INDEX

Contents Page No Contents Page No

1. Real Numbers 01 9. Some Applications of Trigonometry 39

2. Polynomial 04 10. Circles 44

3. Linear Equation in two variables 08 11. Constructions 50

4. Quadratic Equations 12 12. Areas Related to Circles 51

5. Arithmetic Progressions 17 13. Surface Areas and Volumes 54

6. Triangles 22 14. Statistics 60

7. Coordinate Geometry 28 15. Probability 63

8. Introduction to Trigonometry 32

IDEALINDIANSCHOOL

10std QUESTION - BANK

201710 – DEPT. OF MATHEMATICS

MATHEMATICS 2017-18

1. Real Numbers

1. What is the least number divisible by all the natural numbers from 1 to 10 ?

Ans: LCM (1,2,3 , .. .. ,10 )=2520

2. Find the greatest number of 6 digits which is exactly divisible by 24 ,15 and 36 .

Ans: Greatest 6 digit number = 999999

LCM (24 ,15 ,36 )=279Required number =999999−279=999720

3. Find the smallest number that, when divided by 35 ,56 ,and 91 leaves a remainder of 7 in each.

Ans: LCM=(35 ,56 ,91 )=3640

The number required=3640+7=3647

4. Show that cube of a positive integer is of the form 6q+r , where q is an integer.

Ans: Let a be the integer

a=6q+rb=6 , r=0,1,2,3,4,5

(i) r=0⇒a=6q⇒a3=(6q )3=216q3

⇒a3=6 ( 36q3 )⇒a3=6m where m=36 q3

(ii) r=1⇒a=6q+1⇒a3=6m+1 where m=36 q3+36 q2+12q+1

Similarly take r as 2,3,4,5 the form will take as in the before.

5. Prove that

√3√3−√2 a rational number.

Ans:

√3√3−√2

×√3+√2√3+√2

= √9+√6(√3 )2−(√2 )2

= 3+√63−2

=3+√61

=3+√6

6. Find the HCF of 81 and 237 . Express the HCF in the form of 237 p+81q . Find the value

of (2 p+q )

Ans: 237=81×2+75

81=75×1+675=12×6+3

From the 3rd division algorithm,3=75−12×6

237 p+81q=237−81×2−12 (81−75×1 )=237−81×2−12×81+12×75=237−81×2−12×81+12×(237−81×2 )=237=81×2−12×81+12×237−81×24=237×(13 )+81×(−38 )

∴ p=13 , q=−382 p+q=2×13+(−38 )

=−12

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7. Show that cube of any positive integer is of the form 4m, 4m+1and 4m+3 , for some integer.Ans: Let n be any positive integer,

Then it will be of the form 4 q . 4q+1and 4 q+3

Where b=4 , r=0,1,2,3

Case I: r=0⇒n=4 q+0⇒n3=64q3⇒n3=4m where m=16q3

Case II: r=1⇒n=4q+1⇒n3=(4 q+1 )3⇒n3=64q3+48q2+12q+1

⇒n3=4m+1 where m=16q3+12q2+3q

Case III: r=2⇒n=4q+2⇒n3=(4 q+2 )3=64 q3+96q2+48q+8

⇒n3=4m where m=16q3+24 q2+2

Case IV: r=3⇒n=4q+3⇒n3=( 4q+3 )3=64q3+144 q2+108q+27

⇒n3=64q3+144q2+24+3⇒n3=4m+3 where m=16q3+36 q2+6

Hence n is of the form 4m, 4m+1and 4m+3 .

8. A class of 20 boys and 15 girls is divided into n groups so that each group has x boys

and y girls. Find x ,y and n . What values are referred in the class?Ans: The boys and girls divided in to n groups,

The members in each group x+ y=20+15n

x+ y=20+15n

nx+ny=20+15Possible combinations are,

Total number of boys, nx=20⇒n×x=2×10=45=5×4

Total number of girls, ny=15⇒n× y=3×5=5×3∴nx=5×4 and ny=5×3

When n=5 , x−4 , y=3then only its possible.

9. At an international airport, planes take off from five different runways at 3,4,8 ,12and 15 minutes intervals. At 7 :30 a.m., planes took off from all five runways simultaneously. When will five planes take off together again.

Ans: LCM (3,4,8 ,12 ,15 )=120 minutesHence four planes take off together after 2 hours i.e., 9.30

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2. Polynomial

1. If f ( x )=x3+x2−ax+b is divisible by x2−x write the values of a ,b .

Ans: x2−x=0

x (x−1 )=0x=0 ; x=1f ( x )=x3+x−ax+bx=0⇒ f (0 )=03+0−a×0+b⇒0=b∴b=0 …(1)

x=1⇒ f (1 )=13+12−a×1+b⇒0=1+1−a+b⇒a−b=2

From (1),a+0=2

∴a=2

2. If one zero of the polynomial (a2+9 )x2+13 x+6a is reciprocal of the other, find the value of a .

Ans: α=m , β= 1

m

αβ= ca

m× 1m

= 6aa2+9

1×(a2+9 )=6aa2−6 a+9=0

(a−3 )2=0a=3

3. Apply division algorithm find the quotient and remainder on dividing

f ( x )=4 x3−24 x+8 x2+7 andg ( x )=2−2 x+x2.

Ans: f ( x )=4 x3+8x2−24 x+7 ,g ( x )=x2−2x+2deg (q ( x ) )=deg (f ( x ) )−deg (g (x ) )=3−2=1

Hence quotient, q ( x )=ax+b and r ( x )=kBy division algorithm,

f ( x )=g ( x ) q ( x )+r (x )4 x3+8 x2−24 x2+7= (x2−2 x+2 ) (ax+b )+k

=ax3−2ax2+2ax+bx2−2bx+2b+k=ax3+ (b−2a ) x2+ (2a−2b ) x+ (2b+k )

By comparison,

ax 3=4 x3⇒a=4

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(b−2a ) x2=8 x2⇒b−2×4=8⇒b=8+8⇒b=16And 2b+k=7⇒2×16+k=7⇒k=−25

4. If the squared difference of the zeros of the polynomial f ( x )=x2+ px+45 is equal to 144 , find the value of p .

Ans: (α−β )2=144

α 2+β2−2αβ=144(α+β )2−2αβ−2αβ=144(−p )2−4×45=144p2=144+180p=18

5. Find all of the zeros of the polynomialx4−6 x3−26 x2+138 x−35 , if its two zeros are

(2+√3 ) and (2−√3 ) .

Ans: Let f ( x )=x 4−6 x3−26 x2+138 x−35

α=2+√3 and β=2−√3

The divisor,g ( x )=x2− (α+β ) x+αβ

=x2−(2+√3+2−√3 )x+(2+√3 ) (2−√3 )

g ( x )=x2−4 x+1

¿∴ f ( x )=(x2−4 x+1 ) (x2−2 x−35 )The other two can be find by factorising x

2−2x−35=0i .e . , x=7 ,−5

The zeros are 2+√3 ,2−√3 ,7 and −5

6. If α and β are the zeros of the polynomialf ( x )=3 x2−4 x+1 , find a quadratic

polynomial whose zeros are

α2

β and

β2

α .

Ans: 3 x2−4 x+1=0

x= 13,1

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α=13, β=1

α2

β= 1

9×1= 1

9 ,

β2

α=1

3

The equation: f ( x )=k (x2−(α+β ) x+αβ )

f ( x )=k (27 x2−12x−127 )

f ( x )=27x2−12 x−1

7. On dividing p ( x )=5 x4−4 x3+3 x2−2 x+1 by g ( x )=x2+2 , if q ( x )=ax2+bx+c , find a ,band c

Ans: p ( x )=g ( x )×q ( x )+r ( x )

5 x4−4 x3+3 x2−2 x+1=(x2+2 ) (ax2+bx+c )+r (x )

=ax4+bx3+cx2+2ax2+2bx+2c+r ( x )=ax4+bx3+ (c+2a ) x2+2bx+(2c+r ( x ) )a=5b=−4c+2a=3⇒c=−7

8. If ( x+q ) is a factor of two polynomials x2+ px+q and x

2+mx+n , then prove that

q= n−qm−p

Ans:x=−q

f ( x )=x2+ px+q

f (−q )=q2−pq+q=0 …(1)

g (−q )=q2−mq+n=0 …(2)(1 )= (2 )

q2−pq+q=q2−mq+nmq−pq=n−q(m−p ) q=n−q

∴q= n−qm−p

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3. Pair of linear Equation in Two VariablesGraph Based

1. Draw the graphs of the equations x=3 and 2 x− y−4=0 . Also find the area of the quadrilateral formed by the lines and x− axis.

2. Determine graphically coordinates of the vertices of a triangle, the equations of whose

sides arey=x ; y=2 x ; x+ y=6 . Also find the area so formed.

3. The cost of 2kg of potatoes and 3kg of onions on a day was found to be Rs .56 . After a

month cost of 3kg of potatoes and 2kg of onions is Rs . 49 . Solve graphically.Other Problems

4. Solve: 0 .2x+0 . 3 y=1.30 . 4 x+0 .5 y=2.3

Ans: Multiply by each term by 10 ,2 x+3 y=13. . . (1 )4 x+5 y=23. . . (2 )Multiply Eqn 1 by 2,4 x+6 y=26 .. . (3 )Subtract (2) in (3)

x=2 and y=3

5. Solve:

x+12

+ y−13

=8

x−13

+ y+12

=9

Ans:

x+12

+ y−13

=8

⇒3 x+3+2 y−2=48⇒3 x+2 y=47. . . (1 )x−1

3+ y+1

2=9

⇒2x−2+3 y+3=54⇒2 x+3 y=53 .. . (2 )Multiply Eqn (1) by 2 and Eqn (2) by 3, Subtract (2) in (1),x=7 , y=13

6. Solve: 2u+15v=17uv5u+15v=16uv

Ans: 2u+15v=17 uv . . . (1 )

5u+15v=16 uv . . . (2 )

Divide each term by uv , and take

1v=p , 1

u=q

2v+15u

=17⇒2 p+15q=17 . .. (1 )

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5v+15u

=16⇒5 p+15q=16 . .. (2 )

(2 )− (1 )⇒3 p=−1

p=−13

Substituting p in Eqn (1),

2×13+15q=17

15q=17+ 23=51+2

3⇒q=53

45

p=1v= 1

−3⇒ v=−3

q=1u=53

45⇒u=45

53Word Problems:7. A railway half – ticket costs half the full fare but the reservation charge remains the

same on a half ticket as on a full full-ticket. One reserved first class ticket from station A to station B costs Rs. 2125. Also, one reserved first class ticket and one reserved half ticket from A to B costs Rs. 3200. Find the full fare from station A to B and also find the reservation charges for a ticket.

