Lesson: Surface Areas and Volumes

Exercise 13.1 (8)

Question: 1

A plastic box 1.5 m long, 1.25 m wide and 65 cm deep

is to be made. It is opened at the top. Ignoring the

thickness of the plastic sheet, determine:

(i) The area of the sheet required for making the box.

(ii) The cost of sheet for it, if a sheet measuring 1

m

2

costs Rs 20.

Solution

(i)

Since,

The length of plastic box,

1.50ml

The width of plastic box,

1.25mb

The depth of plastic box,

65cm 0.65mh

The area of sheet required for making the box is equal

to the sum of lateral surface area and base area of the

plastic box,

So,

Surface area of the box

Lateral surface area

Area

of the base

2

2

2

2

2 1.50 1.25 0.65 1.50 1.25 m

3.575 1.875 m

[]

5.45m

l b h l b

Therefore, the area of the sheet required for making

the box is

2

5.45 m .

(ii)

Since,

Cost of 1 m

2

of sheet

20Rs

Therefore,

Cost of

2

5.45m

of sheet

20 5.()45 109Rs Rs

Question: 2

The length, breadth and height of a room are 5m, 4m

and 3m respectively. Find the cost of white washing

the walls of the room and the ceiling at the rate of

Rs 7.50 per m

2

.

Solution

Since,

The length of the room

5m

The breadth of the room

4m

The height of the room

3m

As,

The area for white washing is equal to the sum of the

area of four walls and the area of the ceiling.

So,

The required area for white washing is

2 2 2

2

2 5 4 3 5 4 54 20 m 74m

l b h l b

m

Now,

The cost of white washing for

area

2

1m 7.50Rs

Therefore, the total cost of white washing

74 7.50 5 5()5Rs Rs

Question: 3

The floor of a rectangular hall has a perimeter of 250

m. If the cost of painting the four walls at the rate

of Rs 10 per m

2

is Rs 15,000, find the height of the

hall.

[Hint: Area of the four walls

Lateral surface area]

Solution

Since,

The perimeter of rectangular hall

2 250mlb

and,

Total cost of painting the four walls

15000Rs

Now,

The cost of painting for

2

1m area 10Rs

.

Area of four walls

2 2

2 250 l b hm h m

So,

250 10 15000h

or,

2500 15000 h

or,

15000

250

m6

0

m h

Therefore, the height of the hall is 6 m.

Question: 4

The paint in a certain container is sufficient to paint an

area equal to 9.375 m

2

. How many bricks of

dimensions

22.5cm 10cm 7.5cm

can be painted

out of this container?

Solution

Since,

The area that can be painted out of the given container

2

2

2

9.375

9.375 100 100

93750 cm

m

cm

And,

The dimensions of brick

22.5cm 10cm 7.5cm

So,

Total surface area of a brick

2

2

2

2

2

2

2 22.5 10 10 7.5 22.5 7.5 cm

2 225 75 168.7

(

5 cm

2 468.75 cm

93

)

7.5cm

lb bh lh cm

Therefore, the number of bricks that can be painted

out of the given container

93750

937.5

100

Question: 5

A cubical box has each edge measuring 10 cm and

another cuboidal box is 12.5 cm long, 10 cm wide and

8 cm high.

(i) Which box has a greater lateral surface area and by

how much?

(ii) Which box has a smaller total surface area and by

how much?

Solution

(i)

The edge of the given cubical box

10 cm

So, the lateral surface area of the given cubical box

2 2 2

4 10 cm 400 cm

And the lateral surface area of the given cuboid box

2

22

2 2 12.5 10 8 cm

2 22.5 8 cm 360 cm

l b h

Therefore, the lateral surface area of the cubical box is

greater by

22

400 360 cm 40 cm

(ii)

Total surface area of given cubical box

2 2 2

6 10 cm 600 cm

Total surface area of the cuboidal box

2

2

2

2

2

2 12.5 10 10 8 8 12.5 cm

2 125 80 100 cm

2 305 cm

610 cm

lb bh lh

Therefore, the total surface area of the cubical box is

smaller by 10 cm

2

.

Question: 6

A small indoor greenhouse (herbarium) is made

entirely of glass panes (including base) held together

with tape. It is 30 cm long, 25 cm wide and 25 cm

high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

Solution

(i)

Since,

The length of greenhouse,

30cml

The breadth of greenhouse,

25cmb

The height of greenhouse,

25cmh

So,

Total surface area of the greenhouse

2

22

2 2 30 25 25 25 25 30 cm

2 750 625 750 cm 4250 cm

lb bh lh

Therefore, the area of the glass is 4250 cm

2

.

(ii)

The required length of the tape

4 4 30 25 25 cm

4 80 cm 320 cm

l b h

Therefore, the tape is of length 320 cm.

Question: 7

Shanti Sweets Stall was placing an order for making

cardboard boxes for packing their sweets. Two sizes

of boxes were required. The bigger of dimensions 25

cm 20cm 5 cm

and the smaller of dimensions 15

12 cm 5cmcm

. For all the overlaps, 5% of the total

surface area is required extra. If the cost of the

cardboard is Rs 4 for 1000 cm

2

, find the cost of

cardboard required for supplying 250 boxes of each

kind.

Solution

Since,

The dimensions of the bigger box

25cm 20 cm 5 cm

So,

Total surface area of the bigger box

2

2 2 25 20 20 5 25 5 cmlb bh lh

22

2 500 100 125 cm 1450 cm

Now,

The dimensions of the smaller box

15cm 12cm 5 cm

Total surface area of the smaller box

2

2

2

()

2

215 12 12 5 15 5 cm

2 180 60 75 cm

630 cm

lb bh lh

Therefore,

Total surface area of 250 boxes of each kind

2

2

250 1450 630 cm

250 2080 cm

520000cm

Next,

Required extra area

2

5

520000

10

cm

0

2

26000cm

Hence, the required total cardboard area

2 2

520000 26000 cm 546000cm

Total cost of required cardboard

4

5460

00

1000

2184

s

Rs

R

Question: 8

Parveen wanted to make a temporary shelter for her

car, by making a box-like structure with tarpaulin that

covers all the four sides and the top of the car (with

the front face as a flap which can be rolled up).

Assuming that the stitching margins are very small,

and therefore negligible, how much tarpaulin would be

required to make the shelter of height 2.5 m, with base

dimensions

4m 3m

?

