–

2, then a = 3 – 4 1

2 − = 5

∴ S16 = 162

12 ×5 + (16 – 1)× –

2

= 8{ }1510 –

2 =

8 52×

= 20.

22. From figure, ar (∆ABD)

= [ ]1– 5(– 5 – 5) – 4(5 – 7) + 4(7 + 5)

2

= 1

[50 + 8 + 48]2

= 1062

= 53 sq. units.

Also, ar (∆BCD)

= 1

[– 4(– 6 – 5) – 1(5 + 5) + 4(– 5 + 6)]2

= 12

[44 – 10 + 4] =382

= 19 sq. units.

So area of quadrilateral ABCD= ar (∆ABD) + ar (∆BCD)= 53 + 19 = 72 sq. units.

23. Let ABCD is a parallelogram.⇒ AB = DC and AD = BC∴ Consider

AD = AS + DS= AP + DR

{∵ AS = AP, DS = DR}= (AB – PB) + (DC – RC)= (AB – BQ) + (AB – CQ)

{∵ DC = AB and PB = BQ and RC = CQ}= 2 AB – (BQ + CQ)= 2 AB – BC

AD = 2 AB – AD {∵ BC = AD}⇒ 2 AD = 2 AB ⇒ AD = AB∴ ABCD is a rhombus. Hence proved.


24.

∆A’BC’ is required triangle.OR

AP = AQ and BR = BS are required tangents.


25. Let PQ be the tower of height h m.

In right ∆APQ,

tan θ = 4h

... (i)

In right ∆BPQ,

tan (90° – θ) = 9h

cot θ = 9h ... (ii)

Multiplying corresponding sides ofequations (i) and (ii), we get

tan θ × cot θ = 4 9h h×

⇒ 1 = 2

36h

⇒ h2 = 36

∴ h = 6 m.

26. It is given that the point A divides the joinof P(–5, 1) and Q(3, 5) in the ratio k : 1. Sothe coordinate of A are:

3 5 5 1,

1 1k kk k

− + + +

∴ Area of ∆ABC

=12

x1(y2 – y3) + x2(y3 – y1) + x3(x1 – y2)

⇒ ∆ =− + ⋅ + + − − + +

1 3 5 5 1(5 2) 2

2 1 1k kk k + 7

+ − + 5 1

51

kk

=281 21 35 7 3

2 1 1 1k kk k k

−− +− ++ + +

=1 14 662 1

kk

−+

= 7 33

1kk

−+

∴ 7 331

kk

−+

= 2 [... ar(∆ABC = 2]

⇒−+

7 331

kk = ± 2

⇒ 7 332

1kk

−=

+

or 7 33

21

kk

−= −

+⇒ 7k – 33 = 2k + 2 or 7k – 33 = – 2k – 2⇒ 5k = 35 or 9k = 31

⇒ k = 7 or 319

k = .

Therefore, required values of k = 7, 319

.

27. Total area= Area of sector OAB + Area ofsector ODC + Area of ∆OAD+ Area of ∆OBC.

= 90 22 90 22× × 28 × 56 + ×

360 7 360 7× 28 × 56

+ 14

× 56 × 56 + 14

× 56 × 56 m2

= 22 × 56 + 22 × 56 + 14 × 56 + 14 × 56= 56(22 + 22 + 14 + 14) m2

= 56 × 72 m2 = 4032 m2.

ORWe can divide the given figure into twoparts, one part as a rectangle with lengtha = 8 m, breadth b = 4 m and the other oneas a semicircle with radius r = 2 m.

Now, required area= ar(rectangle) + ar(semicircle)

= a × b + 12

πr2

= 8 × 4 + 12

×227

× 22

= 32 + 6.29 = 38.29 m2.

28. Let BPC be the hemisphere and ABCbe the cone standing on the base of thehemisphere (see figure). The radius BO ofthe hemisphere (as well as of the cone)


=12

× 4 cm = 2 cm.

So, volume of the toy = 3 22 13 3

r r hπ + π

= 3 22 1× 3.14 × (2) × 3.14 × (2) × 2

3 3 +

= 23 × 2 1

× 3.14 × 3.143 3

+ = 8 × 3.14

= 25.12 cm3.

Now, let the right circular cylinder EFGHcircumscribe the given solid. The radius ofthe base of the right circular cylinder = HP= BO = 2 cm, and its height isEH = AO + OP = (2 + 2) cm = 4 cm

So, the volume required= Volume of the right circular cylinder

– Volume of the toy= (3.14 × 22 × 4 – 25.12) cm3

= 25.12 cm3

Hence, the required difference of the twovolumes = 25.12 cm3.

SECTION-D

29. Let PQ be the leaning tower.

Let PM = x and QM = y,

In right ∆PMQ,

cot θ = xy

⇒ x = y cot θ .... (i)

Similarly, right triangles AMQ and BMQrespectively provides.

a + x = y cot α ... (ii)and b + x = y cot β ... (iii)Eliminating x from equation (i) and (ii), weget

a = y (cot α – cot θ) ... (iv)Again eliminating x from equations (i) and(iii), we get

b = y (cot β – cot θ) ... (v)Eliminating y from equations (iv) and (v),we get

ab

=cot cot cot cot

α − θβ − θ

⇒ b cot α – b cot θ = a cot β – a cot θ⇒ a cot θ – b cot θ = a cot β – b cot α⇒ cot θ (b – a) = b cot α – a cot β

⇒ b – a =cot cot

cot b aα − β

θHence proved.

ORLet AB = height of building = 30 m

CD = height of boy = 1.5 m∴ AG = AB – GB

= AB – CD {∵ GB = CD}= 30 – 1.5 = 28.5 m

Let the distance travelled by the boy towardsthe building be x m.