Ans: Reservation charge =x , Cost of the ticket be y .The total cost of one reserved first class ticket,x+ y=2125 . .. (1 )The total cost of one first class full and half ticket,

( x+ y )+(x+ 12y )=3200

⇒2x+32y=3200⇒4 x+3 y=6400 .. . (2 )

Eqn . (1 )×4−Eqn . (2 )⇒ y=2100Then, x=25

The reservation charge is Rs . 125 and Rs . 2100

8. The sum digits of a two digit number is 11. The number obtained by interchanging the digits of the given number exceeds that number by 63. Find the number.

Ans: x+ y=11. .. (1 )

63+10 x+ y=10 y+x9 x−9 y=−63Divide each term by 9,x− y=−7 .. . (2 )Eqn . (1 )+Eqn . (2 ) ⇒2 x=4x=2

Then, y=9

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The number is 29 .9. Ramesh travels 760km to his home, partly by train and partly by car. He takes 8

hours if he travels 160km by train and the rest by car. He takes 12 minutes more if he travels 240km by train and the rest by car. Find the speed of the train and car respectively.

Ans: Le the speed of train be xkm /h and car be ykm/h

Option – 1: Time taken in travelling 160km by train=160

xhr

Time taken in travelling (760-160)km by car=600

yhr

Total time taken =8hrs

160x

+600y

=8 . ..(1 )

Option – 2: Time taken in travelling 240km by train=240

xhr

Time taken in travelling (760-240)km by car=520

yhr

Total time taken =8hrs+12min=8 .2hr

240x

+520y

=8 .2. .. (2 )

Take

1x=u , 1

y=v

,

Eqn . (1 ) :160u+600v=8⇒20u+75 v=1 .. . (3 )Eqn . (2 ) :240u+520v=8 .2⇒120u+260 v=4 . 1. .. (4 )

Solving (3) and (4) we get, x=80 , y=100The speed of train and car is 80km/hr and 100km /hr respectively.

10. The boat goes 12km upstream and 40km downstream in 8 hours. It can go 16km upstream and 32km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

Ans: Speed of boat in still water=x km/h

Speed of stream = y km/h

Time taken to cover 12km upstream =12x− y

hrs

Time taken to cover 40km downstream =40x+ y

hrs

Total taken for the journey =8hrs

⇒12x− y

+40x+ y

=8…(1)

Time taken for 16km upstream=16x− y

hrs

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Time taken for 32km downstream=32x+ y

hrs

Total time taken for the journey =8hrs

16x− y

+32x+ y

=8…(2)

Let

1x+ y

=u and

1x− y

=v

From (1), 12v+40u=83v+10u=2 …(3)

From (2), 16v+32u=84 v+8u=2 …(4)

On solving, we get speed of boat is 6km/h and speed of water is 4km/h.

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4. Quadratic Equations

1. Solve for x :x−4x−5

+ x−6x−7

=103; x≠5,7

Ans:

x−4x−5

+ x−6x−7

=103; x≠5,7

( x−4 ) (x−7 )+( x−6 ) ( x−5 )( x−5 ) ( x−7 )

=103

(2 x2−22 x+58 ) 3=10 (x2−12x+35 )6 x2−66 x+174=10 x2−120 x+35010 x2−6 x2−120 x2+66 x+350−174=04 x2−54 x+176=02 x2−27 x+88=0Use formula method or squaring further…

2. Solve for x : 9 x2−6ax+ (a2−b2)=0

Ans: 9 x2−6ax+ (a2−b2)=0

9 x2−6ax+ (a+b ) (a−b )=09 x2−3 (a+b ) x−3 (a−b ) x+(a+b ) (a−b )=03 x (3 x−(a+b ) )+ (a−b ) (3x−(a+b ) )=0(3 x+a−b ) (3 x−(a+b ) )=03 x+a−b=0 or 3 x−a−b=0

x=b−a

3 orx= a−b

3

3. Solve for x :

2xx−3

+ 12 x+3

+ 3x+9( x−3 ) (2x+3 )

=0 , x≠3 ,−32

Ans:

2x (2 x+3 )+ ( x−3 )+ (3x+9 )( x−3 ) (2x+3 )

=0

4 x2+6 x+ x−3+3 x+9=02 x2+5 x+3=02 x2+3 x+2 x+3=0x (2 x+3 )+1 (2 x+3 )=0( x+1 ) (2 x+3 )=0

x=−1 ,−32

4. Find the roots: ax2+a=a2 x+x

Ans: ax2−a2 x−x+a=0

ax 2−a2 x−x+a=0ax ( x−a )−1 ( x−a )=0

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(ax−1 ) ( x−a )=0

x=1a,a

5. Find the values of k for which the equation kx ( x−2√5 )+10has equal roots.

Ans: kx (x−2√5 )+10=0

kx 2−2√5kx+10=0D=b2−4 ac

=(−2√5k )2−4×k×10=20 k2−40kFor real roots, D=0

20k2−40k=0k=0,2

6. Find the value of p for which the quadratic equation

(2 p+1 ) x2−(7 p+2 ) x+(7 p−3 )=0 has equal roots. Also find the roots.

Ans:D=0

⇒0=b2−4ac

(−(7 p+2 ) )2−4×(2 p+1 )×(7 p−3 )=0

7 p2−24 p−16=0p=−4

7,4

7. Two pipes running together can fill a cistern in 3 1

13 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

Ans: Let x and x+3 be the time taken by each pipe to fill the cistern1x+ 1x+3

= 1

3 113

x+3+xx ( x+3 )

=1340

(2 x+3 ) 40=13 (x2+3 x )80 x+120=13 x2+39 x13 x2−41 x−120=0Use quadratic formula: x=5The time taken by by each pipe is 5 minutes and 8 minutes.

8. A train covers a distance of 90km at a uniform speed. Had the speed been 15km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.

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Ans: Time=Distance

Speed ,

The original speed of the train =xkm /hr

The expected speed = (x+15 ) km /hrAccording to the given condition,90x

−90x+15

=3060

90 x+1350−90xx2+15 x

=12

x2+15 x−2700=0⇒ x=45The original speed of the train is 45 km /hr

9. The denominator of a fraction is one more than twice the numerator. If the sum of

the fraction and its reciprocal is 2 16

21 , find the fraction.

Ans:

x2x+1

+ 2 x+1x

=2 1621

x2+4 x2+4 x+12x2+x

=5821

By Cross multiplication,

(5 x2+4 x+1 )21=58 ( 2x2+x )11 x2−26x−21=0

x=−73,3

The fraction is

37

10. A plane left 30 minutes later than the schedule time and in order to reach the destination, 1500km away, in time, it had to increase its speed by 250km/hr from its usual speed. Find its usual speed.

Ans: Let the speed be xkm /h .

The time taken with actual speed=1500

x km/h

The time taken in increased speed=1500x+250 km/h

The difference in the arrival time=30 min

⇒1500x

−1500x+250

=3060

1500x

−1500x+250

= 12

1500 x+1500×250−1500xx ( x+250 )

=12

x2−250 x−750000=0

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( x+10000 ) (x−750 )=0x=750

The speed of the plane is 750 km /hr

11. Thrisha can her boat at a speed of 5km/h in still water. If it takes her 1hr more to row the boat 5.25km upstream than to return downstream, find the the speed of the stream.

Ans: Speed of upstream= (5−x ) km /h

Speed of downstream= (5+ x ) km/h

Time taken for going 5.25km upstream = 5 .25

5−xh

Time taken for going 5.25km downstream= 5 .25

5+xh

The time difference is the journey=1hr5.255−x

−5 .255+x

=1

5 .25 ( 15−x

− 15+x )=1

214 ( 5+x−5+x

(5−x ) (5+x ) )=1

2 x2+21x−50=0

x=2 or x=−252

The speed of the stream is 2km/h

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5. Arithmetic Progression1. Find whether 0 is a term of an AP: 40, 37, 34, 31,….

Ans: an=0 , a=40 , d=−3 , n=?

an=a+(n−1 ) d0=40+ (n−1 ) (−3 )3n=43

n=433

n is not a natural number so, O is not a term of A.P.2. The fifth term of an AP is 1 where as its 31st term is -77. Which term of the AP is -17?

Ans: a5=1⇒a+4 d=1 .. . (1 )

a31=−77⇒a+30 d=−77 .. . (2 )(2 )− (1 )⇒26 d=−78d=−3

a+4 (−3 )=1a=13a17=13+16×(−3 )=13−48=−35

3. Find the value of a25−a15 for the AP: 6,9,12,15….

Ans: a=6 , d=3 ,a25−a15=a+24d−(a+14 d )=10d∴ a25−a14=10×3=30

4. Determine k so that k2+4 k+8 , 2k 2+3k+6 , 3k2+4 k+4 are three consecutive terms

of an AP.

Ans: a2−a1=a3−a2

2k2+3k+6− (k2+4 k+8 )=3k2+4k+4−(2k2+3 k+6 )∴ k=0

5. Find the sum of all odd positive integers.Ans: AP: 1,3,5,7,…

a=1 , d=2 , Sn=?

Sn=n2 [2a+(n−1 )d ]

=n2 [2×1+(n−1 )2 ]

= n2×2n

=n2

6. The eighth term of an AP is half its second term and the eleventh term exceeds one-third of its fourth term by 1. Find the 15th term.

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Ans: a8=

12a2⇒a+7d= 1

2(a+d )⇒ 2a+14d=a+d⇒ a=−13d . . . (1 )

a11=13a4+1⇒a+10d=1

3(a+3d )+1⇒2a+27d=3 . .. (2 )

Substituting (1) in (2), we get,d=3 ,

From (1), a=−13×3=−39

a15=−39+14×3=37. The sum of three numbers of an AP is 27 and their product is 405. Find the numbers.