Solution

The dimensions of the box-like structure

4m 3m 2.5m

Tarpaulin is required for all the four sides and the top

of the car.

So, the required area of the tarpaulin

2

2

2

2

2 4 3 2.5 4 3 m

35 12 m

47 m

l b h lb

Therefore, to make the shelter for car, 47 m

2

tarpaulin

is required.

Exercise 13.2 (11)

Question: 1

The curved surface area of a right circular cylinder of

height

14 cm

is

2

88 cm

. Find the diameter of the base

of the cylinder.

Solution

Let r be the base radius of the right circular cylinder.

The height of the cylinder,

14 cmh

The curved surface area of cylinder

2

2 88 cmrh

So,

22

2 14 88

7

r

or,

88

22

2 14

7

r

or,

1cmr

Therefore,

The diameter of the base of the cylinder

2

2 1cm

2cm

r

Question: 2

It is required to make a closed cylindrical tank of

height 1 m and base diameter 140 cm from a metal

sheet. How many square metres of the sheet are

required for the same?

Solution

Let r be the base radius of the cylindrical tank and h

be the height of the cylindrical tank. The height of the

cylindrical tank,

1mh

.

The base diameter of cylindrical tank,

2 140 cmr

The base radius of cylindrical tank is

70 cm

140

cm

2

0.7 m

r

Therefore, the metal sheet required to make a closed

cylindrical tank is

2

2

2

2m

22

2 0.7 1.7 m

7

7.48 m

r h r

Question: 3

A metal pipe is 77 cm long. The inner diameter of a

cross section is 4 cm, the outer diameter being 4.4 cm

(see Fig. 13.11). Find its

(i) Inner curved surface area,

(ii) Outer curved surface area,

(iii) Total surface area.

Fig. 13.11

Solution

(i)

Let the external radius of the pipe be R, the internal

radius be r and the length of the pipe be h.

Since,

4.4

cm=2.2 cm

2

4

cm 2 cm

2

77 cm

R

r

h

Inner curved surface area

2

2

2

22

2 2 77 cm

7

968 cm

rh

(ii)

Outer curved surface area

2

2

2

22

2 2.2 77 cm

7

1064.

2 cm

8 cm

Rh

(iii)

Now,

Total surface area of the pipe

Inner curved surface

area + Outer curved surface area

Areas of two bases

Therefore,

Total surface area of pipe

22

2

2

2

2

22

968 1064.8 2 4.84 4 cm

7

44

2

2 2 2

2032.8 5.28

032.8 0.84 cm

cm

2038.08c

7

m

rh Rh R r

Question: 4

The diameter of a roller is 84 cm and its length is 120

cm. It takes 500 complete revolutions to move once

over to level a playground. Find the area of the

playground in m

2

.

Solution

Since,

The length of the roller is 120 cm or 1.20 m.

The radius of the cylinder

84

cm

2

42cm

0.42 m

Now,

Total number of revolutions

500

So,

Distance covered by roller in one revolution

Curved surface area of the roller

2

2

2

22

2 0.42 1.20 m

7

3.168 m

rh

Therefore, the area of the playground

2

2

500 3.168 m

1584 m

Question: 5

A cylindrical pillar is 50 cm in diameter and 3.5 m in

height. Find the cost of painting the curved surface of

the pillar at the rate of Rs 12.50 per m

2

.

Solution

The diameter of the pillar is 50 cm.

Therefore, the radius is

25 cm

5

0

0

.

cm

m

2

25

r

The height of the pillar,

3.5mh

Now,

The cost of painting for

2

1m

area of the pillar

12.50Rs

So,

The curved surface area of the pillar

2 rh

2

2

22

2 0.25 3.5 m

7

5.5 m

Therefore, the total cost of painting the curved surface

area of the pillar is

5.5 12.5

68.75

Rs

Rs

Question: 6

Curved surface area of a right circular cylinder is

2

4.4 m

. If the radius of the base of the cylinder is 0.7

m, find its height.

Solution

Let the radius of the base be r and the height of the

cylinder be h.

The curved surface area of a right circular cylinder

2

2 4.4 mrh

So,

22

2 0.7 4.4

7

h

or,

4.4

m 1 m

22

2 0.7

7

h

Therefore, the height of the right circular cylinder is 1

m.

Question: 7

The inner diameter of a circular well is 3.5 m. It is 10

m deep. Find

(i) Its inner curved surface area,

(ii) The cost of plastering this curved surface at the

rate of

per

2

40 mRs

.

Solution

(i)

The inner radius of the circular well,

3.5

m 1.75 m

2

r

The depth of the well,

10mh

Now,

The cost of plastering for

2

1m

area of circular well

40Rs

So,

The inner curved surface area of circular well

2 rh

2

2

22

2 1.75 10 m

7

110 m

(ii)

The cost of plastering for 1 m

2

area of circular well

40Rs

Therefore, the cost of plastering the curved surface

area of the circular well

110 40

4400

Rs

Rs

Question: 8

In a hot water heating system, there is a cylindrical

pipe of length 28 m and diameter 5 cm. Find the total

radiating surface in the system.

Solution

The radius of the pipe

5

cm=2.5 cm 0.025 m

2

r

The length of the pipe

28 mh

Therefore, the total radiating surface in the system

Curved surface area of the pipe

2

2

2

22

2 0.025 28 m

7

4.4 m

rh

Question: 9

Find

(i) The lateral or curved surface area of a closed

cylindrical petrol storage tank that is 4.2 m in diameter

and 4.5 m high.

(ii) How much steel was actually used, if

1

12

of the

steel actually used was wasted in making the tank.

Solution

(i)

The radius of the tank, r

4.2

m=2.1 m

2

The height of the tank,

4.5mh

So,

The curved surface area of the cylindrical tank

2

2

2

22

2 2.1 4.5 m

7

59.4 m

rh

(ii)

Now,

Total surface area of the tank

2

2

22

2 2.1 2.1 4.5 m

7

8m

2

7.12

r r h

Let s be the area of the actual steel used in making the

tank.

Therefore,

1

1 87.12

12

s

or,

12

87.12

11

s

or,

2

95.04 ms

Hence, the area of the steel actually used in making

the tank is

2

95.04 m

.

