CHAPTER - 13
SURFACE AREAS AND VOLUMES
Exercise – 13.1
Q1.
A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is
opened at 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.
Sol:
Length of plastic box,  = 1.5 m
Breadth of plastic box, b = 1.25 m
Deepness of plastic box, h = 65 cm 
(i)
65
m = 0.65 m
100
Area of plastic box (open at top)  2  b  h  b
 21.5  1.25 0.65   1.5  1.25  m2
 2  2.75  0.65  1.5  1.25  m2
 3.575  1.875  m2
 5.45 m2
Area of sheet required for box = Area of plastic box = 5.45 m2
(ii)
Cost of sheet for 1m2 = Rs. 20
Cost of sheet for 5.45 m2 = Rs. 20  5.45  Rs.109
Q2.
The length, breadth and height of a room are 5 m, 4 m and 3 m respectively.
Find the cost of white washing the walls of the room and ceiling at the rate
of Rs. 7.50 per m2 .
Sol:
Length of room,  = 5 m
Breadth of room, b = 4 m
Height of room, h = 3 m
As room is of cuboid shape, and
Area of cuboid = 2b  bh  h
So, area of room = 2b  bh  h
But, area of walls and ceiling of room  2  b h  b
( no white washing on floor)
 25  4 3  5  4 m2
 2  9  3  5  4  m 2
= 54  20  m 2
 74 m2
Cost of white washing for 1m2 =Rs. 7.50
Cost of white washing for 74 m2 = Rs. 7.50  74
= Rs. 555
So, cost of white washing on walls and ceiling is Rs. 555.
Q3.
The floor of a rectangular hall has perimeter 250 m. If the cost of painting
the four walls at the rate of Rs. 10 per m2 is Rs. 15000, find the height of
hall.
Sol:
Let , b and h be the length, breadth and height of hall.
Perimeter of rectangular hall = 250 m
And perimeter of rectangle = 2  b
 2  b  250 m......(1)
As we know, area of four walls = Lateral surface area of cuboid
 2  b h
Area painted at cost Rs. 10 = 1m2
Area painted at cost Rs. 1 =
1 2
m
10
Area painted at Cost Rs. 15000 =
1
 15000 m 2  1500 m2
10
So according to question,
Area of four walls = 1500 m2
 2  b  h  1500 m 2
 250  h  1500
h 
[From (1)]
1500
250
 h  6m
So, height of hall = 6 m
Q4.
The paint in a certain container is sufficient to paint an area equal to 9.375
m2 . How many bricks of dimensions 22.5 cm  10 cm  7.5 cm can be
painted out of this container?
Sol:
Length of brick,  = 22.5 cm
Breadth of brick, b = 10 cm
Height of brick, h = 7.5 cm
Total surface area of brick = 2b  bh  h
 222 .5  10  10  7.5  22 .5  7.5 cm2
 2225 .0  75 .0  168 .75  cm2
 2468 .75  cm2
 937.50 cm2
Let ‘n’ bricks can be painted by the paint in container.
Area of 1 brick = 937.50 cm2
Area of n bricks = 937.50 n cm2
The paint in a container is sufficient to paint an area = 9.375 m2
 9.375  10000 cm2
[ 1m2  10000 cm2 ]
 93750 cm2
 937.50 n cm2  93750 cm2
93750 cm2
n
937 .50 cm2
n
93750  100
 n  100
93750
So, 100 bricks can be painted with the paint in container.
Q5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm
long, 10 cm wide and 8 cm high.
(i) Which box has the greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Sol:
(i) Each edge of cubical box = 10 cm
As, Lateral surface area of cube = 4a 2
2
 Lateral surface area of cubical box = 4  10  cm 2
 4  10  10 cm2
= 400 cm2
And
Length of cuboidal box,  =12.5 cm
Breadth of cuboidal box, b = 10 cm
Height of cuboidal box, h = 8 cm
As, lateral surface area of cuboid = 2  b h
 Lateral surface area of cuboid box  2  b  h
 212 .5  10   8 cm2
 2  22.5  8  cm2
 360 cm2
So, we conclude
L.S.A. of cubical box > L.S.A. of cuboidal box.
[ 400 cm2  360 cm2 ]
L.S.A. of cubical box – L.S.A. of cuboidal box = 400  360  cm2  40 cm2
 L.S.A. of cubical box is greater than L.S.A. of cubiodal box by 40 cm2 .
(ii) Total surface area of cube = 6a 2
 Total surface area of cubical box = 6a 2
 6  10  cm2
2
 6  10  10  cm2
= 600 cm2
Total surface area of cuboid = 2b  bh  h 
 Total surface area of cuboidal box = 2b  bh  h 
 212 .5  10  10  8  12 .5  8 cm2
 2125 .0  80  100 .0  cm2
 2305 .0 cm2
= 610.0 cm2
So, total surface area of cubical box is smaller than total surface area of cuboidal
box.
T.S.A. of cuboidal box – T.S.A of cubical box  610 cm2  600 cm2
 610  600  cm2  10 cm2
 Total surface area of cubical box is smaller by 10 cm 2 .
Q6.
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 glass?
(ii) How much tape is needed for all the 12 edges?
Sol:
Length of greenhouse,  = 30 cm
Breadth of greenhouse, b = 25 cm
Height of greenhouse, h = 25 cm
As, green house is of cuboidal shape & Area of cuboid = 2b  bh  h
 Total surface Area of greenhouse = 2b  bh  h
 230  25  25  25  30  25  cm2
 2750  625  750  cm2
 22125  cm2
 4250 cm2
So, area of glass needed for greenhouse = area of greenhouse
 Area of glass = 4250 cm2
(ii) As tape is used on 12 edges of greenhouse.
12 edges including 4 lengths, 4 breadths and 4 heights.
 Amount of tape is needed for 12 edges  4  b  h
 430  25  25  cm
 480  cm
 320 cm
So, 320 cm tape is needed for its edges.
Q7.
Shanti Sweets Stall was placing an order for making cardboard boxes for
packing their sweets. Two sizes of boxes were required. The bigger of
25cm 20cm 5cm
dimensions
and
smaller
of
dimensions
15cm 12cm  5cm . For all the overlaps, 5% of the total surface area is
required extra. If the cost of the cardboard is Rs. 4 for 1000 cm2 , find the
cost of cardboard required for supplying 250 boxes of each kind.
Sol:
For bigger box,
Length = 25 cm
Breadth = 20 cm
Height = 5 cm
Total surface area of bigger box = 2 b  bh  h
 225  20  20  5  25  5 cm2
 2 500  100  125cm 2
 2725  cm2
 1450 cm2
Area required for overlapping = 5% of T.S.A.

