4

“Section – A”

Q 1. If 7 times the 7th term of an A.P. is equal to 11 times the 11th term, then 18th term is:

a) 0

b) –1

c) 1

d) 2

Q 2. If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k:

a) 1/3

b) –1/3

Guess Paper: 2014Class: X

Subject: Mathematics

Time Allowed: 3Hours Maximum Marks: 90

General Instructions:

1. All questions are compulsory.2. The question paper consists of 34 questions divided into four sections A, B, C and D:

Section A comprises of 8 questions of 1 mark eachSection B comprises of 6 questions of 2 marks eachSection C comprises of 10 questions of 3 marks eachSection D comprises of 10 questions of 4 marks each

3. Question numbers 1 to 8 in Sections A are multiple choice questions where you are to select one correct option out of the given four.

Use of calculators is not permitted.

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c) 2/3

d) –2/3

Q 3. In the figure, RQ is a tangent to the circle wit centre O. If SQ = 6 cm and QR = 4 cm, then OR is:

a) 8 cm

b) 3 cm

c) 2.5 cm

d) 5 cm

Q 4. The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is:

a) 70 cm2

b) 140 cm2

c) 210 cm2

d) 420 cm2

Q 5. If p and q are the roots of the equation x2 – px + q = 0, then

a) p = 1, q = –2

b) p = 0, q = 1

c) p = –2, q = 0

d) p = –2, q = 1

Q 6. In the given figure, the area of the shaded region is:

a) 3π cm2

b) 6π cm2

c) 7π cm2

d) 9π cm2

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Q 7. What is the probability that a leap year has 52 Mondays?

a) 2/7

b) 4/7

c) 5/7

d) 6/7

Q 8. The tops of two poles of height 16 m and 10 m are connected by a wire of length l metres. If the wire makes an angle of 30o with the horizontal, then l is:

a) 26

b) 16

c) 12

d) 10

“Section – B”

Q 9. Find the 11th term from the last term of an A.P.

10, 7, 4, ….., –62

Q 10. Find the real roots for the equation 2 5 3 6 0x x− + =

Q 11. In the given figure, if AB = AC, prove that BE = EC.

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Q 12. Two circles touch externally. The sum of their areas is 130π sq. cm and the distance between their cen-tres is 14 cm. Find the radii of the circles.

Q 13. Show that the following points represent a rhombus (7,3), (3,0), (0, –4) and (4, –1).

Q 14. A bag contains 16 balls out of which x are white. If 8 more white balls are put in the bag, the probability of drawing a white ball will be doubled that if one ball is drawn at random. Find x.

“Section – C”

Q 15. The two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Q 16. If the mth term of an A.P. be 1n

and nth term be 1m

, then show that its (mn)th tern is 1.

Q 17. Tanvi observes that the angle of elevation of a marker on the top of a hill was 30o. She walks for 20 m towards the foot of the hill, along level ground and finds the angle of elevation of the marker to be 60o. How far from the surveyor’s first position been marker placed?

Q 18. The horizontal distance between two towers is 140 m. The angle of elevation of the top of the first tower when seen from the top of the second tower is 30o. If the height of the second tower is 60, find the height of the first tower.

Q 19. Find the points on y-axis which is equidistance from the points A (2,5) and B (3,4).

Q 20. Find the centre of the circle passing through the points (1,7), (7,–1) and (8,6).

Q 21. The wheels of a car of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km/hr?

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Q 22. A well with inner radius 4 m is dug 14 m deep. The earth taken out is spread evenly around the bound-ary of the well to make a circular ring platform of 3 m width. Find the height of platform.

Q 23. A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of vessel is 6 cm. If the sphere is completely submerged in water, find by how much will the surface of water level be raised.

Q 24. A cylinder whose height is two thirds of its diameter, is made from a sphere of radius 4 cm. Calculate the radius of the base of cylinder.

“Section – D”

Q 25. Out of a number of albatross, one – fourth of the numbers are moving on ice sheets, the 1/9 coupled (along) with 1/4th as well 7 times the square root of the number are moving over icy mountains, 56 of them remain revolving over the sea. Find the total number of albatross.

Q 26. Shelly takes 6 days less than the time taken by Sangeetha to finish a piece of work. If both of them can finish the work together in four days, then find the time taken by Sangeetha alone to finish the work.

