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
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