In ∆AEG,

tan 60° =AGGE

⇒ 3 = AGGE

⇒ GE =AG 28.5

=3 3

... (i)


Again in ∆AGC,

tan 30° =AGGC

= AG

GE + x

⇒ 13

=28.5

GE + x

⇒ GE + x = 28.5 × 3

From equation (i),

⇒28.5

+3

x = 28.5 × 3

⇒ 28.5 + 3 x = 28.5 × 3

⇒ 3 x = 85.5 – 28.5

⇒ x =57 3

×3 3

=57

33

= 19 3 m.

30. Let side of 1st square = x mand side of 2nd square = y m

(Assuming y > x)∴ Area of 1st square = x2

And area of 2nd square = y2

According to question,x2 + y2 = 468 ... (i)

Also, perimeter of 1st square = 4xAnd perimeter of 2nd square = 4y

Again according to question,4y – 4x = 24

⇒ y – x = 6⇒ y = 6 + x ... (ii)Using (ii) in (i),

x2 + (6 + x)2 = 468⇒ x2 + 36 + x2+ 12x = 468⇒ 2x2 + 12x – 432 = 0

⇒ x2 + 6x – 216 = 0⇒ x2 + 18x – 12x – 216 = 0⇒ x(x + 18) – 12(x + 18) = 0⇒ (x + 18) (x – 12) = 0⇒ x = – 18 or x = 12As side can’t be negative, so take

x = 12 m∴ y = x + 6 = 18 m.

ORDuring 2 pm to 3 pm the minutes handcomplete one round, i.e., 60 minutes.

At t minutes past 2 pm, the minutes hand

needed 2

– 34t

minutes to show 3 pm.

i.e.,2

– 34t

+ t = 60

⇒2 – 12

4t

+ t = 60

⇒ t2 – 12 + 4t = 240

⇒ t2 + 4t – 252 = 0

⇒ t2 + 18t – 14t – 252 = 0

⇒ t(t + 18) – 14(t + 18) = 0

⇒ (t – 14) (t + 18) = 0

⇒ t – 14 = 0 or t + 18 = 0⇒ t = 14 or t = – 18(Rejected because time cannot be negative)

∴ t = 14.

31. 1st PartLet AP and AQ are tangents drawn from anexternal point A to the circle with centre O.Join OP, OQ and OA.

Now in ∆OPA and ∆OQA,OP = OQ = r (Radius of circle)OA = OA (Common)


∠OPA = ∠OQA = 90°⇒ ∆OPA ≅ ∆OQA (RHS)⇒ AP = AQ (CPCT)

Hence proved.2nd PartUsing above theorem in figure,

AD = AF ... (i)BD = BE ... (ii)CE = CF ... (iii)

As given, AB = AC⇒ AB – AD = AC – AD⇒ AB – AD = AC – AF [Use equation (i)]⇒ BD = CF⇒ BE = CE

[Use equations (ii) and (iii)]Hence proved.

32. Volume of frustum

=13

πh (r12 + r2

2 + r1r2)

=13

×227

× 45 × (282 + 72 + 28 × 7)

=22 15

7×

× 7 × (28 × 4 + 7 + 28)

= 330 × 147 = 48510 cm3

Curved surface area

= π (r1 + r2)l, i.e., π(r1 + r2)2 2

1 2( – )h r r+

= 227

(28 + 7) 2 245 (28 – 7)+

= 110 2025 441+ = 110 × 49.65 cm2

= 5461.5 cm2.

Now, the total surface area of the frustum

= Curved surface area + πr12 + πr2

2

= 5461.5 cm2 +227

(28)2 cm2 +227

(7)2 cm2

= 5461.5 cm2 + 2464 cm2 + 154 cm2

= 8079.5 cm2.

33. Let an A.P. is a, a + d, a + 2d, ....Here, n = 37 which is odd.

So three middle most terms are:th th th1 1 1

– 1 , , 12 2 2

n n n+ + + +

i.e., th th th37 1 37 1 37 1

– 1 , , 12 2 2+ + + +

i.e., a18, a19 , a20

As sum of three middle most term = 225⇒ a18 + a19 + a20 = 225

⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225(d is common difference)

⇒ 3a + 54d = 225i.e., a + 18d = 75 ...(i)Again, sum of last three term = 429

⇒ a35 + a36 + a37 = 429(... n = 37)

⇒ (a + 34d) + (a + 35d)(a + 36d) = 429⇒ 3a + 105d = 429i.e., a + 35d = 143 ...(ii)Solving (i) and (ii), we get a = 3, d = 4

Therefore, A.P. is 3, 7, 11, 15, ........

34. 98

m

Hint: Volume of earth dug out = π23

2

× 14

Volume of embankment = π[(5.5)2 – (1.5)2] × h

⇒94

× 14 = 7× 4 × h

⇒ h = 94

× 147 × 4

= 98

m.

Practice Paper – 4

SECTION-A

1. (D) b2 – 4ac = 0

⇒ c =2

4b

a.

2. (C) an = a + (n – 1)d

⇒103

= – 1 + (n – 1) × 16


⇒– 16

n=

103

+ 1 ⇒ n – 1 = 6 ×133

⇒ n = 27.

3. (A) In ∆OPR,∠POR = 55°, ∠ORP = 90°

∴ ∠OPR = 180° – (55° + 90°) = 35°∴ ∠QPR = 2∠OPR = 2 × 35° = 70°.

4. (B)

x = 2 225 – 7 = ( )( )+25 7 25 –7

= 576 = 24 cm.

5. (A) ar(segment APB)

= ar(sector APBO) – ar(∆AOB)

= °°

60360

× πr2 – 3

4r2

= 3

–6 4

π

r2 sq. units.

6. (C).

7. (A)Volume of n balls= Volume of the cone

⇒ n × 43

× π × 33 = 13

π × 122 × 24

⇒ n × 36π = 144 × 8π⇒ n = 32.

8. (C) Number of cards which are neitherred nor queen = 52 – 26 – 4 + 2 = 24P(a drawn card is neither red nor

queen) = =24 652 13

.