Ans: The three consecutive AP is be a−d ,a ,a+da−d+a+a+d=273a=27⇒a=9(a−d ) (a ) (a+d )=405

⇒(9−d ) (9+d )=405

9 ⇒81−d2=45⇒81−45=d2⇒d2=36∴d=6

The numbers are, 3,9 and 128. Divide 56 into four parts which are in AP such that the ratio of product of its

extremes to the product of means is 5:6. Find the numbers

Ans: The four parts are, a1=a−3d ,a2=a−d ,a3=a+d and a4=a+3d

(a−3d )+(a−d )+(a+d )+ (a+3d )=56⇒4a=56⇒a=14The ratio is 5:6a1×a4

a2×a3=5

6⇒

(a−3d ) (a+3d )(a−d ) (a+d )

=56⇒142−9d2

142−d2 =56

⇒6×142−6×9d2=5×142−5d2

⇒6×142−5×142=−5d2+54d2

⇒ (6−1 )×142=49d2

⇒d2=142

49∴d=2The numbers are, a−3d=14−3×2=8

a−d=14−2=12a+d=14+2=16a+3d=14+3×2=20

9. The sum of first six terms of an AP is 42. The ratio of its 10th term to ints 30th term is 1:3. Calculate the first and the 13th term of the AP.

Ans: S6=42⇒ 6

2[ 2a+5d ]=42⇒2a+5d=14 .. . (1 )

a10 :a30=1:3⇒3 a+27d=a+29d⇒2a=2d⇒a=d . . . (2 )

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Substituting (2) in (1), we geta=2 , d=2

Then, a13=a+12d=2+2×12=26

10. If the pth

term of an AP is q and qth

term of an AP is p . Prove that its nth

term is

( p+q−n )Ans: Let a be the first term and d be the common difference of the given A.P.

Then pth

term =q⇒a+( p−1 )d=q …(1)

q th term =p⇒a+(q−1 )d=p …(2)

(1) – (2)⇒d=−1From (1),

a=( p+q−1 )∴nth Term term =a+(n−1 )d=(p+q−1 )+(n−1 )×(−1 )= (p+q−n )

11. Find the 12th term from the end of the AP: -2, -4, -6, ….-100.Ans: Reverse the AP,

AP:−100 ,−98 ,−96 , . .. .. .. . .. . ,−6 ,−4 ,−2a=−100 , d=−98−(−100 )=2 , a=−2 , an=?a12=a+11d=−100+11×2=−78

12. An AP consists of 37 terms. The sum of three middle most terms is 225 and the sum of the last three terms is 429. Find the AP.

Ans: The mid-term is a19 .

The related three midterms are, a18 , a19 and a20 .

a18+a19+a20=225a+17d+a+18d+a+19d=2253a+54 d=225a+18d=75 .. . (1 )Sum of last three terms,

a35+a36+a37=429a+34d+a+35d+a+36d=4293a+105d=429a+35d=143 .. . (2 )By Equating (1) and (2), we geta=3 , d=4AP: 3,7,11,15,….

13. The sums of n terms of an AP are S1 , S2 , S3 . The first term of each is 5 and their

common difference is 2, 4, 6 respectively. Prove that S1+S3=2S2 .Ans:

S1=n2 [2×5+ (n−1 ) 2 ] S2=

n2 [2×5+ (n−1 ) 4 ] ∴LHS:S1+S3=n ( 4+n )+n (2+3 n )

=n (6+4 n )

17 | P a g e

MATHEMATICS 2017-18

=n2

[10+2n−2 ]

=n2

[ 8+2n ]

= n2×2×[4+n ]

∴S1=n ( 4+n )

∴S2=n (3+2n )

S3=n2 [2×5+(n−1 ) ]

∴S3=n (2+3n )

=n×2×(3+2n )=2×n (3+2n )=2 S3=RHS`

14. The student of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What was the maximum distance she travelled carrying a flag.Ans: The flag is kept in the middle where the 14th flag is placed. She has to place 13 flags

either side of the 14th flag. Since she can carry one flag each time she has to travel the same distance twice.

So the distance travelled for 1st flag =2+2=4 m

2nd flag=4+4=8 m

3rd flag=6+6=12m

Hence the AP is 4,8 ,12 , .. .. .upto 13 flags.a=4 , d=4 , n=13Distance travelled on one side of the 14th flag is ,

Sn=n2 [2a+(n−1 )d ]

S13=132 [2×4+ (13−1 ) 4 ]

=132

[8+48 ]=132

×56=13×28=364m

Total distance travelled on either side=364+364=728mSince the longest distance travelled is when she carried a 13th flag, but while coming

back she was not having flag in her hand. So distance counted is only one way i.e.,26 metres

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MATHEMATICS 2017-18

5. Triangles1. In the figure, P is a midpoint of BC and Q is the mid-point of AP. IF BQ when produced

meets AC at R, prove that RA=1

3CA

.

Ans: Its is given that, R is a midpoint,⇒ AR=RS

Construct PS||BR ,

Now, Q is a midpoint of AP , PS||QR ,By converse of midpoint theorem,S is a midpoint of RC .⇒RS=SCThen, AR=RS=SC

∴ AC=(AR+RC+SC )=3 AR

⇒ AR=13AC

2. Two poles of height ‘a’ metres and ‘b’ metres are ‘p’ metres apart. Prove that the height of a the point of intersection of the lines joining the top of each of pole to the

foot of the opposite pole is given by

aba+b .

Ans: In Δ ABC ,AB||EF⇒ ΔABC|||ΔEFC [AA−Criteria ]ABEF

= BCFC

⇒ ah= pFC

⇒ FC= pha …(1)

In ΔBCD, EF||DC⇒ ΔDCB||||ΔEFB

⇒ DCEF

=BCFB

⇒ bh= pFB

⇒FB= phb …(2)

Adding (1) and (2),

FC+FB= pha

+ phb

p=ph [a+bab ]∴h= ab

a+b3. AD is altitude of an equilateral triangle ABC. On AD as base, another equilateral

triangle ADE is constructed. Prove that ar(ADE):ar(ABC)=3:4.Ans: Given ABC and ADE are equilateral triangles,

Let us consider,AB=CA=BC=a

Altitude of AD=√3

2a

[Proved in the Exercise]

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B

R

P C

A

SQ

F

A

B C

D

Eab

h

p

A

B CD

E

A

CB

D

M

E

L

MATHEMATICS 2017-18

Since the triangles are equilateral, ΔABC|||ΔADE

⇒ar (Δ ADEar (Δ ABC )

= AD 2

BC 2

[ Areas Theorem ]

=(√3

2 a)2

a2 =

34 a

2

a2 =34

Hence, ΔABC:ΔADE=3:44. The perpendicular AD on the base of a triangle ABC intersects BC at D so that

DB=3CD. Prove that 2AB2=2AC2+BC2.

Ans: Given:DB=3CD AD⊥ΒC and

To prove: 2 AB2=2 AC 2+BC 2

Proof: BD+DC=BC3CD+CD=BC

4CD=BC⇒CD=14BC

DB=3CD=34BC

In Δ ACD,AC 2=AD 2+CD2

AC 2=AD 2+BC2

16 …(1)

In Δ ABD ,

AB2=AD2+BD 2

AB2=AD2+ 916BC2

…(2)Subtracting (1) from (2) we obtain

AB2−AC 2= 9

16BC 2− 1

16BC 2

16 ( AB2−AC 2 )=8 BC 2

2 (AB2−2 AC 2 )=8BC 2

2 AB2=2 AC 2+BC2

Hence proved.5. Through the midpoint M of the side CD of a parallelogram ABCD, the line BM is drawn

intersecting AC in L and AD produced in E. Prove that EL=2BL.

Ans: To prove: 2BL=EL

Proof: In ΔBMC and ΔEMD ,∠BMC=∠EMD [V .O . A ]MC=DM [ Given ]∠BCM=∠EMD [A . A ]

20 | P a g e

A

B

C

D

MATHEMATICS 2017-18

ΔBMC≃ΔEMD [ ASA Rule ]⇒B=DE( i )[CPCT ]⇒ AE=AD+DE=BC+BC=2BC . ..( ii )

Now ΔBLC|||ΔELA … [ AA Criterion ]

⇒ BLEL

=BCAE [CPCT ]

∴ BLEL

= BC2BC [ from (i) ]

BLEL

= 12

2BL=EL∴EL=2 BL

6. In the figure, PQR is a right angled triangle. The point S and T

trisects QR. Prove that 8PT2=3PR2+5PS2.

Ans: Given: A ΔPQR in which ∠PQR=90 ° , S and T are the points

trisection of QR .

To prove: 8PT2=3PR2+5PS2

Proof: Let QS=ST=TR=x , then QS=x ,QT=2 x and QR=3 x .

From right triangles PQS ,PQT and PQR,by Pythagoras theorem, we have

PS2=PQ2+QS2 ,PT 2=PQ2+QT 2

And, PR2=PQ 2+QR 2

∴3 PR2+5 PS2−8PT 2=3 (PQ 2+QR 2)+5 (PQ2+QS2)−8 (PQ2+QT2 )=3QR 2+5QS2−8QT2

=3×(3 x )2+5 ( x )2−8×(2 x )2

=27 x2+5x2−32x2=0∴3 PR2+5 PS2−8PT 2=0∴8 PT 2=3 PR2+5 PS2

7. In right-angled triangle ABC,∠B=90 ° . If D is the midpoint of BC, prove that

AC 2=4 AD2−3 AB2 .

Ans: Given: In Δ ABC ,∠B=90 ° , D is the midpoint of BC.

To prove: AC2=4 AD2−3 AB2

Proof: In Δ ABC ,∠B=90 ° ,By Pythagoras theorem,

∴ AC2=AB2+BC2

=AB2+(2BD )2 [∵BC=2BD ]

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P

Q RS T

MATHEMATICS 2017-18

=AB2+4BD 2

=AB2+4 ( AD2−AB2 ) [∵ AB2+BD 2=AD 2 ]AC 2=4 AD2−3 AB2

8. A point D on the side BC of an equilateral triangle ABC such that DC=

14 BC. Prove that

AD2=13CD2.

Ans: Given: In Δ ABC , DB=1

4BC

AB=BC=CA=a

Hence, DC= 3

4BC=3a

4

Draw AE⊥BC

⇒BE=EC=12BC= a

2

Since ABC , is equilateral triangle AE=√3

2a

Hence DE=BE−BD

=a2− a

4=a

4

In Δ AED, by Pythagoras theorem,

AD2=AE2+DE2

=( √32 a)

2

+( a4 )2

=34a2+ a

2

16

=13 a2

1616 AD2=13CD2

9. In ΔABC, ∠ABC=90 ° . AD and CE are two medians drawn

from a A and C, respectively. If AC=5 cm and AD=5 √3

2cm. Find CE.