Q 27. If the pth term of an A.P. is 1q

and the qth term is 1p

. Prove that the sum of the first pq terms is ( )1 12

pq +.

Q 28. Let A (4,2), B (6,5) and C (1,4) be the vertices of ΔABC.

(i) The median from A meets BC at D. Find the coordinates of D.

(ii) Find the co-ordinates of point P on AD such that AP : PD = 2 : 1.

(iii) Find the co-ordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : PF = 2 : 1.

(iv) What do you observe?

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Q 29. Two stations due south of a leaning tower, which leans towards the north, are at distance a and b from its foot. If α and β are the angles of elevations of top of the tower from these stations. Prove that its inclination θ to the horizontal is given by.

cot cotcot b ab aα βθ −

=−

Q 30. A circle touches all four sides of a quadrilateral ABCD. Prove that angles subtended at the centre of the circle by the opposite sides are supplementary.

Q 31. Draw a line segment AB of length 6 cm. Taking A as centre, draw a circle of radius 3 cm and taking B as centre, draw another circle of radius 2 cm. Construct tangents to each circle from the centre of the other circle.

Q 32. PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal (PQ = QR = RS). Semi-circle is drawn on PQ and QS as diameters. Find the perimeter and area of shaded region.

Q 33. A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered in water and its size is such that when it touches the sides of the conical vessel, it is just immersed. How much water will remain in the cone after the overflow of water?

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Q 34. Suppose you drop a die at random on the rectangular region as shown in figure. What is the probability that it will land inside a circle of 70 cm diameter?

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S o l u t i o n s“Section – A”

“Section – B”

Answer 9: Here a = 10, d = –3 & last term = l = –62

( )( )( )

1

or 62 10 1 3or 25

l a n d

nn

= + −

− = + − −

= Thus, there are 25 terms in A.P.

The 11th term from the last term is the 15th term.

15So, (15 1)

10 14 ( 3)32

a a d= + −= + −= −

Hence, the 11th term from the last is –32.

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Answer 10: 22 5 3 6 0Now, 2

5 36

x xa

bc

− + ==

= −=

( )2Then, 5 3 4 2 6

75 4827 0

D = − − × ×

= −= >

Since 0D > , there will be two unequal roots.

Therefore, roots of the given equation are:

( )( )

2

5 3 27

2 2

5 3 3 34

b Dxa

− ±= − − ± = ±

=

1

2

8 3or, 2 34

2 3 34 2

x

x

= =

= =

Hence, the roots of the given quadratic equation are 2 3 and 3 2 .

Answer 11: From the given figure, AB = AC

The lengths of tangents drawn from a point outside the circle are equal.

Therefore,

( )( )( )

... 1

... 2

... 3

AD AF

BD BE

CE CF

=

=

=

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( )( ) ( )

Since, or

or From Eq. 1

or From Eqs. 2 and 3

or

AB ACAD BD AF CFAF BD AF CF

BD CF

BE CE

=+ = +

+ = + = =

Answer 12: If two circles touch each other externally, then the distance between their centres will be the sum of their radii.

Let r1 and r2 be radii of two circles respectively and O and O’ be their centres.

( )

'1 2

1 2

1 2

or 14or 14 1

OO r rr r

r r

= += +

+ =

It also gives that sum of their areas is 130π cm2.

( )

2 21 2

2 21 2

or 130

or 130 2

r rr r

π π π+ =

+ =

( )

( )

2 2 21 2 1 2 1 2

21 2

1 2

Now, 2

or 14 130 2or 33 3

r r r r r r

r rr r

+ = + +

= +

=

( ) ( )( ) ( ) ( )

( )

2 21 2 1 2 1 2

2 21 2

1 2

Now, 4

or 14 4 33 From eq. 1 and 2

or 8 4

r r r r r r

r r

r r

− = + −

− = − × − =

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Solving Equation (1) and Equation (4), we get

Therefore, 1r = 11 cm and 2r =3 cm

Answer 13: Rhombus is a quadrilateral whose all sides are equal.

Let the points be (7,3), (3,0), (0, –4) and (4, –1).

Therefore,

( ) ( )( ) ( )( ) ( )( ) ( )

2 22

2 22

2 22

2 22

2 2 2 2

3 7 0 3 25

0 3 4 0 25

4 0 1 4 25

4 7 1 3 25

Here, or, or, ABCD is a rhombus.