9. (C) tan 30° = 15h

⇒ h = 15 3

×3 3

⇒ h = ×15 1.732

3

∴ h = 8.66 m.

10. (B) OQ ⊥ PR and AB || PR.So, QO perpendicular bisector of AB⇒ ∆AQB is an isosceles triangle.⇒ QO is also angle bisector of ∠AQB.∴ ∠AQB = 2∠BQO

= 2 × (90° – 70°) = 40°.

SECTION-B

11. From figure,∠OSQ = ∠OQS (As OS = OQ)

= ∠OQL – ∠SQL= 90° – 50° = 40°.

and ∠OSR = ∠ORS (As OS = OR)= ∠ORM – ∠SRM= 90° – 60° = 30°

Now, ∠QSR = ∠OSQ + ∠OSR= 40° + 30° = 70°.

OR

False.Let a number ofcircles with centresO1, O2, O3 ....... touchthe line PQ at A. JoinA to O1, O2, O3, ...... .

Therefore, O1A ⊥ PQ,O2A ⊥ PQ, O3A ⊥ PQ, ..... at same point A.

So; all the centres O1, O2, O3, ...... lie on aline perpendicular to PQ but the line doesnot bisect PQ.

12.

Let coordinates of B are (x, y)


∴ Using section formula:

+ +

3 8 3 20,

7 7x y

= (–1, 2)

⇒ +3 87

x= – 1 and

+3 207

y= 2

⇒ 3x = – 15 and 3y = – 6⇒ x = – 5 and y = – 2∴ B(– 5, – 2) is required point.

13. For equal roots,D = 0

⇒ b2 – 4ac = 0⇒ 4(1 + 3k)2 – 28(3 + 2k) = 0⇒ 1 + 9k2 + 6k – 21 – 14k = 0⇒ 9k2 – 8k – 20 = 0⇒ 9k2 – 18k + 10k – 20 = 0⇒ 9k(k – 2) + 10(k – 2) = 0⇒ (k – 2) (9k + 10) = 0⇒ k – 2 = 0 or 9k + 10 = 0

∴ k = 2 or –10

9.

14. nth term, i.e., an

= Sn – Sn – 1

⇒ = n2 + 8n – {(n – 1)2 + 8(n – 1)}= n2 + 8n – (n2 – 2n + 1 + 8n – 8)= 2n + 7

∴ a15 = 2 × 15 + 7 = 30 + 7 = 37.

15. True, because the coordinates of the mid-points of both the diagonals coincide, thatis, the diagonals bisect each other at

1 5,

2 2 .

16. θ =15. 720

radian =lr

θ ∵

=15.7 180 15.7 180

× × = 4520 20 3.14

° °= °π

1801 radian =

° π ∵

Required area OAPB= Area of the circle – Area of minor

sector OAQB

= π × 202 – π × 202 × 45°

360°

= 400π – 50π = 350 × 3.14

= 1099 cm2.

17. As 2πR = 48 and 2πr = 36

⇒ R = 24π

and r = 18π

∴ V = 13

πh(R2 + r2 + Rr)

= 13

π × 11 × 2 2

2 2 2

24 1824 18 ×+ +

π π π

= 113

π × 21332π

= 11 × 444 × 722

= 1554 cm3.

18. Total number of cards

n(S) = 50 – 2 = 48

Let E be the event that the card is notprimes. Prime numbers are: 3, 5, 7, 11, 13,17, 19, 23, 29, 31, 37, 41, 43, 47.

∴ n(E) = 48 – 14 = 34

∴ P(E) = ( )( )E 34 17S 48 24

nn

= = .

SECTION-C

19. (i) Total outcomes = 36E = 6 will not come up either time

⇒ E = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5)(2, 1), (2, 2), (2, 3), (2, 4), (2, 5)


(3, 1), (3, 2), (3, 3), (3, 4), (3, 5)(4, 1), (4, 2), (4, 3), (4, 4), (4, 5)(5, 1), (5, 2), (5, 3), (5, 4), (5, 5)

∴ n(E) = 25


.

(ii) P(6 will come up at least one)

P(E) = 1 – P(E)

= 1 – 2536

= 1136

.

20. Multiplying the equation throughout by5, we get

25x2 – 30x – 10 = 0This is the same as(5x)2 – 2 × (5x) × 3 + 32 – 32 – 10 = 0i.e. (5x – 3)2 – 9 – 10 = 0i.e. (5x – 3)2 – 19 = 0i.e. (5x – 3)2 = 19

i.e. 5x – 3 = ± 19

i.e. 5x = 3 ± 19

i.e. x = ±3 19

5

Therefore, the roots are +3 195

and

3 – 195

.

21. Let first term = a

Common difference = d

Now, ak = a + (k – 1).d ...(i)

Let S1 = Sum of odd terms

⇒ S1 = a1 + a3 + a5 + ........+ a2n + 1

⇒ S1 = + 12

n {a1 + a2n + 1}

= + 12

n {a + a + (2n + 1 – 1)d}

[∵ Using (i)]

= { }12 2

2n

a nd+

+

S1 = (n + 1) (a + nd)

also S2 = Sum of even terms= a2 + a4 + ........ + a2n

= 2n

{a2 + a2n}

= 2n

{a + d + a + (2n – 1).d}

[∵ Using (i)]

= 2n

{2a + 2nd}

= n(a + nd)∴ S1 : S2 = (n + 1) (a + nd) : n(a + nd)

= n + 1 : nHence proved.

ORLet the angles be a – d, a and a + d.Now,

a – d + a + a + d = 180°(Angles sum property of a triangle)

∴ 3a = 180°or a = 60° ...(i)

Also, a – d = 12

(a + d)

(Given condition)2a – 2d = a + d

or a = 3dUsing (i), we get

3d = 60°⇒ d = 20°∴ a – d = 60° – 20° = 40°and a + d = 60° + 20° = 80°Hence, the required angles are 40°, 60°and 80°.

22. 24 sq. units.