Ans: In Δ ABC , AB2+BC 2=AC2 …(1)

In Δ ABD, AB2+( 1

2BC)

2

=AD 2

AB2+ 14BC 2=AD2

…(2)Subtracting (2) from (1) we get,

AB2+BC 2−(AB2+ 14BC 2)=AC 2−AD2

22 | P a g e

A

D CB E

A

D CB

E

MATHEMATICS 2017-18

AB2+BC2−AB2− 14BC 2=AC2−AD2

34 BC

2=52−( 3√52 )

2

=25−454

=100−454

BC 2=554

×43

Substituting in Eqn (1),

AB2+553

=52

=553

−25

AB2=203

In ΔEBC ,CE2=BE2+BC 2

=( 12AB)

2

+BC 2= 14×20

3+55

3=5

3+55

3=60

3CE2=20

CE=2√5cm10. A lamp is 3.3m above the ground. A boy 110cm tall walks away form the base of this

lamp post at a speed of 0.8m/s. Find the length of the shadow of the boy after 4 seconds.

Ans: Let AB be the lamp post and PQ be the boy, where P is the position of the boy after 4 seconds.AP=Distance moved in 4s at 0.8m/s=4×0 .8=3 .2m .PM is the length of the shadow of the boy.

Let PM=xm

In Δ AMB and ΔPMQ , we have,

∠MAB=∠MPQ=90 °∠ AMB=∠PMQ [Common]∴ ΔAMB|||ΔPMQ [AA – Criterion]

⇒ AP+PMPM

= ABPQ

⇒ 3 . 2+xx

=3 . 31 . 1

⇒3 .2+x=3 x⇒2x=3 .2⇒ x=1. 6mThe length of the shadow of the boy after 4 seconds is 1.6m

23 | P a g e

A

Q

M

B

P

MATHEMATICS 2017-18

7. Coordinate Geometry

1. Show that the points (a ,a ) , (−a ,−a )and (−√3a ,√3a ) are the vertices of an equilateral triangle. Also find the area.

Ans: A (a ,a ) , B (−a ,−a ) and C (−√3a ,√3a )

AB=√ (−a−a )2+(−a−a )2=√4 a2+4a2=2√2a

BC=√(−√3a+a )2+ (√3a−a )2=√3a2−2√3a+a2 3+a2+2√3a+a2=2√2a

AC=√(−√3a−a )2+(√3a−a )2=√6a2+2a2=2√2aSince, AB=BC=AC , ABC is equilateral triangle;

Area, A=√3

4×side2=√3

4×(2√2a )2 √3

4×4×2a2=2√3 a2 sq . units

2. If the three vertices of a rhombus, taken in order are (2 ,−1 ) ,(3,4 ) and (−2,3 ) . Find the fourth vertex.Ans: Since diagonals AC and BD bisect each other, P is a midpoint.A (2 ,−1 ) ,B (3,4 ) and C (−2,3 )

P ( x , y )=( x2+ x2

2,y1+ y2

2 )P ( x , y )=( 2−2

2,−1+3

2 )P ( x , y )=(0,1 )P (0,1 ) is midpoint of BD ,

(0,1 )=(3+x2, 4+ y

2 )3+x

2=0⇒3+x=0⇒ x=−3

4+ y2

=1⇒4+ y=2⇒ y=−2

The fourth vertex is D (−3 ,−2 )

3. If the point P ( x , y ) be equidistant from the points A (a+b ,b−a ) and B (a−b ,a+b ) , then

prove that bx=ay .

Ans: A (a+b , b−a ) ,B (a−b , a+b ) , P ( x , y )AP=BPP ( x , y )AP2=BP2

(a+b−x2 )2+(b−a− y )2=(a−b−x )2+(a+b− y )2

a2+b2+x2−2ab−2bx−2ax+b2+a2+ y2−2ab+2ay−2by¿a2+b2+x2−2ab+2bx−2ax+a2+b2+ y2+2ab−2by−2ay

−2bx−2bx=−2ay−2ay−4bx=−4ay

24 | P a g e

yxP , 4,3B

3,2C

D

1,2 A

MATHEMATICS 2017-18

bx=ay4. If the coordinates of the mid-points of the sides of a triangle are (1 .2 ) , (0,1 ) and (2 ,−1 ) .

Find the coordinates of its vertices.

Ans: D (1,2 )=( x1+x2

2,y1+ y2

2 )⇒ x1+x2=2 ; y1+ y2=4 .. . (1 )

E (0 .1 )=( x2+x3

2,y2+ y3

2 )⇒ x2+x3=0 ; y2+ y3=2 . .. (2 )

F (2 ,−1 )=( x1+x3

2,y1+ y3

2 )⇒ x1+x3=4 ; y1+ y3=−2. . . (3 )

Adding (1), (2) and (3),2 x1+2x2+2x3=6 ;2 y1+2 y2+2 y3=4x1+x2+x3=3 ; y1+ y2+ y3=2Substituting (1), (2) and (3) in the above equation,x1=3; y1=0x2=−1; y2=4

x3=3 ; y3=−2

The coordinates are, A (3,0 ) , B (−1,4 ) ,C (3 ,−2 )

5. A point P divides the line segment joining the points A (3 ,−5 ) and B (−4,8 ) such that APPB

= k1 . If P lies on the line x+ y=0 , then find the value of k .

Ans: P ( x , y ) ,A (3 ,−5 ) , B (−4,8 ) ,m1 :m2=k :1

P ( x , y )=( k×(−4 )+1×3k+1

, 8×k+1×(−5 )k+1 )

⇒ x=−4k+3k+1

; y=8k−5k+1

x+ y=0−4 k+3k+1

+ 8k−5k+1

=0

−4k+3+8k−5=0

∴ k=12

6. If the point C (−1,2 ) divides internally the line segment joining the points A (2,5 ) and

B (x , y ) in the ratio 3 :4 , find the value of x2+ y2

Ans: A (2,5 ) ,B ( x , y ) ,C (−1,2 ) and m1=3 ,m2=4

C (−1,2 )=( 3×x+ ( 4×−1 )7

, 3× y+4×57 )

3x−47

=−1; 3 y+207

=2

3 x−4=−7 ;3 y+20=14x=−1 ; y=−2

25 | P a g e

B C

D

E

A

F

MATHEMATICS 2017-18

∴ x2+ y2=(−1 )2+ (−2 )2=1+4=57. The vertices of Δ ABC are A (4,6 ) .B (1,5 ) and C (7,2 ) . A line is drawn to intersect sides

ABand AC at D and E respectively, such that

ADAB

= AEAC

=14 . Calculate the area of

Δ ADE and compare it with the area of Δ ABCAns: From the figure,ADAB

= AEAC

=14

[Given ]

⇒ ABAD

= ACAE

=4

⇒ AD+DBAD

= AE+ECAE

=4

⇒1+ DBAD

=1+ ECAE

=4

⇒ DBAD

= ECAE

=3

⇒ ADDB

= AEEC

=13

⇒ AD :DB=AE :EC=1:3

Hence D , Edivides AB and AC respectively in the ratio 1 :3 .

Coordinates of D and E are,

( 1+121+3

, 5+181+3 )=(13

4,23

4 ) and

( 7+121+3

, 2+181+3 )=(19

4,5)

respectively.

Hence,ar ( ADE )=15

32sq . units

And ar ( ABC )=15

2 sq . units

ar (ADE )ar (ABC )

=

1532

152

= 116

Hence ar ( ADE ) :ar ( ABC )=1 :16

26 | P a g e

6,4A

5,1B 2,7C

D E3

11

3

MATHEMATICS 2017-18

8. Find the height of the parallelogram ABCD if the vertices are A (1 ,−2 ) ,B (−3,2 ) and

D (−4 ,−3 ) .Ans: Let DM=h , be the height of the parallelogram ABCDwhen AB is taken as base.

ar ( ABD )=12×AB×DM

⇒ar (ABD )=12AB×h

…(1)By distance formula,

AB=√ (2−1 )2+ (3+2 )2=√26 …(2)

Vertices of triangle ABC are A (1 ,−2 ) ,B (2,3 ) , D (−4 ,−3 )By area formula,

ar ( ABD )=12 [x1( y2− y3)+x2 ( y3− y1 )+x3 ( y1− y2 )]

ar ( ABD )=12 sq . units …(3)From (1), (2) and (3),

h=2×12√26

=24√26

×√26√26

=24 √2626

=12√2613

units

27 | P a g e

MATHEMATICS 2017-18

8. Introduction to Trigonometry

1. If secθ=5

4 then prove that

tanθ1+ tan2θ

=sinθsecθ

Ans: secθ= 5

4⇒cos θ= A

H= 4

5By Pythagoras theorem,

H2=O2+A2

52=O2+42

⇒O=3

LHS

= tan θ1+ tan2θ

=

OA

1+(OA )2=

34

1+( 34 )

2=

34

1+ 916

=

34

2516

=34×16

25=12

25

RHS

=sinθsecθ

=

OHHA

=

3554

=35×4

5=12

25

∴LHS=RHS

2. If sin ( A+B )=√3

2 and cos ( A−B )=√3

2 , where 0°< A+B≤90 ° and A>B then find A

and B .

Ans: sin ( A+B )=√3

2⇒ sin (A+B )=sin 60 °⇒ A+B=60 ° .. . (1 )

cos ( A−B )=√32

⇒cos (A−B )=cos30 °⇒ A−B=30 ° .. . (2 )

(1 )+(2 )⇒ 2 A=90 °⇒ A=90 °2

⇒ A=45 °

From (1),A+B=60°45°+B=60°B=60 °−45 °B=15 °

3. Given tan ( A−B )=tan A tanB

1+tan A tanB , evaluate for tan15 ° .