AB

BC

CD

DA

AB BC CD DAAB BC CD DA

= − + − =

= − + − − =

= − + − + =

= − + − − =

= = == = =

Answer 14:

( )1

Number of balls in the bag 16If 8 more white balls are kept in the bag then,Total number of balls 16 8 24

Total Events 24Number of white balls 8

8 p P getting a white ball24

The probabi

xx

=

= + =∴ =

= ++

∴ = =

( )2

1 2

lity of getting a white ball, if the number of balls were 16 is:

or, p P white ball16

It is given that p 2p8or 2

24 16or 4

x

x x

x

= =

=+

= ×

=

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“Section – C”

Answer 15:

2

Let the tens digit be .18Then unit digit

18 Number 10

18And number after interchanging the digits is 10

18 18 10 10 63

10 18

x

x

xx

xx

x xx x

xx

=

∴ =+

+

∴ + − × + = +

⇒

( )( )

2

2 2

2

2

180 63

10 18 180 63 9 63 162 0 7 18 0

9 2 0 9, 2

But the digit cannot be negative,

and number 10 18 1

xx

x x xx x

x xx x

x

x

+− =

⇒ + − − =

⇒ − − =

⇒ − − =

⇒ − + =

∴ =−

= + =180 99

92

× +

=

Answer 16: Let,First termCommon difference

ad

==

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

Then,1th term 1 ... i

1th term 1 ... ii

Subtracting Eq. ii from Eq. i , we get1 1 1 1

1

Su

m a m dn

n a n dm

d m nn mm nd m nmn

dmn

= + − =

= + − =

⇒ − − − = −

−⇒ − =

∴ =

( )

( )

( ) ( )

( )

1bstituting in Eq. i , we get

1 1 1

1 1 1

1

Now th term 11 1 1

dmn

a mmn n

an mn n

amn

mn a mn d

mnmn mn

=

+ − × =

⇒ + − =

⇒ =

=+ −

= + −

1=

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( ) ( )

( ) ( )

( ) ( )

( ) ( )

( )

Then,1th term 1 ... i

1th term 1 ... ii

Subtracting Eq. ii from Eq. i , we get1 1 1 1

1

Su

m a m dn

n a n dm

d m nn mm nd m nmn

dmn

= + − =

= + − =

⇒ − − − = −

−⇒ − =

∴ =

( )

( )

( ) ( )

( )

1bstituting in Eq. i , we get

1 1 1

1 1 1

1

Now th term 11 1 1

dmn

a mmn n

an mn n

amn

mn a mn d

mnmn mn

=

+ − × =

⇒ + − =

⇒ =

=+ −

= + −

1=

Answer 17: Let M be the position of the marker on the hill MC as shown in the figure.

Let A and B be first and second positions respectively of surveyor.

o

o

206030

AB mMBCMAC

=

∠ =

∠ =

Let cm

mRequired distance m

BC xMC h

AM y

=== =

o

o

In , tan 60

or 3

In , tan 30

1or 20 33 1or 20 3

or 10

MCMBCBC

h xMCMACACh

x

xx

x

∆ =

=

∆ =

=+

=+

= m

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o

o

In , tan 60

or 3

In , tan 30

1or 20 33 1or 20 3

or 10

MCMBCBC

h xMCMACACh

x

xx

x

∆ =

=

∆ =

=+

=+

= m

Answer 18: Let the height of the first tower be h and second tower is 60 m.

That is, AB = h m and CD = 60 m

And AP = (h – 60) m

Distance between two towers = DB = CP = 140 m.

Now,

oIn , tan 30

1 60or 1403

140or 603

140or 603

140.83 m

APACPCPh

h

h

∆ =

−=

− =

= +

=

Hence, height of the first tower is 140.83 m.

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Answer 19: Let there be a point P (0,y) on y–axis which is equidistant from A (2,5) and B (3,4).

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2 2 2 22 1 2 1 3 2 3 2

2 2 2 2

2 2

2 2

2 0 5 3 0 4

4 25 10 9 16 8 10 8 9 16 4 25 2 4

PA PB

x x y y x x y y

y y

y y y yy y y y

y

⇒ =

⇒ − + − = − + −

⇒ − + − = − + −

⇒ + + − = + + −

⇒ − − + = + − −⇒ − =−⇒

( ) 2

Hence, Point is 0, 2 .y =

Answer 20: Let the centre of the circle passing through A (1,7), B (7,–1) and C (8,6) be O (x, y), therefore radius = r = OA = OB = OC.