Question: 10

In Fig. 13.12, you see the frame of a lampshade. It is

to be covered with a decorative cloth. The frame has a

base diameter of 20 cm and a height of 30 cm. A

margin of 2.5 cm is to be given for folding it over the

top and the bottom of the frame. Find how much cloth

is required for covering the lampshade.

Solution

The radius of the frame

20

cm 10 cm

2

r

The height of the frame is:

30cm 2 2.5cm 35cmh

[as a margin of 2.5 cm

is to be given for folding over the top and bottom of

the frame]

Therefore, the cloth required for covering the

lampshade

Curved surface area of lampshade

2

2

2

22

2 10 35 cm

7

2200 cm

rh

Question: 11

The students of a Vidyalaya were asked to participate

in a competition for making and decorating penholders

in the shape of a cylinder with a base, using cardboard.

Each penholder was to be of radius 3 cm and height

10.5 cm. The Vidyalaya was to supply the competitors

with cardboard. If there were 35 competitors, how

much cardboard was required to be bought for the

competition?

Solution

The radius of the penholder,

3cmr

The height of the penholder,

10.5cmh

So,

The area of cardboard required by one competitor

Curved surface area of one penholder

area of the

base

22

2

22 22

2 3 105 3 cm

7

2

.

7

rh r

2

2

198

198 cm

7

1584

cm

7

Therefore, the total area of cardboard required for 35

competitors

2

2

1584

35 cm

7

7920 cm

Exercise 13.3 (8)

Question: 1

Diameter of the base of a cone is 10.5 cm and its slant

height is 10 cm. Find its curved surface area.

Solution

The base radius of cone,

10.5

cm 5.25 cm

2

r

and,

The slant height of cone

10cml

Therefore,

The curved surface area of the cone

2

2

22

5.25 10 cm

7

165 cm

rl

Question: 2

Find the total surface area of a cone, if its slant height

is 21 m and diameter of its base is 24 m.

Solution

The base radius of cone,

24

m=12 m

2

r

and,

The slant height of cone,

21ml

Therefore, the total surface area of the cone

2

2

2

22

12 21 12 m

7

22

12 33 m

7

1244.57 m

r l r

Question: 3

The curved surface area of a cone is 308 cm

2

and its

slant height is 14 cm.

Find

(i) Radius of the base and

(ii) Total surface area of the cone

Solution

(i)

The curved surface area of cone

2

308 cm

and,

The slant height of cone

14cml

Let the base radius of cone be r.

So,

308r

or,

22

14 308

7

r

or,

308

cm 7 cm

22

14

7

r

(ii)

Therefore, the total surface area of the cone

2

2

2

22

7 14 7 cm

7

22 21 cm

462cm

r l r

Question: 4

A conical tent is 10 m high and the radius of its base is

24 m. Find

(i) Slant height of the tent.

(ii) Cost of the canvas required to make the tent, if the

cost of 1 m

2

canvas is Rs 70.

Solution

The base radius of the conical tent,

24 mr

and,

The height of the conical tent,

10mh

Let the slant height of the conical tent be l.

So,

2 2 2

l h r

or,

2 2

l h r

or,

2 2

10 24 l

or,

100 576 l

or,

26ml

Therefore,

The slant height of the conical tent is 26 m.

(ii)

The required area of canvas to make the conical tent =

Curved surface of the cone

rl

2

22

24 26 m

7

2

13728

m

7

As,

The cost of

2

1 m

area of canvas

70Rs

Therefore, the total cost of required canvas

13728

70

7

137280

Rs

Rs

Question: 5

What length of a 3 m wide tarpaulin will be required

to make a conical tent of height 8 m and base radius 6

m? Assume that the extra length of material that will

be required for stitching margins and wastage in

cutting is approximately 20 cm (Use π = 3.14).

Solution

The base radius of conical tent,

6mr

The height of the conical tent,

8mh

Let the slant height of the conical tent be l.

So,

2 2

l rh

22

86l

100l

10ml

Therefore,

The curved surface area of the conical tent

2

3.14 6 10 m rl

2

188.4 m

Now,

The breadth of tarpaulin

3m

Let required length of tarpaulin sheet be x.

As,

Wastage in cutting is 20 cm.

So,

0.2 m 3 188.4x

or,

0.2 62.8x

63 mx

Therefore, 63 m long tarpaulin will be required.

Question: 6

The slant height and base diameter of a conical tomb

are 25 m and 14 m respectively. Find the cost of

white-washing its curved surface at the rate of

Rs 210 per 100 m

2

.

Solution

The base radius of conical tomb,

14

m 7 m

2

r

The slant height of conical tomb,

25ml

So,

The curved surface area of conical tomb

2

2

22

25 7 m

7

550 m

rl

The cost of white- washing for 100 m

2

area of conical

tomb

210Rs

Therefore,

Total cost of white-washing the conical tomb

210

550 1155

100

Rs Rs

Question: 7

A joker’s cap is in the form of a right circular cone of

base radius 7 cm and height 24 cm. Find the area of

the sheet required to make 10 such caps.

Solution

The base radius of the cone,

7 cmr

and,

The height of the cone,

24 cmh

Let the slant height of the cone be l.

So,

22

l h r

or,

22

24 7l

625l

25ml

Therefore,

The required area of sheet for one cap

Curved

surface of the cone

2

2

22

7 25 cm

7

550 cm

rl

Thus, the required area of sheet for 10 caps

2 2

550 10 cm 5500 cm

Question: 8

A bus stop is barricaded from the remaining part of the

road, by using 50 hollow cones made of recycled

cardboard. Each cone has a base diameter of 40 cm

and a height of

1m

. If the outer side of each of the

cones is to be painted and the cost of painting is

Rs 12

per m

2

, what will be the cost of painting all these

cones? (Use π = 3.14 and take

1.04 1.02

)

Solution

Since,

The base radius of the cone

40

cm 20 cm 0.2 m

2

r

and,

The height of the cone,

1mh

Let the slant height of the cone be l.