Hint: PQ = 26 ; QR = 26 ;

RS = 26 ; SP = 26⇒ PQRS is a rhombus.

As PR = 4 2 ; QS = 6 2 .

∴ PR ≠ QS⇒ diagonals are unequal.⇒ PQRS is not a square.


Area of PQRS = 12

× (Product of diagonals)

= 24 sq. units.

23. We are given a circlewith centre O, anexternal point T andtwo tangents TP andTQ to the circle,where P, Q are the points of contact (seefigure). We need to prove that

∠PTQ = 2∠OPQLet ∠PTQ = θAs we know that tangent segmentsdrawn from an external point to the circleare equal. So, TP = TQ

⇒ TPQ is an isosceles triangle.

Therefore, ∠TPQ = ∠TQP = 12

(180° – θ)

= 90° – 12

θ

Also, OP ⊥ PT,i.e., ∠OPT = 90°So, ∠OPQ = ∠OPT – ∠TPQ

= 90° – 190 –

2 ° θ

= 12

θ = 12

∠PTQ

This gives ∠PTQ = 2 ∠OPQ

Hence proved.OR

We are given a chord AB of a circle and twotangents CP and CQ drawn at the points Aand B respectively, which intersect at C.

We need to prove that∠CAB = ∠CBA and ∠PAB = ∠QBA

We know that pair of tangents drawn froman external point to a circle are equal inlength.∴ CA = CBIn ∆CAB, ∠CAB = ∠CBA ...(i)

(∵ CA = CB)Now, ∠CAB + ∠PAB = 180° (Linear pair)⇒ ∠PAB = 180° – ∠CAB ...(ii)Similarly, ∠QBA = 180° – ∠CBA⇒ ∠QBA = 180° – ∠CAB ...(iii)

[Using (i)]From (i), (ii) and (iii), we have∠CAB = ∠CBA and ∠PAB = ∠QBA

Hence proved.

24. Let A be the point on the bridge. From A,the angles of depression of the banks Cand D are 30° and 45° respectively.

Also let AB be the height of bridge whichis 30 m.Let us find the width CD of the river.In ∆ABC,

tan 30° = ABBC

⇒13

=30BC

∴ BC = 30 3 m ... (i)

Again in ∆ABD,

tan 45° = ABBD

⇒ 1 =30BD

∴ BD = 30 m ... (ii)Adding (i) and (ii), we get

BC + BD = 30 3 + 30

∴ CD = 30( 3 + 1)

= 30(1.732 + 1)= 30 × 2.732= 81.96 m.


25.

As PT = PT′ and QS = QS′ are required tangents.

26. Let us mark the fourunshaded regions as I,II, III and IV (see figure)Area of I + Area of III

= Area of ABCD – Area of twosemicircles of each of radius 5 cm

= 2110 × 10 – 2 × × × 5

2 π

cm2

= (100 – 3.14 × 25) cm2 = 21.5 cm2

Similarly, Area of II + Area of IV = 21.5 cm2

So, area of the shaded design

= Area of ABCD – Area of (I + II + III + IV)

= (100 – 2 × 21.5) cm2

= (100 – 43) cm2 = 57 cm2.

27. Let coordinate of P be (x, y)

Given: Points A and B are (3, 4) and (5, – 2).

Also, PA = PB

⇒ PA2 = PB2

⇒ (x – 3)2 + (y – 4)2 = (x – 5)2 + (y + 2)2

⇒ x2 – 6x+9 + y2 – 8y+16 = x2 – 10x + 25+ y2 + 4y + 4

⇒ x – 3y – 1 = 0 ...(i)Now, area of ∆PAB = 10

⇒ 12

x1(y2 – y3) + x2 (y3 – y1) + x3(y1 – y2)

= 10

⇒ ( ) ( ) ( )+ + +14 2 3 –2 – 5 – 4

2x y y = 10

⇒ 6x – 6 – 3y + 5y – 20 = ± 20⇒ 6x + 2y – 26 = ± 20⇒ 6x + 2y – 46 = 0 or 6x + 2y – 6 = 0⇒ 3x + y – 23 = 0 ...(ii)or 3x + y – 3 = 0 ...(iii)Solving (i) and (ii), we get x = 7, y = 2Solving (i) and (iii), we get x = 1, y = 0∴ Required point is (7, 2) or (1, 0).

28. Radius of the hemispherical tank

= 32

m

Volume of the tank = 32 22 3

× ×3 7 2

m3

= 9914

m3


Also, tn = t1 + (n – 1)d⇒ c = a + (n – 1) (b – a)⇒ c – a = (n – 1) (b – a)

⇒ ––

c ab a

= n – 1

⇒ n = ––

c ab a

+ 1 = ––

c a + b – ab a

∴ n = +c 2–

b – ab a

.

As we know Sn = 2n

(a + l) where a is first

term and l is last term.

i.e., Sn =

( – 2 )–2

b c cb a+

. (a + c)

∴ Sn =( )( – 2 )

2( – )a c b c a

b a+ +

Hence proved.30. Let the speed of the stream be x km/h.

Therefore, the speed of the boatupstream = (18 – x) km/h and the speedof the boat downstream = (18 + x) km/h.The time taken to go upstream

= =distance 24speed 18 – x

hours.

Similarly, the time taken to return downstream

= 24

18+ x hours.

According to the question,

24 24–

18 – 18+x x= 1

i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)i.e., x2 + 48x – 324 = 0Using the quadratic formula, we get

x = ± + ±

=2– 48 48 1296 – 48 3600

2 2

= ±– 48 60

2 = 6 or –54

So, the volume of the water to be emptied

= ×1 992 14

m3

= 9928 × 1000 litres =

9900028

litres

Since, 257

litres of water is emptied in 1

second.

So,99000

28 litres of water will be emptied

in 9900028

×725

seconds, i.e., in 16.5 minutes.