Ans: tan (60 °−45 ° )=tan 60°×tan 45 °

1+ tan 60 ° tan 45 °

tan15 °=√3× 1

√2

1+√3× 1√2

=

√3√2

1+ √3√2

=

√3√2

√2+√3√2

=√3√2

× √2√2+√3

= √3√2+√3

28 | P a g e

MATHEMATICS 2017-18

= √3√2+√3

×√2−√3√2−√3

=3+√6

4. Prove that:

cot (90 °−θ ) . sin (90 °−θ )sin θ

+cot 40 °tan 50 °

−( cos2 20 °+cos2 70 ° )=1

Ans: LHS=

cot (90 °−θ ) . sin (90 °−θ )sin θ

+cot 40 °tan 50 °

−( cos2 20 °+cos270° )

=cot (90°−θ ) .sin (90 °−θ )cos (90 °−θ )

+ tan50 °tan50 °

−(sin270 °+cos2 70° )

=cot (90°−θ ) . tan (90 °−θ )+1−1=1=RHS

5. Evaluate:

cos58 °sin 32 °

+sin 22 °cos 68°

−cos38 ° cosec52°tan 18 ° tan 35 ° tan60 ° tan 72 ° tan55 °

Ans:

cos58 °sin 32 °

+sin 22 °cos 68°

−cos38 ° cosec52°tan 18 ° tan 35 ° tan60 ° tan 72 ° tan55 °

=cos58°cos58°

+cos68 °cos68 °

−cos 38° sec38 °cot 72 ° tan 35°×√3×tan72 ° cot 35°

=1+1− 1√3

=2− 1√3

=2−√33

=6−√33

6. If a2 sec2θ−b2 tan2θ=c2

prove that sin2θ= c

2−a2

c2−b2.

Ans: a2sec2θ−b2 tan2θ=c2

a2

cos2θ−b

2 sin2θcos2θ

=c2

a2−b2 sin2θcos2θ

=c2

a2−b2 sin2θ=c2cos2θa2−b2 sin2θ=c2 (1−sin2θ )a2−b2 sin2θ=c2−sin2θc2 sin2θ−b2 sin2θ=c2−a2

sin2θ (c2−b2)=c2−a2

sin2θ= c2−a2

c2−b2

7. Prove the following:

(i)(1+

1tan2 A ) .(1+

1cot2 A )= 1

sin2 A−sin4 A

Ans: LHS=( tan2 A+1

tan2 A )(cot2A+1cot2A )

29 | P a g e

MATHEMATICS 2017-18

=(sec2 Atan2A )(cosec2 A

cot2A )=(1

cos2Asin2 A

cos2 A)(

1sin2A

cos2 Asin2A

)=( 1

sin2 A )( 1cos2 A )= 1

sin2 A (1−sin2 A )= 1

sin 2A−sin4 A

(ii) √ 1+cosθ1−cosθ

=cosec θ+cot θ

Ans: LHS=√ 1+cosθ

1−cosθ=√ 1+cosθ

1−cosθ×

1−cosθ1+cosθ

=√ (1−cosθ )2

1−cos2θ

√ (1−cosθ )2

sin2θ=1−cosθ

sin θ= 1

sin θ−cosθ

sin θ=cosec θ−cotθ

=RHS

(iii) sec4θ−sec2θ=tan4θ+ tan2θ

Ans: LHS =sec4θ−sec2θ

=( sec2θ )2−sec2θ=(1+ tan2θ )2−( 1+ tan2θ )=1+ tan4 θ+2 tan2θ−1− tan2θ=tan4 θ+ tan2θ= RHS

(iv)

tan3θ−1tanθ−1

=sec2θ+ tanθ

Ans: LHS =tan3θ−1

tan θ−1=

( tanθ−1 ) (tan2θ+tan θ+1 )( tan θ−1 )

=( tan2θ+1+ tan θ )=(sec2θ+ tanθ ) = RHS

(v) (1+cot θ−cosecθ ) (1+ tanθ+secθ )=2

Ans: LHS = (1+cot θ−cosecθ ) (1+ tanθ+secθ )

=(1+cosθsin θ

− 1sin θ )(1+sinθ

cosθ+ 1

sec θ )=(sin θ+cos θ−1

sin θ )(cos θ+sin θ+1cos θ )

=(sinθ+cos θ )2−12

sinθ=sin2θ+sin2θ+2 sin θ cosθ−1

sin θ cosθ=1+2sin θcosθ−1

sin θ cosθ

=2 sin θ cosθsinθ cos θ

=2= RHS

(vi) sin6 A+cos6=1−3 sin2 A cos2A

Ans: LHS =sin6 A+cos6 A

=( sin2 A )3+(cos2A )3

=( sin2 A+cos2 A ) (sin4 A−2 sin2A cos2 A+cos4A )

30 | P a g e

MATHEMATICS 2017-18

=1×(( sin2A+cos2)2−2sin2 A cos2A−sin2A cos2 A )=1−3 sin2 A cos2A = RHS

(vii)

tanθ+secθ−1tanθ−secθ+1

=1+sin θcos θ

Ans:

LHS=tan θ+secθ−1tan θ−secθ+1

=

sin θcosθ

+ 1cosθ

−1

sin θcosθ

− 1cosθ

+1

=sinθ+1−cosθsinθ−1+cosθ

==sin θ−cosθ+1sin θ+cosθ−1

×sin θ+cosθ+1sin θ+cosθ+1

=(sin θ+1+cosθ ) (sinθ+1−cosθ )

(sin θ+cosθ )2−12=

(sin θ+1 )2−cos2θsin2θ+cos2θ−2sin θcosθ−1

=sin2θ+1+2 sin θ−1+sin2θ2 sin θ cosθ

=2sin2θ+sin θ2sin θ cosθ =

2sin θ (1+sinθ )2sin θcosθ

=1+sin θcosθ = RHS

(viii)

sin A+cos Asin A−cos A

+sin A−cos Asin A+cos A

= 21−2cos2 A

= 22sin2 A−1

Ans: LHS=sin A+cos A

sin A−cosA+sin A−cos A

sin A+cosA

=(sin A+cos A )2+(sin A−cos A )2

sin2A−cos2 A=sin2 A+cos2 A+2sin A cos A+sin2 A+cos2A−2sin A cos A

(1−cos2 A )−cos2A

= 2(1−cos2 A )−cos2A

= 21−2cos2 A

And

= 2sin2A−( 1−sin2 A )

= 22sin2 A−1

8. If 7sin2 A+3cos2 A=4 , show that tan A= 1

√3

Ans: 7 sin2 A+3cos2 A=4

Divide each term by cos2A

7 sin2 Acos2A

+3 cos2Acos2A

= 4cos2 A

7 tan2A+3=4 sec2 A7 tan2A−4 sec2A+3=0

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7 tan2A−4 (1+ tan2A )+3=07 tan2A−4−4 tan2 A+3=03 tan2A−1=0

tan2A=13

tan A= 1√3

9. If cosθ+sinθ=√2 cosθ , show that cosθ−sin θ=√2 sin θ

Ans: cosθ+sinθ=√2cosθ

Squaring both the sides,

(cosθ+sinθ )2=(√2cosθ )2

cos2θ+sin2θ+2sin θcosθ=2cos2θ2 sin θ cosθ=2cos2θ−cos2θ−sin2θ2 sinθ cosθ=cos2θ−sin2θ2 sin θ cosθ= (cosθ+cosθ ) (cosθ−sin θ )2sin θ cosθ=√2cosθ (cosθ−sinθ )2sin θ cosθ√2cosθ

=cosθ−sin θ

cosθ−sin θ=√2 sin θ10. If sin ( A+B )=1 and cos ( A−B )=1 , 0 °< A+B<90 ° , A≥B find A and B .

Ans: sin ( A+B )=1

sin ( A+B )=sin 90°A+B=90 ° …(1)

cos ( A−B )=1cos ( A−B )=cos 0°A−B=0 ° …(2)From (1) and (2),A=B=45 °

11. Find acute angles A and B, if sin ( A+2B )=√3

2 and cos ( A+4 B )=0 , A>B .

Ans: sin ( A+2B )=√3

2⇒sin ( A+2 B )=sin 60 °⇒ A+2B=60 ° .. . (1 )

cos ( A+4 B )=0°⇒cos (A+4 B )=cos90 °⇒ A+4B=90 ° .. . (2 )From (2)-(1), 2B=30 °B=15 °∴ A=30 °

12. If tan A+sin A=m and tan A−sin A=n , prove that (m2−n2 )2=16mn .

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Ans: LHS=(m2−n2 )2

=( (tan A+sin A )2−( tan A−sin A )2)2

=( (tan A+sin A+ tan A−sin A ) ( tan A+sin A−tan A+sin A ) )2

= (2 tan A×2sin A )2= (4 tan A sin A )2

=16 tan2 A sin2 ARHS=16mn=16 ( tan A+sin A ) (tan A−sin A )=16 tan2 A−sin2 A

=16×(sin2 Acos2A

−sin2A )=16 sin2A ( 1

cos2 A−1)

=16 sin2A ( 1−cos2Acos2 A )

=16sin2A (sin2Acos2 A )

=16×sin2Atan2A

×sin2 A

=16 tan2 A sin2 ALHS=RHS

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9. Some Applications of Trigonometry1. Two men on either side of the cliff 80m high observes the angles of elevation of the

top of the cliff to be 30 °and 60 ° respectively. Find the distance between the two men.

Ans: tan60 °=80

x⇒√3=80

x⇒ x=80

√3m

tan30 °=80y

⇒ 1√3

=80y

⇒ y=80√3m

x+ y=80√3

+80 √3

=80( 1√3

+√3)=80( 1+3

√3 )=320

√3

=320√33 m

The men are standing 184.7m apart.2. A man standing on a deck of a ship, which is 10m above the water level, observe the

angle of elevation of the top of a hill as 60 ° and the angle of depression of the base of

the hill is 30 ° . Calculate the distance of the hill from the ship and height of the hill.

Ans: In Δ ADC ,

tanθ= ADDC⇒

tan30°=10DC

⇒ 1√3

=10DC

⇒DC=10√3m

InΔ ABE ,

tan60 °= BEAE

⇒√3= BE10√3

⇒BE=10√3×√3⇒BE=30m

Distance of the hill is 17.32m and height of the hill is 30m.3. Two vertical poles of different heights are standing 20m

away from each other on the same level of the ground.

The angle of elevation of one pole at the foot of the other is 60 ° and the angle of

elevation of the top of the other pole at the foot of the first is30 ° . Find the difference between the heights of the two towers.

Ans: In Δ ABC ,

tan30 °= ABBC

⇒ 1√3

= AB20

⇒ AB=20√3m

In ΔBDC ,

tan60 °= DCBC

⇒√3=DC20

⇒DC=20√3

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DB Cyx

m80

60 30

A

D

B

C

D30

E

A

D

B C60 30

MATHEMATICS 2017-18

The difference between the heights=DC−AB

=20 √3−20√3

=17 . 32m4. A man on a cliff observes a boat, at an angle of depression30 ° , which is sailing

towards the shore to the immediately beneath him. Six minutes later, the angle of

depression of the boat is found to be 60 ° . Assuming that the boat sails at a uniform speed, determine how much more time it will take to reach the shore.