So,

2 2

l h r

or,

2 2

1 0.2 l

or,

1.04 l

or,

1.02 ml

As,

Rate of painting

12 Rs

per

2

m

Therefore,

Curved surface area of one cone

2

2

3.14 0.2 1.02 m

0.64056 m

rl

Now,

Curved surface area of 50 cones

2

2

50 0.64056 m

32.028 m

Thus,

Total cost of painting all the cones

32.028 12

384.34

Rs

Rs

EXERCISE 13.4 (9)

Question: 1

Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Solution

(i)

10.5 cmr

Surface area of the sphere

2

4 r

22

2

2

22

4 (10.5) cm

7

22

4 10.5 10.5 cm

7

1386 cm

(ii)

5.6 cmr

Surface area of the sphere

2

22

2

2

4

22

4 (5.6) cm

7

22

4 5.6 5.6 cm

7

394.24 cm

r

(iii)

14 cmr

Surface area of the sphere

2

22

2

2

4

22

4 (14) cm

7

22

4 14 14 cm

7

2464 cm

r

Question: 2

Find the surface area of a sphere of diameter:

(i) 14 cm

(ii) 21 cm

(iii) 3.5 m

Solution

(i)

14

cm 7 cm

2

r

Therefore,

Surface area of the sphere

2

22

2

2

2

4

22

4 (7) cm

7

22

4 7 7cm

7

88 7 cm

616 cm

r

(ii)

21

cm 10.5 cm

2

r

Therefore,

Surface area of the sphere

2

22

2

2

4

22

4 (10.5) cm

7

22

4 10.5 10.5cm

7

1386 cm

r

(iii)

Since,

3.5

m 1.75 m

2

r

Therefore,

Surface area of the sphere

2

22

2

2

4

22

4 (1.75) m

7

22

4 1.75 1.75 m

7

38.5 m

r

Question: 3

Find the total surface area of a hemisphere of radius

10 cm. (Use

3.14

)

Solution

The radius of hemisphere,

10cmr

Therefore,

Total surface area of the hemisphere

2

3 r

2

2

2

2

3 3.14 10 cm

3 3.14 100 cm

942 cm

Question: 4

The radius of a spherical balloon increases from 7 cm

to 14 cm as air is being pumped into it. Find the ratio

of surface areas of the balloon in the two cases.

Solution

Case I:

The radius of spherical balloon,

7 cmr

So,

Surface area of the balloon

2

4 r

2

4 7 7 cm

Case II:

The radius of spherical balloon,

14 cmR

Therefore,

Surface area of the balloon

2

4 r

2

4 14 14 cm

Hence,

The required ratio of the surface areas of the balloon

in two cases

4 π 7 7

4 π 14 14

1

4

=1:4

Question: 5

A hemispherical bowl made of brass has an inner

diameter of 10.5 cm. Find the cost of tin-plating it on

the inside at the rate of Rs 16 per 100 cm

2

.

Solution

The radius of hemispherical bowl,

10.5

cm 5.25cm

2

r

So,

The inner surface area of the bowl

2

2 r

2

2

44 0.75 5.25 cm

173.25 cm

22

22

2 (5.25) cm

7

As,

Cost of tin plating for

2

100 cm

area of bowl

16Rs

Therefore,

Total cost of tin plating the bowl

16

Rs 173.25 Rs 27.72

100

Question: 6

Find the radius of a sphere whose surface area is 154

cm

2

.

Solution

Since,

Surface area of the sphere

2

4 r

2

2

22

154 4

7

154 7

22 4

7

3.5

2

r

r

r

Therefore, the radius of the sphere

3.5cm.

Question: 7

The diameter of the moon is approximately one fourth

the diameter of the earth. Find the ratio of their surface

areas.

Solution

Let diameter of the earth be 2r.

So,

The radius of the earth

r

As,

The diameter of the moon

2

42

rr

Therefore,

Radius of the moon

4

r

Now,

The surface area of the moon

2

4

4

r

The surface area of the earth

2

4 r

Hence,

The required ratio of the surface areas

2

2

1

4

22

16

4 4 4

r

r

rr

Question: 8

A hemispherical bowl is made of steel, 0.25 cm thick.

The inner radius of the bowl is 5 cm. Find the outer

curved surface area of the bowl.

Solution

Since,

The inner radius of the hemispherical bowl,

5cmr

And,

The thickness of the steel

0.25cm

So,

The outer radius of the hemispherical bowl,

5 0.25 cm 5.25 cmR

Therefore,

The outer curved surface area of the hemispherical

bowl

22

22

2 (5.25) cm

7

2

173.25 cm

Question: 9

A right circular cylinder just encloses a sphere of

radius r (see Fig). Find

(i) Surface area of the sphere,

(ii) Curved surface area of the cylinder,

(iii) Ratio of the areas obtained in (i) and (ii).

Solution

(i)

The radius of the sphere

r

As,

The right circular cylinder just encloses the sphere.

So,

The base radius of the cylinder

r

And,

The height of the cylinder

2r

Therefore,

The surface area of the sphere

2

4 r

(ii)

Since,

The curved surface area of the cylinder

2 rh

2

2 2 4r r r

(iii)

Therefore,

The required ratio of the areas

2

2

4πr 1

4πr 1

EXERCISE 13.5 (9)

Question: 1

A matchbox measures

4cm 2.5 cm 1.5cm

. What

will be the volume of a packet containing 12 such

boxes?

Solution

Since,

The dimensions of match box

4cm 2.5 cm 1.5cm

So,

Volume of one matchbox

bh

3

3

4 2.5 1.5 cm

15cm

Therefore,

Volume of a packet containing 12 match boxes

33

12 15 cm 180 cm

Question: 2

A cuboidal water tank is 6 m long, 5 m wide and 4.5

m deep. How many litres of water can it hold?

3

1m 1000l

Solution

Since,

The length of water tank,

6ml

The width of water tank,

5mb

The depth of water tank,

4.5mh

So,

Volume of the water tank

bh

3

3

6 5 4.5 m

13 m

5

Therefore,

The capacity of water tank

135 1000

litres (Since

3

1m 1000

litres)

135000

litres

Hence,

The water tank can hold 135000 litres of water.

Question: 3

A cuboidal vessel is 10 m long and 8 m wide. How

high must it be made to hold 380 cubic metres of a

liquid?

Solution

The length of cuboidal vessel,

10ml

The width of cuboid vessel,

8 mb

As,

Volume of cuboid vessel

capacity of cuboidal vessel

3

380 m

Now,

Volume of a cuboid

Length

Breadth

Height

Therefore,

Height

Volume of cuboid

Length Breadth

380

m

10 8

4.75 m

Hence,

The height of the cuboidal vessel is 4.75 m.