ORVolume of the wall= 24 m × 0.4 m × 6 m

= 2400 cm × 40 cm × 600 cm= 2400 × 40 × 600 cm3

Volume of the mortar

= 1

10× Volume of the wall

= 240 × 40 × 600 cm3

Let the number of bricks used in theconstruction be N.Now, Volume of the wall = Volume of themortar + Volume of N bricks⇒ 2400 × 40 × 600 = 240 × 40 × 600 + N × 25 × 16 × 10

⇒ N = × × ×

× ×240 40 600 (10 – 1)

25 16 10= 1440 × 9

⇒ N = 12960Hence, 12960 bricks are used.

SECTION-D

29. Given: In an A.P., t1 = a, t2 = b and tn = c.

We have to show that Sn =( + )( + – 2 )

2( – )a c b c a

b aCommon difference d = t2 – t1 = b – a.


Since x is the speed of the stream, itcannot be negative. So, we ignore the rootx = – 54. Therefore, x = 6 which gives thespeed of the stream as 6 km/h.

ORHint:

x2 + (x + 2)2 = 290⇒ 2x2 + 4x – 286 = 0⇒ x2 + 2x – 143 = 0

(x + 13) (x – 11) = 0⇒ x = 11∴ Numbers are 11, 13.

31. Proof: We are given a circle with centreO, a point P lying outside the circle andtwo tangents PQ and PR on the circlefrom P(see figure). We are required toprove that PQ = PR.

For this, we join OP, OQ and OR. Then∠OQP and ∠ORP are right angles,because these angles are between the radiiand tangents. Now in right triangles OQPand ORP,

OQ = OR(Radii of the same circle)

OP = OP (Common)Therefore, ∆OQP ≅ ∆ORP (RHS)This gives PQ = PR (CPCT)

Consider: AB + CD= (AS + BS) + (DQ + CQ)

= (AP + BR) + (DP + CR)(Using above theorem)

= (AP + DP) + (BR + CR)∴ AB + CD = AD + BC. Hence proved.

32. Let r1 = radius of top of frustum= radius of base of cone = 3 m

r2 = radius of base of frustum = 10 mFG = h = height of frustum = 24 mAF = H = height of cone = 28 – 24 = 4 m

∴ L = slant height of cone

= + = + =2 2 2 21 H 3 4 5 mr

l = slant height of frustum

= − +2 22 1( )r r h

= + = +2 27 24 49 576 = 25 m

Quantity of canvas required= C.S.A. of cone + C.S.A. of frustum= πr1 L + π(r1 + r2) l= π(3 × 5 + 13 × 25)

= 22 22

× (15 325) = × 3407 7

+

= 1068.57 m2 (approx.)

33. Let AB = 40 m = Height of towerCD = h = Height of lighthouse

∠CAE = 30°; ∠CBD = 60°

In ∆CBD, tan 60° =CDDB

⇒ 3 =CE + DE

DB


⇒ 3 =CE + 40

DB

(∵AB = DE = 40 m)

3 = +CE 40DB DB

...(i)

Also in ∆AEC,

tan 30° =CEAE

⇒ 13

=CEDB

...(ii) (∵ AE = DB)

Using (ii) in (i), we have

3 = +1 40DB3

⇒40DB

= − =1 23

3 3

⇒ DB = 20 3 m

From equation (ii),

13

=CE

20 3

⇒ CE = 20 m∴ h = CE + DE

= 20 + 40⇒ h = 60 m.Also distance of foot of tower from top oflighthouse = CB.

∴ sin 60° =CDCB

⇒3

2=

60CB

⇒ CB = ×120 33 3

⇒ CB = 40 3 m.

ORIn figure, O is the centre of the balloon.OP = R, ∠PAQ = θ, ∠OAB = φ.

Let the height of the centre of the balloonbe h. Thus OB = hIn right-angled triangle AOP,

sin ∠OAP = OPAO

⇒ sin2θ

= R

AO

⇒ AO = R cosec 2θ ...(i)

Also, in right-angled triangle AOB,

sin φ = OBAO

⇒ sin φ = AO

h

⇒ AO = h cosec φ ...(ii)From equation (i) and (ii), we get

h cosec φ = R cosec 2θ

∴ h = R sin φ cosec 2θ

Hence proved.34. Since the inner diameter of the glass = 5 cm

and height = 10 cmSo, the apparent capacity of the glass

= πr2h= 3.14 × 2.5 × 2.5 × 10 cm3

= 196.25 cm3.


But the actual capacity of theglass is less by the volume ofthe hemisphere at the base ofthe glass.

i.e., it is less by 23

πr3

= 23

× 3.14 × 2.5 × 2.5 × 2.5 cm3

= 32.71 cm3

So, the actual capacity of the glass

= apparent capacity of glass

– volume of the hemisphere

= (196.25 – 32.71) cm3

= 163.54 cm3.

Practice Paper–5

SECTION-A

1. (A) +

21 1 5–

2 2 4k = 0

⇒ 12

k = 5 1

–4 4

⇒ k = 2.

2. (B) The given A.P. is – 11, – 8, – 5, ........, 49.To find 4th term from last, a = 49, d = – 3.∴ Required term = 49 + (4 – 1) × (– 3)

= 49 – 9 = 40.

3. (A) In ∆OPQ,PQ2 = 132 – 52

⇒ PQ = 12 cm

∴ ar( PQOR) = 2 × ar(∆POQ)

= 2 × 12

× 5 × 12

= 60 cm2.

4. (A) Area of a triangledrawn in a semicircleis directly proportionalto its height.

∴ar(largest ∆ABC) = 12

× 2r × r

= r2 sq. units.

5. (D) 1

2

VV

= 6427

⇒

31

32

4343

r

r

π

π=

3

343

⇒ 1

2

rr

= 43

⇒2

122

rr

= 169

∴ 1

2

SS

=2

122

44

rr

ππ

⇒2

122

rr

= 169

i.e., S1 : S2 = 16 : 9.