Ans: tan30 °= x

AB⇒ 1

√3= xAB

⇒ AB=√3x .. . (1 )

tan60 °= x+ yAB

⇒√3= x− yAB

⇒=√3 AB=x+ y .. . (2 )

Substituting (1) in (2),

√3×√3 x=x+ y3 x=x+ y2 x= y . . . (3 )Boat takes 6 minutes to travel a distance of y units.As the speed is same, so the time taken by boat to travel x units is half the time taken to

travel y units.So the time taken to travel the distance to travel x units is minutes.

So total time taken by boat =3+6=9minutes.

5. Angle of elevation of a jet plane from a point A on the ground is 60 ° . After a flight of

15 seconds, the angle of elevation changes to 30 ° . If the jet plane is flying at constant

height of 1500√3m. find the speed of the plane.

Ans: In Δ APB ,

tan60 °= BPAB

⇒√3=1500√3AB

⇒ AB=1500m

In Δ ACQ,

tan30°= CQAC

⇒ 1√3

=1500√3AC

⇒ AC=1500√3×√3⇒ AC=4500m

BC=4500−1500=3000mThe plane travels 3000m in 15 seconds,

Hence speed of the plane=3000

15=200m /s

.6. A tower stands vertically on a bank of a canal. From

a point on the other bank, directly opposite to the tower, the angle of elevation of the top of the tower

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A

D30

A

P

B C

60

30

Q

m31500

MATHEMATICS 2017-18

is 60 ° . From another point 20m away from this point on the line joining this point to

the foot of the tower, the angle of elevation of the top of the tower is 30 ° . Find the height of the tower and the width of the canal.

Ans: In Δ ABD ,

tan30°= OA⇒

1√3

= h20+x ⇒

h=20+x√3

. . . (1 )

In Δ ABC ,

tan60 °= hx⇒

√31

=hx⇒√3 x=h . .. (2 )

From (1) and (2),20+x√3

=√3 x⇒20+ x=3 x⇒3 x−x=20⇒ x=202

⇒ x=10m

h=√3×10=10√3=10×1 . 732=17 .32mThe width of canal is 10m and height of the tower is 17.32m

7. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-

staff of height h . At a point on the plane, the angles of elevation of the bottom of the

bottom and the top of the flag staff are α and β respectively. Prove that height of the

tower is

h tan αtan β− tan α .

Ans: In ΔOAB ,

tanα= ABOA

tanα= yx

x= ytanα

x= y cot α …(1)

In ΔOAC ,

tan β= y+hx

x= y+htan β

x=( y+h )cot β …(2)From (1) and (2),

y cot α=( y+h ) cot β( y cot α− y cot β )=hcot βy (cot α−cot β )=hcot β

y= hcot βcot α−cot β

=

htan β

1tanα

− 1tan β

= h tan αtan β− tan α

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h

C

B

O

y

x

MATHEMATICS 2017-18

Hence proved.

8. An aeroplane when flying at a height of 4000m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes

from the same point on the ground are 60 ° and 45 ° respectively. Find the vertical distance between the aeroplane at that instant.

Ans: In Δ AOPand Δ AOQ ,

tan60 °= OPOA and

tan 45°=OQOA

√3=4000OA ;

1=OQOA

OA=4000√3 ; OQ=OA

OQ=4000√3 m

Vertical distance PQ between the aeroplanes is given by,PQ=OP−OQ

PQ=4000−4000√3

PQ=1690 .53 m

9. The angle of elevation of a cloud from a point 60m above the a lake is 30 ° and the

angle of depression of the reflection of cloud in the lake is 60 ° . Find the height of the cloud.Ans: Let AB be the surface, C be the position of the cloud and C’ be its reflection.

In ΔCMP ,

tan30 °= CMPM

1√3

= hPM

PM=√3h …(1)

In ΔPMC ' ,

tan60 °= C ' MPM

√3=C ' B+BMPM

√3=h+6+60PM

PM=h+120√3 …(2)

From (1) and (2),

37 | P a g e

O

m4000

P

Q

A

6045

A

C

M

B

P

'C

60

30

m60

60h

m60

h

MATHEMATICS 2017-18

√3h=h+120√3

h=60m

Hence, CB=CM +MB=h+60=60+60=120mHeight of the cloud form the surface of the lake is 120 m.

10.A boy whose eyelevel is 1.3m from the ground, spots a balloon moving with the wind in a horizontal line at some height from the ground. The angle of elevation of the

balloon from the eyes of the boy at an instant is 60 ° . After 2 seconds, the angle of

elevation reduces to 30 ° . If the speed of the wind is 29√3m/ s then find the height of the balloon from the ground.

Ans: Let AB be the position of the boy, AX is a horizontal ground. Let C and D be the position of the balloon.

Distance covered in 2 seconds =29√3×2=58m∴CD=58√3 mLet BP =xm.

BN=(BP+PN )=(BP+CD )=(x+58√3 )mAnd DN=CP

From the right ΔBND ,DNBN

=tan 30°

DNBN

= 1√3

DN( x+58√3 )

= 1√3

DN=( x+58 √3 )

√3 …(1)

In ΔBPCCPBP

= tan 60 °

CPx

=√3 CP=x √3 m …(2)

Now,

DN=CP⇒

(x+58√3 )√3

=x√3

⇒ (3 x−x )=58√3⇒2 x=58√3⇒ x=29√3 mFrom (2)

CP=(29√3×√3 )=87 m∴Height of the balloon from the ground=CL=CP+PL=CP+AB=87+1 . 3=88 .3mHeight of the balloon form the ground is 88.3m.

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A

B

C D

L M X

6030x

358

MATHEMATICS 2017-18

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10. Circles1. In the given figure, OP is equal to diameter of the circle. Prove that ABP is an

equilateral triangle

Ans: OA=OB=r ,OP=2 rΔ AOP and ΔBOP ,

AP=AP [Tangent from an externat point ]OP=OP [Common side ]∠OAP=∠OBP [ 90 ° ]Δ AOP≃ΔBOP [RHS ]∠ APO=∠BPO [ CPCT ]In Δ AOP ,∠ APO=θ

sin θ=OAOP

sin θ= r2 r

= 12

sin θ=sin 30°θ=30°∠ APB=2∠ APO=2×30 °=60 °

In Δ ABP , AP=BP∠PAB=∠ ABP=x∠P+∠A+∠B=180°

60°+ x+ x=180 °x=60 °∠P=∠A=∠B=60°Δ APB is a equilateral triangle.

2. If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A

as shown in the figure. Prove that ∠BAT =∠ACB .

Ans: CA⊥AT [Radius is perpendicular to tangent]⇒∠CAB=90°∠BAC+∠BAT=90 ° . .. (1 )Since AC is a diameter, ∠ ABC=90 ° [Angle in the semicircle]

In Δ ABC ,∠ ACB+∠ABC +∠BAC=180 °∠ ACB+∠BAC=180 °−∠ ABC=180 °−90 °=90 °∠ ACB+∠BAC=90 ° . . . (2 )From (1) and (2),∠BAC+∠BAT =∠ ACB+∠BAC∴∠BAT =∠ ACBHence proved.

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O P

A

B

O

A T

B

C

MATHEMATICS 2017-18

3. In the given figure, tangent PQ and PR are drawn from an external point P to a circle

with centre O, such that ∠RPQ=30 ° . A chord RS is drawn parallel to the tangent PQ.

Find ∠RQS .

Ans:PQ=PR [Tangents drawn from an external point]

And PQR is an isosceles triangle

⇒∠RQP=∠QRP∠RPQ+∠QRP+∠RQP=180° [Angle sum property of a triangle]

∠RQP=75°∠RQP=∠QRP=75°∠RQP=∠RSQ=75 °[ Angles in alternate Segment Theorem states that angle between chord and tangent is  equal to the angle in the alternate segment]

It is given that RS||PQ

∴ ∠RQP=∠SRQ=75° [Alternate angles]∠RSQ=∠ SRQ=75 °∴QRS is also an isosceles triangle∠RSQ+∠ SRQ+∠RQS=180 ° [Angle sum property of a triangle]150 °+∠RQS=180 °∠RQS=30 °

4. In figure, AB is a chord of length 16 cm of a circle of

radius 10 cm. The tangents at A and B intersect at P . Find the length of the PA .

Ans: AB=16cm , AL=BL=8cm

In ΔOLB , we have

OB2=OL2+LB2

102=OL2+82

OL=6 cm

Le PL=x and PB= y . Then OP=x+6 cm

In Δ' s PLB and OBP

PB2=PL 2+BL2 and OP2=OB2+PB2

⇒ y2=PL2+64 and ( x+6 )2=100+ y2

By simplification, x=32

3 cm and y=40

3 cm

Hence, PA=PB=40

3 cm

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O

P

R

Q

S

OL

B

A

P

MATHEMATICS 2017-18

5. If a , b , care the sides of a right triangle, where c is the hypotenuse, then prove that the radius r of the circle which touches the side of the triangle is given by

r= a+b−c2

Ans: Let the triangle be ABC , where AB=c ,BC=a , and AC=b .

Let D , E ,F be the points where the circle touches the triangle ABC .By Pythagorean property,

AB2=AC2+BC2

⇒ c2=a2+b2

Since tangents from an external points are equal,AE=AF , CD=CE and BD=BF

We observe that CD=OE and CE=OD⇒CD=r and CE=rHence,⇒ AF=AE and BD=BF⇒ AF=AC−CE and BD=BC−CD⇒ AF=b−r and BD=a−r

⇒ AF+BF=(b−r )+ (a−r )⇒ AB=a+b−2 r⇒ c=a+b−2 r

∴r=a+b−c2

6. The radii of two concentric circles are 13cm and 8cm. AB is the diameter of the bigger circle. BF is the tangent of the smaller circle touching it at D. Find the length of AD.

Ans: From the figure, AB is a diameter and ∠ AEB is a angle in the semicircle.

⇒∠ AEB=90 ° and OD⊥BE ,

And BD=DE

⇒OD||AEIn Δ AEB , O and D are the midpoints, by midpoint theorem,

OD=12AE

⇒ AE=2×8=16 cm

In ΔODB ,OB2=OD2+BD2

132=82+BD 2

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O

A

B D

a

b

c

A

F

C

B

E

Drr

rO

MATHEMATICS 2017-18

BD=√105 cmIn Δ AED ,

AD2=AE2+ED2

AD2=162+(√105 )2=256+105=361AD=19 cm

7. In figure two tangents AB and AC are drawn to a circle

with centre O such that ∠BAC=120 ° . Prove that OA=2 AB .