Question: 4

Find the cost of digging a cuboidal pit 8 m long, 6 m

broad and 3 m deep at the rate of Rs 30 per m

3

.

Solution

The length of the cuboidal pit,

8ml

The breadth of the cuboidal pit,

6mb

The depth of the cuboidal pit,

3 mh

So,

Volume of the cuboidal pit

bh

3

3

8 6 3 m

144 m

As,

The cost of digging

30Rs

per m

3

Therefore,

The cost of digging the cuboidal pit

144 30 4320Rs Rs

Thus,

The total cost of digging the cuboidal pit is

4320Rs

.

Question: 5

The capacity of a cuboidal tank is 50000 litres of

water. Find the breadth of the tank, if its length and

depth are 2.5 m and 10 m respectively.

Solution

Since,

The length of the cuboidal tank,

2.5ml

The depth of the cuboidal tank,

10mh

Volume of cuboidal tank

Capacity of cuboidal tank

50000

litres

litres

33

1

50000 m 1m 1000

1000

3

50 m

Therefore,

Volume of cuboid

Breadth

Length Depth

50

m

2.5 10

2m

Thus,

The breadth of the cuboidal tank is 2 m.

Question: 6

A village, having a population of 4000, requires 150

litres of water per head per day. It has a tank

measuring

20 m 15 m 6 m

. For how many days will

the water of this tank last?

Solution

The dimensions of tank

20 m 15m 6 m

So,

Capacity of the tank

bh

3

3

20 15 6 m

1800m

Required water per head per day

150

litres

So,

Required water for

4000

persons per day

4000 150

3

3

4000 150

m

1000

600 m

Therefore,

Number of days of water available

Capacity of tank

Total required water per day

1800

600

3

Thus,

The water will last for 3 days.

Question: 7

A godown measures

40 m 25m 10 m

. Find the

maximum number of wooden crates each measuring

1.5 m 1.25 m 0.5 m

that can be stored in the

godown.

Solution

The dimensions of godown

40 m 25 m 10 m

So,

Volume of the godown

33

40 25 10 m 10000m

Now,

The dimensions of a crate

1.5m 1.25 m 0.5 m

Therefore,

Volume of 1 crate

33

1.5 1.25 0.5 m 0.9375m

Number of crates that can be stored in the godown

Volume of the godown

Volume of 1crate

10000

0.9375

10666

Question: 8

A solid cube of side 12 cm is cut into eight cubes of

equal volume. What will be the side of the new cube?

Also, find the ratio between their surface areas.

Solution

The side of a solid cube

12 cm

So,

Volume of the cube

3

12 12 12 cm

Now,

Volume of the cube cut out of the solid cube

3

3

33

1

12 12 12 cm

8

216cm

6 cm

This implies that the side of the new cube

6cm

Therefore,

Ratio of the corresponding surface areas

2

2

2

2

6 side

6 side

6 12

66

4

1

Therefore,

The ratio of the surface areas of bigger cube to that of

a smaller cube is 4 : 1.

Question: 9

A river 3 m deep and 40 m wide is flowing at the rate

of 2 km per hour. How much water will fall into the

sea in a minute?

Solution

The rate of water flow of river

2km

per hour

The volume of water flowing into the sea in one hour

= Volume of the cuboid

bh

3

3

2000 40 3 m

240000m

Therefore,

The volume of water that fall into the sea in a minute

3

3

240000

m

60

4000 m

EXERCISE 13.6 (8)

Assume

22

7

, unless stated otherwise.

Question: 1

The circumference of the base of a cylindrical vessel

is 132 cm and its height is 25 cm. How many litres of

water can it hold?

cm

3

1000 1l

Solution

Let the base radius of the cylindrical vessel be r cm

and the height of the cylindrical vessel is 25 cm.

Circumference of the base of cylindrical vessel

132

cm

So,

2 132r

Or,

22

2 132

7

r

Or,

cm

cm

132 7

r

2 22

21

Now,

Volume of the cylindrical vessel

2

rh

3

3

22

21 21 25

7

34650

cm

cm

Therefore,

The capacity of cylindrical vessel

litres

34650

1000

Thus,

The cylindrical vessel can hold 34.65 litres of water.

Question: 2

The inner diameter of a cylindrical wooden pipe is 24

cm and its outer diameter is 28 cm. The length of the

pipe is 35 cm. Find the mass of the pipe, if 1 cm

3

of

wood has a mass of 0.6 g.

Solution

The length of the cylindrical wooden pipe

35

cm

The inner radius of the cylindrical wooden pipe,

1

24

c 2cm

2

mr

The external radius of the cylindrical wooden pipe,

R=

c cmm

2

1

2

8

4

So,

Volume of the wood used in making the pipe

=Volume of the pipe

= Volume of the external cylinder – Volume of the

internal cylinder

22

Rh r h

22

R r h

3

3

3

cm

cm

cm

22

22

14 12 35

7

22

26 2 35

7

5720

Now,

Mass of 1 cm

3

volume of wood

0.6g

Therefore,

Mass of cylindrical wooden pipe

5720 0.6 g

1 1000 kg kg

5720 0.6

1000

3.432 kg

×

g

Question: 3

A soft drink is available in two packs:

(i) a tin can with a rectangular base of length 5 cm and

width 4 cm, having a height of 15 cm and

(ii) a plastic cylinder with circular base of diameter 7

cm and height 10 cm. Which container has greater

capacity and by how much?

Solution

(i)

The capacity of tin can

bh

5 4 15

cm

3

cm

3

300

(ii)

The capacity of plastic cylinder

2

rh

cm

22 7 7

3

10

7 2 2

×××

cm

3

385

Therefore,

The plastic cylinder has a greater capacity by

cm

3

385 300 85

Question: 4

If the lateral surface of a cylinder is

cm

2

94.2

and its

height is 5 cm, then find

(i)Radius of its base

(ii) Its volume. (Use

3.14

)

Solution

(i)

Let the base radius of the cylinder be r and the height

of the cylinder is 5 cm.