( ) ( ) ( ) ( ) ( ) ( )

( )

2 2 2

2 2 2 2 2 2

2 2 2 2

1 7 7 1 8 6 Taking first two equal, we get

2 1 14 49 14 49 2 1 3 4 0 ... i

Taking last

OA OB OC

x y x y x y

x x y y x x y yx y

⇒ = =

⇒ − + − = − + + = − + −

⇒ − + + − + = − + + + +

⇒ − =

( )( ) ( )

( )

2 2 2 2

two equals, we get 14 49 2 1 16 6 12 36 7 25 ... ii

Solving Eqs. and , we get 4 and 3

Centre is 4,3

x x y y x x y y yx y

i ii x y

⇒ − + + + + = − + + = +

⇒ + =

= =

∴

Answer 21: Speed 66 km/hr, 10min66Distance travelled in 10min 10 km60

11 km 11 1000 m 11,00,000 cmDiameter of the wheel 80 cm

radius 40 cmDistan

t= =

= ×

= =× ==

⇒ =

( )ce covered in one revolution Circumference of the wheel

2ð 40 80ð cm The number of revolutions in travelling 11,00,000 cm

=

= =

⇒1100000 110000 7 4375

80ð=

8 22= × =

×

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Answer 22: Inner radius of the well, r = 4 m.

2

2

3

Outer radius of circular platform 4 3 7 mDepth of well 14 m

Width of platform 3 mVolume of the earth taken out Volume of well

Volume of the cylinderð22 4 147

704 m Volume of circular platform

R

r h

= + = =====

=

= × ×

= =

( )( )

2 2

2 2

Volume of circular platform in the form of ring Area of the Base Height

704ð

22 704 7 47

h 6.

R r h

h

= ×

⇒ = − ×

⇒ = − ×

∴ = 78 m Height of platform 6.78 m∴ =

Answer 23: ( )( )

Diameter of sphere 6 cm Radius of sphere 3 cm let

Radius of cylinder 6 cm let et the height of water level raised cmThen, the volume

r

RL h

=

∴ =

=

=2

2 3

2 3

2 3

of water thus raisedðNow, Volume of this water Volume of sphere

4 R34 R34 6 33

R h

h r

h r

h

π π

==

⇒ =

⇒ =

⇒ = ×

⇒ 1 cmh =

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( )( )

Diameter of sphere 6 cm Radius of sphere 3 cm let

Radius of cylinder 6 cm let et the height of water level raised cmThen, the volume

r

RL h

=

∴ =

=

=2

2 3

2 3

2 3

34 R34 6 33

R h

h r

h r

h

π π

==

⇒ =

⇒ =

⇒ = ×

⇒ 1 cmHence, the surface of water level will be raised by 1 cm.

h =

Answer 24:

( )3

Radius of sphere 4 cm4 Volume of sphereð 43

Let radius of base of the cylinder be cm. Diameter 2

2Thus, Height of cylinder diameter3

rr

=

=

∴ =

= ×

( )

( )

2 3

2 3

32

2 4 2 let3 3

Volume of cylinder Volume of sphere4 ð ð34 3

4 4 43 3

rr h

r h R

r h R

rr

= × =

=

⇒ =

⇒ =

⇒ × =

⇒ 4 cmr =

“Section – D”

Answer 25: Let total number of albatross be .According, to the given information.

7 564 9 4

7 564 9 4

x

x x x x x

x x xx x

+ + + + =

⇒ − − − = +

4

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36 9 4 9 7 5636

14 7 5636

7 7 5618

126 1008 7

7 126 1008 0Let

x x x x x

x x

xx

x x

x x

− − −⇒ = +

⇒ = +

⇒ + =

⇒ + =

⇒ − − =

( )( )

2

2

2

2

7 126 1008 0 18 144 0 24 6 144 0 24 6 0 24, 6

x yy y

y yy y y

y yyy

=

⇒ − − =

⇒ − − =

⇒ − + − =

⇒ − + =

⇒ = −

⇒ = [ ]24 Since, negative value not acceptable 24 24 576

Hence, the total number of albatross is 556.x⇒ = × =

Answer 26: Let Sangeetha alone take x days to finish the work, then Shelly alone will finish the work in( )6x − days.

1 So Sangeetha one day work

1and Shelly's one day work6

1 1Thus, Sangeetha and Shelly's one day work6

Both together complete the work in 4 days.