Ans: We know that Δ AOB≃Δ AOC [Prove]

∠BAO=∠CAO=120 °2

=60°

In Δ AOB,∠ ABO=90° [Radius is perpendicular to tangent]

cos60 °= ABOA

12= ABOA

∴OA=2 AB8. In the figure, O is the centre of the circle and

BCD is tangent to it at C. Prove that ∠BAC+∠ ACD=90 °Ans: In ΔOACOA=OC [Radius of the circle]⇒∠OAC =∠OCACD is a tangent to the given circle,∠OCD=90 ° [Radius is perpendiculars to tangent]⇒∠OCA +∠ACD=90 °⇒∠OAC +∠ACD=90°⇒∠BAC +∠ACD=90 ° [∵∠OAC and ∠BAC are same angles ]

9. In the figure ABC is right triangle at A, with AB=6cm and AC=8cm. A circle with centre O has been inscribed inside the triangle. Calculate the value of r , the radius of the inscribed circle.

Ans: In Δ ABC ,∠ A=90 °

BC 2=CA2+AB2

BC=82+62

BC=√100BC=10 cm

Since A ,B and C are the external point for the circle,

AD=AF=x , BD=BE= y and CF=CE=z

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O

A

B

120

C

A B

O

D

E

F

C

O

A

BD

E

MATHEMATICS 2017-18

AB+BC+AC=6+10+8x+z+x+ y+ y+z=24x+ y+z=12 …(1)

Since y+z=BC=10cmx+10=12 [From (1)]x=12−10∴ x=2 cm

Since∠ AFO=∠ADO=90°∴ AFOD is a square,Therefore radius is 2cm.

10. Two concentric circles if radii a and b (a>b ) are given. Find the length of the chord of the larger circle in which touches the smaller circle.

Ans: Let O be the common centre of the circles and ABbe the tangent to the smaller circle

of radius b and chord of the circle bigger circle of radius a , which touches the smaller

circle at C .

Join OA and OC . Then, OA=a and OC=b .

Now, OC⊥ AB and OC bisects AB .

In Δ ACO ,

OA 2=OC2+AC2 [By Pythagoras theorem]

AC=√OA 2−OC2=√a2−b2

But, AB=2 AC

Hence, AB=2√a2−b2

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A

Ba

Ob

MATHEMATICS 2017-18

11. Construction1. Construct an isosceles triangle whose base is 8cm and altitude 4cm and then another

triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle.2. Draw a right triangle in which sides are of length 8cm and 6cm. Then construct another

triangle whose sides are 3:4 times the corresponding sides of the first triangle.

3. Draw a triangle ABC in which AC=AB=4.5cm and ∠ A=90 ° . Draw a triangle similar to ABC with its sides equal to 5:4 times the original triangle.

4. Draw a pair of tangents to a circle of radius 4.5cm, which are inclined to each other at an

angle of 45 ° .

5. Draw a right triangle ABC in which AB=6cm, BC=8cm and ∠B=90 ° . Draw BD perpendicular from B on AC and draw a circle passing through the points B, C and D. Construct tangents from A to this circle.

6. Draw two concentric circles of radii 3cm and 5cm. Construct a tangent to the smaller circle from a point on the larger circle. Also measure its length.

7. Draw a triangle with a side 7cm, base angles 45 ° and 105 ° . The construct a triangle whose sides are 4:3 times their corresponding sides of the triangle.

8. Construct a triangle ABC in which AB=5cm. ∠B=60 ° altitude CD = 3cm. Construct a triangle AQR similar to triangle ABC is 1.5 of that of the corresponding side of triangle ABC.

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12. Ares Related To Circles1. A circle with radius 2 is placed against a right angle . Another

smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

Ans: Diagonal of square is2√2 (Pythagoras Theorem)

Also Diagonal of square is 2+r+√2 r

Equating, r=6−4√22. Find the area of the shaded design given in the figure, where ABCD

is square of side 10cm and semicircles are drawn with each side of the square as diameter. Ans: Area of I + Area of III

= Area of ABCD – Areas of two semicircles of each of radius 5 cm

(10×10−2×12×π×52)= (100−3 .14×25 )=100−78 .5=21. 5cm2

Similarly, Area of II + Area of IV =21 .5c2

So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)

=100−2×21 .5=100−43=57cm2

3. Two circles touch internally. The sum of their areas is 116 cm2and distance between their centres is 6cm. find the radii of the circles.Ans: Let a circle with center O and radius R.

let another circle inside the first circle  with center o' and radius r . Area of 1st circle + area of 2nd circle = 116 cm²π⇒ πR² + πr² = 116π ⇒ π(R² + r²)  = 116π⇒ R² + r² =116 --------------------(i)Now,Distance between the centers of circles = 6 cmi.e, R - r = 6⇒  R = r + 6  -------------------(ii)From Eqn (i) & (ii),(r + 6)² + r² = 116⇒ r² + 12r +36 + r² =116⇒ 2r² +12r +36 -116 = 0⇒ 2r² +12r - 80 = 0⇒ r² +6r - 40 = 0⇒ r² +10r - 4r - 40 = 0⇒ r(r + 10) - 4(r + 10) = 0⇒ (r + 10)(r - 4) = 0 hence  r = 4 cm

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MATHEMATICS 2017-18

r ≠ -10 cm        {∵ length can't be -ve}Therefore radii of the circles are r = 4 cm ,R = 4 + 6 = 10 cm.

4. A wire is bent in the form of a square, encloses a area of 121 sq.cm. If the same wire is bent in the form of a circle, find the area of the circle.Ans: Area of square = 121 cm2

So, side of square = 11 cmAnd perimeter of the square = 4 x 11 cm = 44 cmthus, the length of the wire is 44 cm.Now, the wire is bent to form a circle. Circumeference of the circle = 44 cm2  r = 44πr = 44 x 7 / 2 x 22 = 7 cmThus, radius of the circle is 7 cmTherefore Area of Circle = πr2= x π 72 = 154cm2.

5. In the given figure, diameter AB is 12cm long. AB is trisected at points P and Q. Find the area of the shaded region. Ans: AP=PQ=QB

AP+PQ+QB=12 cm (Trisected)AP=PQ=QB=4 cm (Trisected)3AP=3PQ=3QB=12 cmNow,AQ=2AP=8 cm &PB=2QB=8 cmR=AQ/2=4 cmSimilarly, r=AP/2=2 cmTherefore,

Area of shaded region= 2×(12 R²-π 1

2 r²)=37.7 cm²π

6. Two circle touch externally. The sum of their areas is 130π sq.cm. and the distance between their centres is 14cm. Find the radii of the circles.Ans: Since the two circles touch each other externally so, the distance between their

centers = sum of their radii. Let the center of one circle be x cm. Then the radius of the other circles is (14 - x) cm. Therefore, Sum of their areas = xπ 2 + (a) Since the two circles touch each other externally so, the distance between their centers = sum of their radii. Let the center of one circle be x cm. Then the radius of the other circles is (14 - x) cm. Therefore, Sum of their areas = xπ 2 + (14 - x)π 2 = 130 (given) πx2 + 196 - 28 x + x2 = 130; x2 - 14 x + 33 = 0 (x - 3) (x - 11) = 0 x = 3 or 11 Therefore, 14 - x = 11 or 3 Hence, the radii of the two circles are 11 cm and 3 cm respectively.

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A PQ B

MATHEMATICS 2017-18

7. An elastic belt is placed round the rim of a pulley of radius 5cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P., 10cm from O, Find the length of the belt that is in contact with the rim of pulley. Also find the shaded area.Ans: Length of the tangents drawn from an external point to a

circle are equal. OA = 5 cmOP = 10 cmTriangle OAP is a right angled triangle. Therefore, OP2 = OA2 + AP2

⇒ 102 = 52 + AP2

⇒ 100 = 25 + AP2

⇒ AP2 = 75

⇒ AP = √75=5√3 cm

∴ Length of the belt that is in contact with the rim of the pulley =5√3 cm

Area of the right angled triangle OAP =

12 x base x height

=12×5√3×5=25

2 √3cm2

Area of the right angled triangle OBP

252 √3

cm2

Total area of the triangles = 25√3cm2.

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Ocm5

B

A

Pcm5

Q

MATHEMATICS 2017-18

13. Surface Areas and Volumes1. The volumes of two right circular cones are in the ratio 3:5 and their heights are in

the ratio 5:3. Find the ratio between their radii.

Ans: V 1 :V 2=3 :5 , h1 : h2=5 :3

V 1

V 2=

13πr1 h2

13πr 2h2

⇒ 35=r1

r2×5

3⇒r1

r2= 9

25

The ratio of the radii is 9 :25 .2. A sphere of diameter 12cm is dropped in to a right cylindrical vessel, partly filled

with water. If the sphere is completely submerged in water, the water level in the

cylindrical vessel rises by 3 5

9 cm. Find the diameter of the cylinder.

Ans: r1=

122

=6cm,h=3 5

9=32

9 ,r2=?

Volume of the sphere = Volume of water raised in the cylindrical vessel43πr

13=πr22h

4×63=r22×

329

r22=81

r2=9cm

Diameter of the cylinder is 18cm .3. Water is flowing at the rate of 3km/hr, through a circular pipe of 20cm internal

diameter into a circular cistern of diameter 10cm and depth 2m. in how much time will the cistern be filled?

Ans: Length of the water column in one hour =3 km=3000m

Length of water column in 1min=3000

60=50m

Radius of the pipe =20

2=10=0 .1m

Volume of water column in 1min=πr2h=π×0 .1×0 .1×50

=12πm3

Volume of cistern =π×52×2=50 πm3

Time taken to fill the water=Volume of cistern

Volume of water column in one min

=50 π12π

=50×2=100 min

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4. A hemispherical tank full of water emptied by a pipe at the rate of 3 4

7 litres per second. How much time will it take to make the tank half empty, if the tank is 3m diameter?

Ans: Radius of hemispherical tank=3

2m

Volume of tank == 2

3πr3 =2

3×22

7×3

2×3

2×3

2=99

14m3=99000

14l

Volume of half the tank =99000

14× 1

2=99000

28 l

Water emptied in 1 sec =25

7l

Required time =99000

28×25

7 =990 sec5. A cone of radius 4cm is divided into two parts by drawing a plane through the

midpoint of its axis and parallel to its base. Compare the volumes of two parts.