The lateral surface of cylinder

cm

2

94.2

So,

2 94.2rh

Or,

2 3.14 5 94.2r

Or,

94.2

2 3.14 5

r

Therefore,

The base radius of the cylinder

3

cm

(ii) Volume of the cylinder

2

rh

cm

23

3.14 3 5

cm

3

141.3

Question: 5

It costs Rs 2200 to paint the inner curved surface of a

cylindrical vessel 10 m deep. If the cost of painting is

at the rate of Rs 20 per m

2

, find

(i) Inner curved surface area of the vessel,

(ii) Radius of the base,

(iii) Capacity of the vessel.

Solution

(i)

The inner curved surface area of the cylindrical vessel

Total cost of painting the cylindrical vessel

Rate of painting

2

2

m

m

2200

20

110

(ii)

Let the base radius of the cylindrical vessel be r and

the height of the cylindrical vessel is 10 m.

So,

2 110rh

Or,

22

2 10 110

7

r

Or,

.

110 7

2 22 10

7

4

175

r

×

××

Therefore,

The base radius of the cylindrical vessel

1.75

m

(iii)

Capacity of the cylindrical vessel

2

rh

m

3

1.75 1.75 10

22

()

7

m

3

96.25

Question: 6

The capacity of a closed cylindrical vessel of height 1

m is 15.4 litres. How many square metres of metal

sheet would be needed to make it?

Solution

The capacity of a closed cylindrical vessel

15.4

litres

m

3

1

15.4

1000

()

m

3

0.0154

(

m

3

1 1000

liters)

Let the base radius of the cylindrical vessel be r and

the height of the cylindrical vessel is 1 m.

So,

Volume

2 2 2

1r h r r

2

0.0154r

Or,

2

22

0.0154

7

r

Or,

2

0.0154 7

0.0049

22

r

Or,

m0.0049 0.07r

Therefore, the base radius of cylindrical vessel

0.07

m

Metal sheet required to make the cylindrical vessel

= Total surface area of the cylindrical vessel

2

22rh r

2 r h r

m

2

22

2 0.07 (1 0.07)

7

m

2

44 0.01 1.07

m

2

0.4708

Question: 7

A lead pencil consists of a cylinder of wood with a

solid cylinder of graphite filled in the interior. The

diameter of the pencil is 7 mm and the diameter of the

graphite is 1 mm. If the length of the pencil is 14 cm,

find the volume of the wood and that of the graphite.

Solution

The diameter of the graphite cylinder

1

mm

1

10

cm

So,

The radius of the graphite cylinder

1

20

cm

The length of the graphite cylinder

14

cm

Therefore,

Volume of the graphite cylinder

2

rh

cm

cm

3

3

22 1 1

14

7 20 20

0.11

Now,

The diameter of the pencil

7

m

7

m

10

cm

The radius of the pencil

cm

7

20

And,

The length of the pencil

14

cm

Hence,

Volume of the pencil

2

rh

cm

3

22 7 7

14

7 20 20

cm

3

5.39

Finally,

Volume of the wood = Volume of the pencil –

Volume of the graphite

m c

3

5.39 0.11

cm

3

5.28

Question: 8

A patient in a hospital is given soup daily in a

cylindrical bowl of diameter 7 cm. If the bowl is filled

with soup to a height of 4 cm, how much soup can the

hospital prepare daily to serve 250 patients?

Solution

The diameter of the cylindrical bowl

7

cm

So,

The radius of the cylindrical bowl

7

2

cm

The height of the bowl

4

cm

Therefore,

Capacity of the bowl= Volume of the bowl

2

rh

3

cm

22 7 7

4

7 2 2

cm

3

154

Now,

The required volume of soup needed for 250 patients

cm

3

250 154

cm

3

38500

38.5

The hospital will have to prepare 38.5 litres of soup

daily to serve 250 patients.

EXERCISE 13.7 (9)

Assume

22

7

, unless stated otherwise.

Question: 1

Find the volume of the right circular cone with

(i) Radius 6 cm, height 7 cm

(ii) Radius 3.5 cm, height 12 cm

Solution

(i)

Radius,

6r

cm

Height,

7h

cm

So,

Volume of a right circular cone

2

1

3

rh

cm

3

1 22

667

37

cm

3

264

(ii)

Radius,

3.5r

cm

Height,

12h

cm

So,

Volume of the right circular cone

2

1

3

rh

cm

3

1 22

3.5 3.5 12

37

cm

3

154 .

Question: 2

Find the capacity in litres of a conical vessel with

(i) Radius 7 cm, slant height 25 cm

(ii) Height 12 cm, slant height 13 cm

Solution

(i)

Radius,

7r

cm

Slant height,

25

cm

Let the height of the conical vessel be h cm.

So,

2 2 2 2 2

25 7 625 49 576hr

Therefore,

576 24h

cm

Volume of the conical vessel

2

1

3

rh

cm

3

1 22

7 7 24

37

cm

3

1232

The capacity of the vessel (in litres)

1232

1.232

1000

l l

(ii)

Height,

12h

cm

Slant height,

13l

cm

Let the base radius of the cone be r cm.

So,

2 2 2 2 2

13 12 169 144 25rh

Therefore,

cm25 5r

Volume of the conical vessel

2

1

3

rh

cm

3

1 22

5 5 12

37

cm

3

2200

7

Finally,

Capacity of the vessel (in litres)

2200 1

7 1000

11

35

l

l

Question: 3

The height of a cone is 15 cm. If its volume is 1570

cm

3

, find the radius of the base. (Use

3.14

)

Solution

Height of the cone,

15h

cm

Let the base radius of cone be r cm.

Volume of the cone

1570

cm

3

So,

2

1

1570

3

rh

Or,

2

1

3.14 15 1570

3

r

Or,

2

11

1570 100

3.14 5

r

Therefore,

100 10r

Thus,

The base radius of cone is 10 cm.

Question: 4

If the volume of a right circular cone of height 9 cm is

48 π cm

3

, find the diameter of its base.

Solution

Here,

Height of the right circular cone,

9h

cm

Let the base radius of the cone be r cm.

As,

Volume of the right circular cone

48

cm

3

,

So,

2

1

9 48

3

r

or,

2

3 48r

or,

2

48

16

3

r

Therefore,

cm16 4r

The diameter of the base of the right circular cone

28r

cm

Question: 5

A conical pit of top diameter 3.5 m is 12 m deep.

What is its capacity in kilolitres?