One day work of bot

x

x

x x

=

=−

= +−

∴1h4

=

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( )2

2

Comparing two works for one day, we get1 1 4

66 1

6 4

14 24 0 12 2 24 0

x xx xx x

x xx x x

x x

+ =−

− +⇒ =

−

⇒ − + =

⇒ − − + =

⇒ −( ) ( )( )( )12 2 12 0

12 2 0 12 or 2

But cannot be less than 6.Hence, Sangeetha alone can finish the work in 12 days.

x

x xx

x

− − =

⇒ − − =

∴ =

Answer 27:

p

q

From the given question,1 t

1 ( 1) (i)

1 t

(

q

a p dq

p

a q

=

⇒ + − =

=

⇒ + −11) (ii)

Solving Eqs (i) and (iii), we get [from Eq (i) Eq (ii)]1 1[( 1) ( 1)] [ 1 1]

( )

dp

p qd p q d p qq p pq

p qd p qpq

=

−−

− − − = − ⇒ − − + =

−⇒ − =

∴1

Substituting the value of 'd' in Eq. (i), we get1 1 ( 1)

dpq

a ppq q

=

+ − × =

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1 1

1

Now for the sum of first pq terms we have1 1 , ,

[2 (2n

papq pq q

apq

a d n pqpq pqnS a n

⇒ + − =

∴ =

= = =

= +1) ]

2 1 ( 1)2

2 1 12

1 2

pq

d

pqS pqpq pq

pqpq pq

pq pqpq

−

⇒ = + − ×

= + − +

=

1 (2

pq 1)= +

Answer 28:

1 2 1 2

Median is a line segment joining the vertex of the triangle to mid-pointof opposite side.(i) is the Mid-point of

coordinates of point , 2 2

6

D BCx x y yD + +

⇒ =

+=

1 5 4 7 9, ,2 2 2 2

(ii) is on such that : 2 :17 91 4 2 2 1 22 2 coordinates of ,

1 2 1 211 11 ,3 3

P AD AP OD

P

+ =

=

× + × × + ×⇒ =

+ + =

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2 1 1 2

1 2

4 1 2 4 5(iii) Coordinate of , ,32 2 2

is on BE such that : 2 :1

Coordinates of

51 6 2 112 2 1 3

and

E

Q BQ QEm x m xQ

m m

+ + = =

=+

⇒ =+

× + ×= =

+

2 1 1 2

1 2

52 2 3 112 2 1 3

11 11 Point ,3 3

11 11Similarly coordinates of on ,3 3

(iv) We came to conclusion that medians of a tria

m y m ym m

Q

R CF

× + ×+= = =

+ +

∴ =

=

ngle meet at one point. This point iscalled centroid of triangle and centroid divides medians in the ratio 2 :1.

Answer 29: Let , and represent the leaning tower, and the two given points (see figure) , ,

CD A BAC a BC b

DCMDAM a

θ= =

∠ =∠ =

.

In

tan

tan

cot

DBMLet DM hand CM x

DCMDMCMhxx h

β

θ

θ

θ

∠ ===

∆

=

=

⇒ = (i)

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( )

In , tan ;

tan

cot cot cot cot cot cot [From Eq. (i)]

DMDAMAM

ha x

a x ha h x h ha h

α

α

αα α θα θ

∆ =

⇒ =+

⇒ + =⇒ = − = −

⇒ =−

( )

(ii)

In , tan tan cot

cot cot cot cot cot [From Eq. (i)]

DM hDBM b x hBM b x

b h x h hb h

β β β

β β θβ θ

∆ = ⇒ = ⇒ + =+

⇒ = − = −

⇒ =−

( )( )

( )

(iii)

cot cotDividing Eq. (ii) by Eq. (iii), we get;

cot cot cot cot cot cot cot cot cot cot cot cot cot

cot cot cot

hab h

a a b b b a b ab a b a

b ab a

α θβ θ

β θ α θ θ θ θ βθ α β

α βθ

−=

−

⇒ − = − ⇒ − = −

⇒ − = −

−=

−

Answer 30: Given: In figure, a circle ( , ) touches all the four sides of a quadrilateral at , , and .