6. (D) Let O be the centre of the circle and PAand PB are two tangents inclined at 60°.In right-angled ∆OAP,

∠OPA = 30°

APBOPA = OPB =

2∠ ∠ ∠

∵

∴ tan 30° = OAAP

⇒13 =

3AP

⇒ AP = 3 3 cm.

7. (D) Each pair of opposite sides of aquadrilateral circumscribing a circlesubtends complementary angles at thecentre of the circle.∴ ∠AOB + ∠COD = 180°⇒ ∠COD = 180° – 125°

= 55°.

8. (C) Let the girl bought n tickets.

∴6000

n= 0.08

⇒ n = 480.


9. (B) Let the sun’s elevation = θ.

⇒ tan θ = ABBC

=6

2 3 = 3

= tan 60°

⇒ θ = 60°.

10. (D) As XY ⊥ AB and CD || XY so CD ⊥ AB.

⇒ CD is bisected by AB at M.

⇒ CD = 2 DM.

In ∆OMD,

DM = 2 2OD – OM

= 2 25 – 3

(... r = 5 cm, OM = AM – AO = 8 – 5 = 3 cm)∴ DM = 4 cmTherefore, CD = 2 × 4 = 8 cm.

SECTION-B

11. Side of square = a = 7.5 × 2 = 15 cm

Area of shaded part = Area of square – Area of circle= a2 – πr2

= 152 – 3.14 × (7.5)2

= 225 – 176.625= 48.375 cm2.

12. Let first term = a1 and commondifference = d.

⇒ 7a7 = 11a11

⇒ 7(a1 + 6d) = 11(a1 + 10d)

⇒ 7a1 – 11a1 = 110d – 42d

⇒ a1 = – 17d

⇒ a1 + 17d = 0

∴ a18 = 0.

13. 6x2 – 2x – 2 = 0Comparing the coefficients of like powersof the given equation with ax2 + bx + c = 0,

we get a = 6, b = – 2 , c = – 2

D = b2 – 4ac = 2 + 48 = 50Quadratic formula:

x = 2– – 4

2

b b ac

a

±

∴ x = ±2 5012

= ±2 5 212

∴ x = 6 212

or x = – 4 2

12

i.e., x = 2

2 or x =

2–

3

Hence, required roots are 2

2 and

2–

3.

OR

Consider the equation,2x2 – 3x – 5 = 0

⇒ 2x2 – (5 – 2)x – 5 = 0(Spliting middle term)

⇒ 2x2 – 5x + 2x – 5 = 0⇒ x(2x – 5) + 1 (2x – 5) = 0⇒ (2x – 5) (x + 1) = 0⇒ 2x – 5 = 0 or x + 1 = 0

i.e., x =52

or – 1

Thus, 52

and – 1 are the required roots.

14. We are given three points are A(– 4, 0),B(4, 0) and C(0, 3).

Using distance formula, we find thelengths of segments (sides) as

AB = +2 28 0 = 8;

BC = +2 24 3 = 5

CA = +2 24 3 = 5

⇒ BC = CA ≠ AB⇒ ∆ABC is an isosceles triangle.


15. 2 20– –,7 7

Hint: Use Section formula

x = 2 1lx mxl m

++

: y = 2 1ly myl m

++

where l : m = 3 : 4.

16. Sample space (all possible outcomes)when rolling a pair of dice is given by:

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 1), (2, 2), (2, 3), (2, 4) (2, 5), (2, 6),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}.

∴ n(S) = 36

Favourable outcomes are: (1, 3), (2, 2),(2, 6), (3, 1), (3, 5), (4, 4), (5, 3), (6, 2),(6, 6). These are 9 in counting.

Hence, the required probability

= 9

36 =

14

.

17. Volume of the remaining solid

= Volume of the cube

– Volume of the cone.

= l3 –13

πr2h

= 73 –13

×227

× 32 × 7

= 343 – 66 = 277 cm3.

18.

Extend AB and CD to meet at P.Since the pair of tangents drawn from anexternal point to a circle are equal,i.e., BP = DP and AP = CP.So, AP – BP = CP – DPi.e., AB = CD. Hence proved.

SECTION-C

19. Number of cards with numbers from 2to 101

= 101 – 2 + 1 = 100∴ n(S) = 100(i) Let E1 be the event of drawing a card

having an even number.Even numbers are : 2, 4, 6, ....., 100,which are 50.∴ n(E1) = 50

Now, P(E1) = 1(E ) 50 1= =

(S) 100 2nn

.

(ii) Let E2 be the event of drawing a cardhaving a square number. Square numbersare: 4, 9, 16, 25, 36, 49, 64, 81, 100.

∴ n(E2) = 9 ∴ P(E2) = 2(E ) 9(S) 100

nn

= .

OR

Number of non-defective bulbs= 24 – 6 = 18

P(First bulbs is not defective)

= Number of non-defective bulbs

Total number of bulbs

=1824

= 34

If the first selected bulb is defective, thenthe number of remaining defective bulbs= 6 – 1 = 5 and the remaing total numberof bulbs = 24 – 1 = 23.

∴ P(Second bulb is defective) = 5

23.

20. Given equation:abx2 + (b2 – ac)x – bc = 0

Comparing it withAx2 + Bx + C = 0, we get

A = ab; B = b2 – ac; C = –bc

Quadratic formula:

x = 2–B B – 4 AC

2A

±

= 2 2 2– ( ) ( ) ( )

2

b – ac b – ac – 4ab – bc

ab

±


= 2 4 2 2 22

2

ac – b b + ab c + a c

ab

±

= 2 2 2( )

2

ac – b b + ac

ab

±=

2( )2

2ac – b b +acab

±

⇒ x = 22acab

or x = 2– 2

2bab

⇒ x = cb

or x = –ba .