Solution

The diameter of the top of the conical pit,

2 3.5r

m

So,

The radius of the top of the conical pit,

1.75r

m

Now,

The depth of the conical pit,

12h

m

Therefore,

Volume of the conical pit

2

1

3

rh

m

3

1 22

1.75 1.75 12

37

m

3

38.5

(

m

3

11

kilolitres)

Thus,

The capacity of conical pit

38.5

kilolitres.

Question: 6

The volume of a right circular cone is 9856 cm

3

. If the

diameter of the base is 28 cm, find

(i) Height of the cone

(ii) Slant height of the cone

(iii) Curved surface area of the cone

Solution

(i)

The diameter of the base of the right circular cone,

2 28r

cm

The base radius of the right circular cone,

28

2

r

cm

14

cm

Let the height of the right circular cone be h cm.

Volume of the right circular cone

9856

cm

3

2

1

9856

3

rh

or,

1 22

14 14 9856

37

h

Finally,

cm

9856 3 7

48

22 14 14

h

Thus,

The height of the right circular cone

48

cm

(ii)

14r

cm and

48h

cm

Let l be the slant height of the right circular cone.

So,

2 2 2 2 2

48 14 2304 196 2500l h r

Therefore,

2500 50l

Thus,

The slant height of the right circular cone

50

cm.

(iii)

14r

m and

50l

cm.

Now,

Curved surface area of the right circular cone

r

cm

2

22

14 50

7

cm

2

2200

Question: 7

A right triangle ABC with sides 5 cm, 12 cm and 13

cm is revolved about the side 12 cm. Find the volume

of the solid so obtained.

Solution

After revolving the right triangle ABC about the side

AB (

12

cm), a right circular cone is formed as shown

in the figure.

Volume of obtained solid right circular

cone

2

1

3

rh

cm

3

1

25 12

3

cm

3

100

Question: 8

If the triangle ABC in the Question 7 above is revolved

about the side 5 cm, then find the volume of the solid

so obtained. Find also the ratio of the volumes of the

two solids obtained in Questions 7 and 8.

Solution

After revolving the right triangle ABC about the side

cm5,BC

a right circular cone is formed as shown

in the figure,

Therefore,

Volume of solid right circular cone

2

1

3

rh

cm cm, = cm

3

1

12 12 5 5 12

3

h r

cm

3

240

The ratio of corresponding volumes of both cones

100 :240

5:12

Question: 9

A heap of wheat is in the form of a cone whose

diameter is 10.5 m and height is 3 m. Find its volume.

The heap is to be covered by canvas to protect it from

rain. Find the area of the canvas required.

Solution

The diameter of the base of the cone,

2 10.5r

m

So,

The base radius of the cone,

5.25r

m

Now,

The height of the cone

3

m

Therefore,

Volume of the heap (cone)

2

1

3

rh

m

3

1 22

5.25 5.25 3

37

m

3

86.625

Now,

Let the slant height of the heap be ℓ.

2

2 2 2 2

3 5.25 9 27.5625 36.5625hr

Or,

36.5625 6.0467

The required canvas to protect wheat from rain =

Curved surface area of heap

r

m

2

22

5.25 6.0467

7

m

2

99.77

Exercise 13.8 (10)

(Assume

22

7

, unless stated otherwise.)

Question: 1

Find the volume of a sphere whose radius is

(i) 7 cm

(ii) 0.63 cm

Solution

(i)

The radius of the sphere,

7r

cm

So,

Volume of the sphere

3

4

3

r

cm

3

4 22

777

37

cm

3

4312

3

cm

3

1437.33

(ii)

The radius of the sphere,

0.63r

m

Volume of the sphere

3

4

3

r

m

3

4 22

0.63 0.63 0.63

37

m

3

1.05

Question: 2

Find the amount of water displaced by a solid

spherical ball of diameter

(i) 28 cm

(ii) 0.21 m

Solution

(i) The diameter of the spherical ball,

2 28r

cm

So,

The radius of the spherical ball,

28

2

r

cm

14

cm

Amount of water displaced by the solid spherical ball

= Volume of the spherical ball

3

4

3

r

cm

3

4 22

14 14 14

37

cm

cm

3

3

34496

3

11498.67

(ii)

The diameter of the spherical ball,

2 0.21r

m

The radius of the spherical ball,

m 0.105 m

0.21

2

r

Amount of water displaced by the solid spherical ball

= Volume of the spherical ball

3

4

3

r

m

3

4 22

0.105 0.105 0.105

37

m

3

0.0049

Question: 3

The diameter of a metallic ball is 4.2 cm. What is the

mass of the ball, if the density of the metal is 8.9 g per

cm

3

?

Solution

The diameter of the metallic ball,

2 4.2r

cm

So,

The radius of the metallic ball,

2.1r

cm

Therefore,

Volume of the metallic ball

3

4

3

r

cm

3

4 22

2.1 2.1 2.1

37

cm

3

38.808

Now,

The density of the metal is 8.9 g per cm

3

.

Thus,

The mass of the metallic ball

38.808 8.9 345.39gg

Question: 4

The diameter of the moon is approximately one -

fourth the diameter of the earth. What fraction of the

volume of the earth is the volume of the moon?

Solution

Let the diameter of the moon be 2r.

Then,

The radius of the moon

r

The diameter of the moon is one-fourth the diameter

of the earth,

So,

The diameter of the earth

8r

Therefore,

The radius of the earth

4r

Thus,

The volume of the moon

3

4

3

r

and,

The volume of the earth

3

4

(4 )

3

r

3

4

64

3

r

As a result,

The volume of the earth is 64 times the volume of the

moon.

Question: 5

How many litres of milk can a hemispherical bowl of

diameter 10.5 cm hold?

Solution

The diameter of hemispherical bowl,

2 10.5r

cm

So,

The radius of hemispherical bowl,

cm=5.25 cm

10.5

2

r

Volume of the bowl

3

2

3

r

cm

3

2 22

5.25 5.25 5.25

37

cm

3

303.1875

litres0.3

Therefore, the hemispherical bowl can hold 0.3 litres

of milk.

Question: 6

A hemispherical tank is made up of an iron sheet 1 cm

thick. If the inner radius is 1 m, then find the volume

of the iron used to make the tank.

Solution

Let R cm and r cm be the external and internal radii of

the hemispherical vessel respectively.