Ans: let the height of the cone be H and the radius be R . This cone is divided into two parts through the mid-point of its axis.

Therefore AQ=1

2AP

Since QD||PC

∴ triangle AQD is similar to the triangle APCBy the condition of similarity

QDPC

= AQAP

= AQ2 AQ

QAR

=12⇒QD=R

2

Volume of the cone,ABC=1

3πR2H

volume of the frustum = volume of the cone ABC - volume of the cone AED

=13πR2H−1

3π (R2 )

2

(H2 )=1

3πR2H− 1

3π R

2H8

=

13πR2H (1−1

8 )Volume of the cone,AED=

78×1

3πR2H

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∴ Volume of the part taken out/ volume of the remaining part of the cone

=

18×1

3πR2H

78×1

3×πR2H =1

7

6. An agricultural field is in the form of a rectangle of length 20m and width 14m. A 10m deep well of diameter 7m is dug in a corner of the field and the earth taken out of the well is spread evenly over the remaining part of the field. Find the rise in its level.

Ans: r=3. 5m ,h=10m, l=20m ,b=14m, H=?Volume of the well = Volume of the rectangular field around the well

πr2h=lbH

227

×3.5×3. 5×10=20×14×H

H=1 . 375m7. A hollow sphere of internal and external radii 4cm and 8cm respectively melted into

a cone of base radii 8cm. Calculate the height of the cone.

Ans: R=8cm, r1=4 cm ,r2=8cm ,h=?

Volume of the hollow sphere = Volume of the cone13π (R3−r

13)=13πr

23 h

4 (83−43 )=82h4 (512−64 )=82 h

h=4×4488×8

h=28cmHeight of the cone is 28cm

8. Water is flowing at the rate of 2.52km/h through a cylindrical pipe into a cylindrical tank; the radius of the base is 40cm. If the increase in the water level of water in the tank, in half an hour is 3.15m, find the internal diameter of the pipe.

Ans: Increase in the level of water in half an hour =3 .15m=315 cm

Radius of water tank, r=40cm

Volume of the water tank after 30minutes=πr2h=π×40×40×315=50400 π cm3

Rate of flow of water=2 .52 km/hr

Length of water column in 30mins = 2 .52×30

60=1 . 26km=126000 cm

Let the internal diameter of the pipe be d .

Volume of the water that flows through the pipe in 30min−π×d

2×126000

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Vol. of water that flow though pipe in 30mins = Vol. of water in tank after 30mins

π×d2×126000=504000 π

d=4cmThus the internal diameter of the pipe is 4 cm .

9. A hemispherical depression is cut from one face of the cuboidal wooden block such

that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Ans: r= l

2,side=l

Surface area of the remaining wooden block

=TSA of wooden block + CSA of hemisphere - πr2

=6l2+2π ( l2 )2

−π ( l2 )2

=6l2+2π l4

2−π l

2

4

=6l2+π l2

4

= l2

4(24+π )

10. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9cm.

Ans: h=9cm ,r=9

2

Volume of the cone=1

3πr2h

=13×22

7×9

2×9

2×9

=33×8114×2

=95. 46cm3

11. A building is in the form of a cylinder surmounted by a hemispherical vaulted dome

and contains 41 19

21m3

of air. If the internal diameter of the dome is equal to the total height above the floor, find the height of the building.

Ans: Let R be the radius, and h be the height of the cylindrical portion.

Height of the building, R+h=2R=d internal diameter.

This gives, h=R .Volume = Volume of cylindrical portion + Volume of hemispherical portion

=πR2h+( 4 πR3 /3 )

2=880

21

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MATHEMATICS 2017-18

=227

×(R2h+ 2R3

3 )=88021

=R2(h+ 2R3 )=440

3

R2(R+ 2R3 )=440

3R3=88R=4 .45mTotal Height of the building, =R+h=2 R=8.9m

12. A container, open from the top, made up of a metal sheet is in the form of a frustum of a cone of height 16cm with radii of its lower and upper ends as 8cm and 20cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs. 15 per litre and the cost of metal sheet used, if the cost Rs. 5 per 100cm2.

Ans: h=16 cm ,r1=20cm , r2=8cm

Volume of the container= 1

3πh (r12+r22+r1r2)

=13π×16 (202+82+20×8 )

=10445 .76 cm3

=10445 . 761000

=10 .44576 litreCost of 1litre milk=Rs. 20

Cost of 10 .44576 litre=20×10 .44576

Rs . 208 . 91=Rs. 20913. A bucket is in the form of a frustum of a cone and holds 28.490 litres of water. The

radii of the top and bottom are 28cm and 21cm respectively. Find the height of the bucket.

Ans: Volume of the bucket=1

3πh (r12+r22+r1r2)

28 . 490=13×22

7×h (282+212+28×21 )

h=28490 cm

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MATHEMATICS 2017-18

14. Statistics

1. The mean of the following data is 50. Find the values of x and y if the total frequency is 25

C – I 0-20 20-40 40-60 60-80 80-100 TotalF 17 f 1 32 f 2 19 120

Ans:

Class( f i ) Frequency Mid-Valueui=

x i−Ah

f iui

0-2020-4040-6060-80

80-100

17f i32f 2

19

1030507090

-2-1012

-34

-f 1

0f 2

38

∑ f i=68+ f 1+ f 2 ∑ f iui=4− f 1+ f 2

68+ f 1+ f 2=120f 1+ f 2=52 …(1)

Mean, x=A+

∑ f iui∑ f i

×h

50=50+4−f 1+f 2

120×20

f 1−f 2=4 …(2)

On solving (1) and (2) we get, f 1=28 and f 2=24

2. The age wise participation of students in the annual function of a school is shown in the following distribution.Ages 5-7 7-9 9-11 11-13 13-15 15-17 17-19No.of students

x 15 18 30 50 48 x

Find the missing frequencies when the sum of frequencies is 181. Also, find the mode of the data.Ans:

Ages No. of students ∑ f i=2 x+161181=2 x+161

x=181−1612

=202

=10

Use formula and simplify.

5-79-11

11-1313-1515-1717-19

x

1518305048x

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MATHEMATICS 2017-18

3. Recast the following cumulative frequency table in the form of an ordinary frequency table and find the median.No. of days

L.T. 5 L.T. 10

L.T. 15

L.T. 20

L.T. 25

L.T. 30

L.T. 35

L.T. 40

L.T. 45

No. of students

29 224 465 582 634 644 650 653 655

Ans: No. of days

No. of students Frequency Less Than type cumulative frequency

L.T. 5L.T. 10L.T. 15L.T. 20L.T. 25L.T. 30L.T. 35L.T. 40L.T. 45

0-55-10

10-1515-2020-2525-3030-3535-4040-45

291952411175210632

29224465582634644650653655

∑ f i=655 ,

Since

6552 belongs to cu.fr 465 of the class interval 10-15, therefore 10-15 is the

median class.l=10h=5c f=224f=241

Median=l+( n2−c f

f )×h=10+( 6552

−465

241 )×5=121. 147

4. If median of the following frequency distribution is 46, find the missing frequencies.Variable

10-20 20-30 30-40 40-50 50-60 60-70 70-80 Total

F 12 30 f 1 65 f 2 25 18 229

Ans: f 1+ f 2+150=∑ f if 1+ f 2=229−150f 1+ f 2=79Since median is 46, median class is 40-50,

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MATHEMATICS 2017-18

l=40 , h=10 , f=65 and C f=12+30+ f 1=42+ f 1 , ∑ f i=229

Median=l+(∑ f i

2−C f

f )×h46=40+( 229

2−(42+ f 1 )

65 )×10

f 1=33 . 5f 1+ f 2=79f 2=79−33 .5f 2=44 . 5

5. The following table gives the production yield per hectare of wheat of 100 farms of a village:

Production 50-55 55-60 60-65 65-70 70-75 75-80No. of

farms2 8 12 24 38 16

Draw less than and more than ogive, and hence find the median.

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MATHEMATICS 2017-18

15. Probability1. All black faced cards are removed from a pack of 52 cards. The remaining cards are well

shuffled and the na card is drawn at random. Find the probability of of getting a,a. Face card c. Red cardb. Black card d. King e. Face card

2. Two different dice are tossed together. Find the probability that,a. The number of each die is evenb. The sum of the numbers appearing on the two sides dice is 5

3. What is the probability that a leap year has 53 Sundays.Ans: 1 year = 365 days. A leap year has 366 days

A year has 52 weeks. Hence there will be 52 Sundays for sure.52 weeks =  days366 – 364 =2 daysIn a leap year there will be 52 Sundays and 2 days will be left.These 2 days can be:Sunday, Monday; Monday, Tuesday; Tuesday, Wednesday; Wednesday, ThursdayThursday, Friday; Friday, Saturday; Saturday, SundayOf these total 7 outcomes, the favourable outcomes are 2.

Hence the probability of getting 53 days 4. Two players Sangeetha and Reshma, play a tennis match. It is known as the probability of

winning the match by Sangeetha is 0 . 62 . What is the probability of winning the match by Reshma?

5. Find the probability that a leap year selected at random will contain 53 Sundays.6. At a fete cards bearing numbers 1 to 500, one on each card, are put in a box. Each player

selects one card at random and that card is not replaced. If the selected card bears a number which is a perfect square of an even number the player wins the prize. What is the probability that the first player wins a prize.

7. A jar contains 24v marbles some are green and others are blue. If a marble is drawn at

random form the jar, the probability that it is green is 2 :3 . Find the number of blue marbles in the jar.

8. A piggy bank contains hundred 50 paise coins, fifty Rs. 1 coin, twenty Rs. 2 coins and ten Rs. 5 coins. If it is equally likely that one of the coins will fall out when the bank is tunred upside down, find the probability that the coin which fell.(i) Will be a 50 pise coin (ii) will be of value more that Rs. 1(iii) will be value less than Rs. 5 (iv) will be a Rs. 1 or Rs. 2 coin

9. On the disc shown below, a player spins the arrow twice. The fraction

ab is formed, where

a is the number of the sector where the arrow stops after the first spin and b is the number of sector where the arrow stops after the second spin. On every spin each of the numbered sector has a n equal probability of being the sector on which arrow stops. What

is the probability that the fraction

ab is greater than 1 .

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