21. Sn = 23 5

+2 2n n

Substituting n = n – 1, we get

Sn–1 = 32

(n – 1)2 + 52

(n – 1)

= 32

(n2 – 2n + 1) + 52

(n – 1)

= 32

n2 – 3n + 32

+ 52

n – 52

= 32

n2 – 12

n – 1

nth term of an A.P. is given byan = Sn – Sn – 1

= 32

n2 +52

n – 32

n2 +12

n + 1

= 3n + 1∴ a31 = 3 × 31 + 1 = 94.

ORCommon difference = a3 – a2 = a2 – a1⇒ 3k2 + 4k + 4 – (2k2 + 3k + 6)

= 2k2 + 3k + 6 – (k2 + 4k)∴ k2 + k – 2 = k2 – k + 6⇒ 2k = 8 ⇒ k = 4∴ Common difference = 42 + 4 – 2 = 18Hence, the given terms are:42 + 4 × 4, 42 + 4 × 4 + 18 and 42 + 4 × 4 + 2 × 18i.e., 32, 50 and 68.

22. We are giventhat AB = 8 cm;BC = 10 cm andCA = 12 cm

Let AD = x, then AF = x;BD = AB – AD = 8 – x = BEEC = BC – BE

= 10 – 8 + x = 2 + x = CF.AF = AC – CF

= 12 – 2 – x = 10 – xHere, we observe that

AF = x and AF = 10 – x∴ x = 10 – x ⇒ x = 5 cmHence, AD = 5 cm; BE = 8 – x = 3 cm;

CF = 2 + x = 7 cm.

23. Use ASP:∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°⇒ ∠C = 180° – 150°= 30°⇒ ∠C = 30°.

∆A′BC′ ~ ∆ABC such that ′A B

AB=

BC A C 4BC AC 3

′ ′ ′= = .


24. Circumcentre of a triangle is equidistantfrom the vertices of the triangleLet O (x, y), be the circumcentre of the given∆ABC.∴∴∴∴∴ OA = OB = OCi.e., OA2 = OB2 = OC2

Taking OA2 = OC2, we get

(x – 8)2 + (y – 6)2 = (x – 2)2 + (y + 2)2

⇒ x2 – 16x + 64 + y2 – 12y + 36

= x2 – 4x + 4 + y2 + 4y + 4

⇒ – 12x – 16y = 8 – 100

⇒ 12x + 16y = 92

⇒ 3x + 4y = 23 ...(i)

Also taking OB2 = OC2,

(x – 8)2 + (y + 2)2 = (x – 2)2 + (y + 2)2

⇒ (x – 8)2 = (x – 2)2

⇒ x – 8 = ± (x – 2)

But x – 8 = (x – 2) as – 8 ≠ – 2

∴ x – 8 = – (x – 2)

i.e., 2x = 10

i.e., x = 5

Substituting x = 5 in equation (i), we get

3 × 5 + 4y = 23

i.e., 4y = 23 – 15 = 8

i.e., y = 2Circumradius = OA

= ( ) ( )+2 2– 8 – 8x y

= ( ) ( )+2 25 – 8 2 – 6

= +9 16 = 5

Hence, the circumcentre is (5, 2) and thecircumradius is 5 units.

25. Given chord is AC. Exterior angleAOC = 270°. ACDA is the major segment.∴ ar(segment ACDA)

= ar(sector AOCDA) + ar(∆AOC)

= 270°360°

× π × (20)2 + 12

× 20 × 20

= 34

× 3.14 × 400 + 10 × 20

= 942 + 200

= 1142 cm2.

26. Area of square = (side)2 = (4)2

= 16 cm2

Area of each quadrant = 14

πr2

= 14

× 3.14 × (1)2

= 3.144

cm2

∴ Area of 4 quadrants = 4 × 3.144

= 3.14 cm2

Also area of the circle drawn in middle ofthe square = πr2

= 3.14 ×22

2

= 3.14 cm2


27. In the figure drawn here, let AB be a wallof height h m and AC be a ladder 15 mlong that makes an angle of 60° with thewall.

In right-angled ∆ABC,

cos 60° =ABAC

⇒ 12

= 15h

⇒ 2h = 15

∴ h = 152

= 7.5 m.

28. Let the required ratio be λ : 1

We will use section formula,

– 3 = – 2 – 5

+1λ

λ ; p =

– 4 + 3+1

λλ

i.e., – 3λ – 3 = – 2λ – 5 ; p = – 4 + 3

+1λ

λ

i.e., λ = 2 ; p = – 4 + 6

3=

23

Required ratio is 2 : 1 and p = 23

.

SECTION-D

29. r = AO = BO = CO = 3.5 cm

CD = 15.5 cm

OD = CD – CO

= 15.5 – 3.5 = 12 cm

Now, area of shaded part = Area of square– Sum of areas of 4 quadrants – Area ofcircle drawn in centre.

= 16 – 3.14 – 3.14= 16 – 6.28 = 9.72 cm2

OR

Side of ∆ABC = a = 10 cm

Radius of each circle = r = 102

= 5 cm

ar(ABC) = 3

4a2 =

34

× 102

= 25 3 = 25 × 1.732

= 43.30 cm2

Area of each sector situated in ∆ABC

= 260°×

360°rπ =

16

× 3.14 × 52 cm2.

Area of the shaded region

= ar(ABC) – 3 × area of any onesector

= 43.30 – 3 × 16

× 3.14 × 25

= 43.30 – 39.25 = 4.05 cm2.


l = AD = 2 2OD AO+

= +144 12.25

= 12.5 cm.

Now, total surface area

= C.S.A. of the cone ABD + C.S.A. of the hemisphere ACB

= πrl + 2πr2 = πr(l + 2r)

=227

× 3.5 × (12.5 + 7)

= 22 × 0.5 × 19.5 = 214.5 cm2.

OR

Given: r1 = 20 cm, r2 = 8 cm and h = 16 cm

Volume of the container

V = ( )2 21 2 1 2

13

h r r r rπ + +

= ( )2 21× 3.14 × 16 20 8 20 8

3+ + ×

= ( )1× 50.24 × 400 64 160

3+ +

= 13

× 50.24 × 624 = 10450 cm3

(approx) ∴ Capacity of container

= 104501000

l = 10.45 l

Cost of milk = Volume of the container × Rate per litre

= 10.45 × ` 15= ` 156.75.