Given,

m1r

and,

m1.01R

(Thickness of iron sheet

cm1 .01

m)

Therefore,

Volume of iron used to make the tank = External

volume – Internal volume

R

33

22

33

r

33

2

()

3

Rr

3 3 3

2 22

[(1.01) (1) ]

37

m

m

3

44

(1.030301 1)

21

m

3

44

0.030301

21

m

3

0.06348

Question: 7

Find the volume of a sphere whose surface area is

cm

2

154

.

Solution

Let the radius of the sphere be r cm.

The surface area of sphere

154

cm

2

So,

2

4 154r

Or,

2

22

4 154

7

r

Or,

2

154 7

12.25

4 22

r

Therefore,

cm12.25 3.5r

Volume of the sphere

3

4

3

r

cm

3

4 22

3.5 3.5 3.5

37

cm

cm

3

3

539

3

179.67

Question: 8

A dome of a building is in the form of a hemisphere.

From inside, it was white - washed at the cost of Rs

498.96. If the cost of white- washing is Rs.2.00 per

square metre, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome.

Solution

(i)

Inside surface area of the dome

Total cost of white washing

Rate of white washing

2

2

498.96

2.00

249.48

m

m

(ii)

Let the radius of the dome be r.

Surface area of the dome

2

2 r

So,

2

22

2 249.48

7

r

Or,

2

249.48 7

39.69

2 22

r

Or,

39.69 6.3r

Therefore,

Volume of the air inside the dome

Volume of the

dome

3

2

3

r

m

3

2 22

6.3 6.3 6.3

37

3

m523.9

Question: 9

Twenty seven solid iron spheres, each of radius r and

surface area S are melted to form a sphere with surface

area S'. Find the

(i) Radius r' of the new sphere,

(ii) Ratio of S and S'.

Solution

(i)

Volume of 27 solid sphere of radius

3

4

27

3

rr

Let the volume of the new sphere of radius be r′.

So,

3 3

44

27

33

rr

Or,

3

33

27 3r r r

Therefore,

3rr

(ii)

Required ratio of the surface areas

22

2

2

1

' ' 9 9

2

2

4

4

3

s r r

sr

r

r

r

Question: 10

A capsule of medicine is in the shape of a sphere of

diameter 3.5 mm. How much medicine (in mm

3

) is

needed to fill this capsule?

Solution

The diameter of the spherical capsule,

2 3.5r

mm

So,

The radius of the spherical capsule,

1.75r

mm

Therefore,

Medicine required for filling the capsule = Volume of

spherical capsule

3

4

3

r

3

3

mm

mm

4 22

1.75 1.75 1.75

37

22.46

××

Exercise 13.9 (3)

Question: 1

A wooden bookshelf has external dimensions as

follows: Height = 110 cm, Depth

25cm

, Breadth

85 cm

(See the given figure). The thickness of the

plank is 5 cm everywhere. The external faces are to be

polished and the inner faces are to be painted. If the

rate of polishing is 20 paise per cm

2

and the rate of

painting is 10 paise per cm

2

, find the total expenses

required for polishing and painting the surface of the

bookshelf.

Solution

External length of bookshelf,

85 cml

External breadth of bookshelf,

25cmb

External height of bookshelf,

110cmh

So,

External surface area of bookshelf without the front

face of the bookshelf

2

2

2

2

85 110 2 85 25 25 110 cm

9350 9750 cm

19100 cm

lh lb bh

Now,

Area of the front face

2

2

2

85 110 75 100 2 75 5 cm

1850 750 cm

2600 cm

Therefore,

Required area to be polished

2

2

19100 2600 cm

21700 cm

As,

Cost of polishing

2

1cm area 0.20Rs

Hence,

Total cost of polishing

21700 0.20

4340

Rs

Rs

The length (l), breadth (b) and height (h) of each row

of the book shelf are

75 cm, 20 cm

and

30 cm

respectively,

Next,

Required area to be painted in one row

2

2

2

2

2 75 30 20 75 30 cm

4200 2250 cm

6450 cm

l h b lh

So,

Area to be painted in 3 rows

2

2

3 6450 cm

19350 cm

As,

Cost of painting 1 cm

2

area

0.10Rs

Therefore,

Cost of painting

2

19350 cm

area

19350 0.1

1935

Rs

Rs

Total expenses for polishing and painting

4340 1935 6275Rs Rs

Hence,

Total cost for polishing and painting the surface of the

bookshelf is Rs 6275.

Question: 2

The front compound wall of a house is decorated by

wooden spheres of diameter 21 cm, placed on small

supports as shown in the given figure. Eight such

spheres are used for this purpose, and are to be painted

silver. Each support is a cylinder of radius 1.5 cm and

height 7 cm and is to be painted black. Find the cost of

paint required if silver paint costs 25 paise per cm

2

and

black paint costs 5 paise per cm

2

.

Solution

The part of the sphere that is resting on the sphere will

be subtracted while finding the cost of silver paint,

So,

Required surface area to be silver painted

8

(Curved surface area of the sphere

area of

circle on which sphere is resting)

2 2 2

8 4 10.5 , 1.5()R r cm R cm r cm

22

2

2

()

(

8 4 10.5 2.25 cm

8 2.25 cm

8 438.

1)

75

4

cm

4

Therefore,

The cost of silver painting at the rate of 25 paise per

cm

2

22

8 438.75 0.25

7

2757.86

Rs

Rs

Now,

Required surface area to be black painted

8

Curved surface area of the cylinder

2

2

82

22

8 2 1.5 7 cm

7

528 cm

rh

Hence,

The cost of black painting at the rate of 5 paise per

cm

2

528 0.

20

05

6.4

Rs

Finally,

Total cost of painting

2757.86 26.40

2784.26

Rs

Rs

Question: 3

The diameter of a sphere is decreased by 25%. By

what percent does it’s curved surface area decrease?

Solution

Let the diameter of the sphere be d.

So,

The surface area of the sphere

2

2

d

4d

2

Now,

After decreasing the diameter of the sphere by 25%,

let the new diameter be

d

.

75 3

100 4

d

dd

Therefore,

The new surface area

2

4

2

d

2

2

2

13

4

24

9

4

64

9

16

d

d

d

As a result,

Decrease in surface area

2

9

1

16

d

2

7

16

d

Percentage decrease in the curved surface area of the

sphere

0

0

2

0

0

2

0

0

0

0

Decrease in surface area

100

Initial surface area

71

100

16

700

16

43.75

d

d