To Prove: 80 and 180Proof: Since and

O rABCD P Q R S

AOB DOCAOD BOC

AP AS

∠ +∠ =

∠ +∠ =

are tangents to circle from point A, therefor centre Olies on the angel bisector of

1or 2

Similarly BO, CO and DO are angle bi

PASSAO PAO

DAO BAA BAO

∠⇒ =∠

∠ =∠ =∠

( )

sectors of , and respectively. Weknow that sum of angles of a triangle is 180 . Therefore in

180 AOB 180

1 AOB 1802

B C DAOB

AOB OBA OABOBA OAB

AB

∠ ∠ ∠

∆

∠ +∠ +∠ =

⇒ ∠ = − ∠ +∠

⇒ ∠ = − ∠

12

C BAD + ∠

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( )

[ ]

1 180 (i)2

Similarly1 180 (ii)2

Adding Eqs (i)

AOB ABC BAD

DOC BCD ADC

∴ ∠ = − ∠ +∠

∠ = − ∠ +∠

( ) ( )

( )

and (ii), we get1 180 18021 360 21 360 360 1802

180

AOB DOC ABC BAD BCD ADC

AOB DOC ABC BCD ADC BAD

AOB DOC

AOB DOC

∠ +∠ = − ∠ +∠ + − ∠ +∠

⇒ ∠ +∠ = − ∠ +∠ +∠ +∠

⇒ ∠ +∠ = − × =

⇒ ∠ +∠ =

Similarly 180AOD BOC∠ +∠ =

Answer 31: Step of Construction:

(i) Draw AB = 6 cm.

(ii) Taking A as centre draw a circle (A, 3 cm).

(iii) Taking B as centre draw another circle (B, 2 cm).

(iv) Bisect the line segment AB. Let M be the mid-point of AB.

(v) Taking M as centre and radius equal.

To AM or MB draw third circle passing through A and B and intersecting the circle (A, 3 cm) at P and Q.

Also, intersecting the circle (B, 2 cm) at R and S.

(vi) Join AR, AS, BP and BQ.

AR and AS are tangents from A to circle (B, 2 cm).

Also BP and BQ are tangents from B to circle (A, 3 cm).

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Answer 32: Diameter PS = 2 6 = 12 cm PS = PQ + QR + RS = 12 cm Also PQ = QR = RS = 4 cm P = Perimeter of Shaded region

×

1 = Circumference of circle (diameter PS)21 + Circumference of circle (diameter QS)21 + Circumferenc2

1 2 3

1 2 3

e of circle (diameter PQ)

1 P = [2 2 2 ]21 = 2 ( )2

= (6 4 2) 12 377.717

Area of shaded por

r r r

r r r

cm

π π π

π

π π

∴ + +

× + +

+ + = =

tion = Area of semi-circle with diameter PS Area of semi-circle with diameter QS + Area of semi-circle with diameter

−

2 2 21 2 3

2 2 21 2 3

2 2 2

PQ1 1 1 Area = 2 2 21 = ( )21 22 (6 4 2 )2 711

r r r

r r r

π π π

π

∴ − +

− +

= × − +

= 24 37.71 . .7

sq cm× =

Answer 33: The sphere is just immersed in the cone filled of water. Hence upper most point of the sphere and centre of base of the cone will be same, i.e. lie on point P.

Let ABC represent the cone and O is center of sphere touching the cone at M and N.

∴ PB = Radius of base of the cone

= 6 cm

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and PC = height of cone = h = 8 cm

Now in ∆PBC, we have

BC2 = PB2 + PC2 = 62 + 82 = 100

∴ BC = 10 cm.

Now, we know that tangents from a external point to a circle are equal.

∴ BP = BN = 6 cm 10 6 = 4 cm.

Let the radius of the sphere cm cm

NC BC BNON x

OP ON x

⇒ = − = −= =

⇒ = =∴

O

2 2 2

2 2 2

(8 ) cmNow in , we have ( 90 )

(8 ) 4

OC PC OP xONC BNO

OC ON NCx x

= − = −

∆ ∠ =

∴ = +

⇒ − = +⇒

3

3 3

3 cm radius of sphere4 Volume of sphereð34 22 3 cm3 71 Volume of cone3

x

r

= =

∴ =

= × ×

= 2

2

ð

1 22 × (6) × 83 7

Volume of remaining water Volume of cone Volume of sphere1 22 3 7

R h

=

∴ = −

= × 2 3

3

4 226 8 (3)3 7

188.57 cm

× × − × ×

=

Answer 34:

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