30. Let the tap of larger diameter takes t hoursto fill the tank. So the smaller one will take(10 + t) hours on the same work.∴ The part of the tank filled by larger tap in

1 hour = 1t

And the part of the tank filled by the smaller

tap in 1 hour = +

110 t

So, the part of the tank filled by both the taps

simultaneously in 1 hour = 1t

+ +

110 t

...(i)

But it is given that the taps together fill the

tank in 3

98

=758

hrs.

So, the part of the tank filled by both the taps

simultaneously in 1 hour = 1

758

= 8

75hrs

...(ii)From the results (i) and (ii), we have

++

1 110t t

= 875

10( 10)

t tt t

+ ++

= 8

75

⇒ 750 + 150t = 8t2 + 80t

⇒ 8t2 – 70t – 750 = 0,


i.e., 4t2 – 35t – 375 = 0

D = (– 35t)2 – 4 × 4 × (– 375) = 1225 + 6000

∴ D = 7225± = ± 85

∴ t = 35 85

2 4±×

= 120

8 or

– 508

∴ t = 1208

= 15 hrs

(Time cannot be negative so t = – 50

8is rejected.)

And t + 10 = 25 hrs.Thus, time taken by the larger tap = 15 hrs.and time taken by the smaller tap = 25 hrs.

31. DA and DC are two tangents from D to thecircle with centre F.∴ ∠ADF = ∠CDF ...(i)Similarly,∴ ∠BEF = ∠CEF ...(ii)As l || m and DE is transversal,∴ ∠ADE + ∠BED = 180°

(Cointerior angles)∴ (∠ADF + ∠EDF) + (∠BEF + ∠DEF) = 180°∴ (∠EDF + ∠EDF) + (∠DEF + ∠DEF) = 180°⇒ 2 ∠EDF + 2 ∠DEF = 180°⇒ ∠EDF + ∠DEF = 90° ...(iii)

In ∆EDF,

⇒ ∠EDF + ∠DEF + ∠DFE = 180°

(Angle sum property of a triangle)

⇒ 90° + ∠DFE = 180°[Using (iii)]

∴ ∠DFE = 90°.

Hence proved.

32. Let us take LHS of the given equation

– 4 + (– 1) + 2 +......+ x

Here, 2nd term – 1st term = 3rd term – 2nd term= 3. So, this is the sum of A.P. with commondifference d = 3 and the first term a = – 4.

Let this A.P. contains n terms.

Then, using an = a + (n – 1)d, we get

x = – 4 + (n – 1) 3

⇒ n – 1 = + 43

x

⇒ n = + 73

x

Now, LHS = + 73

x(– 4 + x)

[Using Sn = 2n

(a + l)]

= 2– 4 – 28 76

x x x+ +

= +2 3 – 28

6x x

Hence, the given equation becomes

+2 3 – 286

x x= 437

⇒ x2 + 3x – 2650 = 0

x2 + 53x – 50x – 2650 = 0

⇒ (x + 53) (x – 50) = 0

⇒ x = – 53 or x = 50

But x = – 53 is not possible because the A.P.has no negative term after the second term.

Therefore, x = 50.

OR

Ruchi has 13 flags to be fixed at one sideof her, each at R1, R2,......., R3 (see figure)and the same no. on the opposite side ofit such that the flags are on the straightpassage.


Let initially Ruchi is at the point R withher books and 27 flags.

Ruchi fixed one flag at the point R also

RR1 = R1R2 = ..... = R12R13 = 2 m

Fixing one flag at R1, distance covered byRuchi

= RR1 + R1R

= 2 + 2 = 4 m

Fixing one flag at R2, distance covered byRuchi = RR2 + R2R

= 4 + 4 = 8 m and so on.

Therefore distance covered by Ruchi to fixall flags on one side of R.

= 4 + 8 + 12 + ....... to 13 term

= 4(1 + 2 + 3 +.......to 13 terms)

= 4 × 13 × 14

2 = 364 m

Similarly, distance covered by Ruchi to fixall flags on other side of R = 364 m.

Hence, total distance covered

= 364 + 364 = 728 m

Maximum distance travelled by Ruchicarrying a flag = RR3

= 2 × 13 = 26 m.

33. Let XP = x and PQ = y

∴ YM = x; PM = 40 m; QM = y – 40

In right-angled triangle QYM,

tan 45° = QMYM

⇒ 1 = – 40yx

⇒ x = y – 40 ...(i)

In right-angled triangle QXP,

tan 60° = QPXP

⇒ 3 = yx

⇒ x = 3

y...(ii)


y – 40 = 3

y

⇒ 3 y – 40 3 = y

⇒ ( 3 – 1)y = 40 3

⇒ y = 40 3

3 – 1

⇒ y = 40 3

3 – 1 ×

++

3 13 1

y = 40(3 3)

3 – 1+

⇒ y = 20 (3 3)+ ...(iii)

i.e. PQ = 20 (3 3)+ metres

Again in right-angled triangle QXP,

sin 60° = QPXQ

⇒3

2=

20(3 3)XQ

+

⇒ XQ = 20(3 + 3) × 2

3

⇒ XQ = 40( 3 1)+

Hence, PQ = 20(3 3)+ metres and

XQ = 40( 3 1)+ metres.


34. h = 16 cmr2 = 8 cmr1 = 20 cm

l = 2 21 2( – )h r r+ = 2 216 12+

= 256 144+ = 400 = 20 cm.

∴ Area of sheet used to make bucket

= C.S.A. of frustum + Area of base

= πl(r1 + r2) + πr22

= π[l(r1 + r2) + r22]

= π[20 × 28 + 64]

= 3.14 × 624

= 1959.36 cm2

∴ Cost = 1959.36100

× 15

= ` 293.90