5
 1450 cm2
100

725
cm2
10
 72.5 cm2
Total area of bigger box including extra area for overlapping
= Extra Area + T.S.A. of Bigger box
 72 .5  1450  cm2
 1522.5 cm2
So, Area of cardboard for 1 bigger box = 1522.5 cm2
Area of cardboard for 250 bigger boxes  1522 .5  250  cm2  380625 cm 2
For smaller box,
Length = 15 cm
Breadth = 12 cm
Height = 5 cm
Total surface area of smaller box  2b  bh  h
 215  12  12  5  15  5  cm2
 2180  60  75  cm2
 2315 cm 2
 630 cm2
Area required for overlapping = 5 % of T.S.A.

5
 630 cm2
100

315
cm2
10
 31.5 cm2
Total area of cardboard required for small box including extra area
= extra area + T.S.A. of small box
= 31 .5  630  cm2
 661.5 cm2
Area of cardboard for 1 small box = 661.5 cm2
Area of cardboard for 250 small boxes = 661 .5  250  cm2
 165375.0 cm2
 165375 cm2
Total cardboard required for supplying 250 boxes of each kind
= cardboard required for 250 big boxes+ cardboard required for 250 small boxes
= 380625  165375  cm2  546000 cm2
Cost for 1000 cm2 cardboard = Rs. 4
Cost for 1 cm2 cardboard = Rs.
4
1000
Cost of 546000 cm2 cardboard = Rs.
4
 546000 = Rs. 2184
1000
Hence the cost of cardboard required for supplying 250 boxes of each kind
is Rs. 2184
Q8.
Parveen wanted to make a temporary shelter for her car, by making a box –
like structure with tarpaulin that covers all its four sides and 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 4 m  3 m ?
Sol:
Length of shelter,  = 4 m
Breadth of shelter, b = 3 m
Height of shelter, h = 2.5 m
Now, tarpaulin is only needed for four walls and top.
Tarpaulin required  2  b h  b
 2  4  3  2.5  4  3 m 2
 2  7  2.5  12  m 2
 35 .0  12  m 2
 47 .0 m 2
 47 m 2
So, 47 m2 tarpaulin is required for shelter.
Exercise - 13.2
Q1.
The curved surface area of a right circular cylinder of height 14 cm is 88
cm2 . Find the diameter of the base of the cylinder.
Sol:
Let r be the radius of cylinder.
Height of cylinder, h  14 cm
Curved surface area of cylinder = 88 cm2
& curved surface area of cylinder = 2rh
 2rh  88 cm2
 2
22
 r  14  88
7
 88  r  88
Divide by 88 on both sides, we get:

88
88
r
88
88
 r  1cm
Diameter = 2 radius
 2  1 cm  2 cm
So, diameter of base of cylinder is 2 cm.
Q2.
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 meters of sheet are
required for the same?
Sol:
Height of cylinder, h = 1m
Diameter of base of cylinder, d = 140 cm

140
m
100

14
m
10
1


1 cm  100 m


Radius of base of cylinder, r 
diameter
2
14
14
7
 10 m 
m
m
2
10  2
10
 It is for a closed cylinder, so sheet required for it is equal to its total surface
area.
 Sheet required = Total surface area of cylinder
 2r r  h

22 7  7

 2 

 1 m2

7 10  10 

 2
22 7  7  10  2

m
7 10  10 
22 7 17  2

 2 

 m
7 10 10 

 2  22  17  2

m
 10  10 
 748  2

m
 100 
 7.48 m2
So, 7.48 m2 sheet is required.
Q3.
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 figure 13.11). Find its
(i) Inner curved surface area
(ii) Outer curved surface area
(iii) Total surface area
Fig.13.11
Sol: (i) For inner curved surface area:
Height (length) of pipe, h = 77 cm
Inner radius of pipe, r =
Diameter (inner) d

2
2

4
cm  2cm
2
Inner curved surface area of pipe = 2rh
22


 2  77 cm2
= 2
7


 968 cm2
So, 968 cm2 is the inner curved surface area of pipe.
(ii) For outer curved surface area,
Outer diameter of pipe, D = 4.4 cm
Outer radius of pipe, R =

D
2
4.4
cm  2.2 cm
2
Outer curved surface area of pipe  2Rh
22


 2
 2.2  77 cm2
7


 2  22  2.2  11cm2
2
= 1064.8 cm
So, 1064.8 cm2 is the outer curved surface area of pipe.
(iii)
Total surface area:
As pipe is empty from inside
So total surface area of pipe
= outer curved area + Inner curved area + area of both bases

= 2RH  2rh  2 R 2  r 2



22

2.22  22  cm2
 1064 .8  968  2 
7


[From (i) , (ii)]
22

4.84  4 cm2
 1064 .8  968  2 
7


22


 1064 .8  968  2 
 0.84  cm2
7


 1064 .8  968  2  22  0.12  cm2
 1064 .8  968  5.28  cm2
 2038.08 cm2
So, Total surface Area of pipe is 2038.08 cm2
Q4.
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 m2 .
Sol:
Diameter of roller = 84 cm
Radius of roller, r 
diameter
2

84
cm  42 cm
2

42
m
100
[ 1 cm 
1
m]
100
Height (length) of roller, h = 120 cm

120
m
100
[ 1 cm 
1
m]
100
Now roller is of cylindrical shape
Curved surface area of cylinder = 2rh
Curved surface area of roller = 2rh
 2
22 42 120 2


m
7 100 100

2  22  6  12 2
m
10  100

3168 2
m
1000
Area covered in 1 revolution =
3168 2
m
1000
Area covered in 500 revolutions =
3168
 500 m2
1000
 1584 m2
So, Area of playground = 1584 m2
Q5.
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 m2 ?
Sol:
Height of pillar, h = 3.5 m
Diameter of pillar, d = 50 cm
diameter
2
Radius of pillar, r 
50
cm
2

 25 cm 
25
m
100
[ 1 cm 
1
m]
100
Pillar is cylindrical.
 Curved surface area of pillar
= Curved surface area of cylinder = 2rh


 C.S.A. of pillar =  2 
22 25 35  2


m
7 100 10 
 1 11  5  1 2

m
 1 10  1 

55 2
m  5.5 m2
10
Cost of painting 1m2 area  Rs.12.50
Cost of painting 5.5 m2 area = Rs. 12.50  5.5 = Rs. 68.750 = Rs. 68.75
So, cost of painting curved surface of pillar is Rs. 68.75.
Q6.
Curved surface area of a right circular cylinder is 4.4 m2 . If the radius of the
base of the cylinder is 0.7 m, find its height.
Sol:
Let h be the height of cylinder.
Radius of cylinder, r = 0.7 m
Curved surface area of cylinder = 2rh
But we have given:
Curved surface area of cylinder = 4.4 m2
So, 2rh  4.4 m2
2
22
 0.7  h  4.4
7
Multiply by 7 on both sides, we get:
2
22
 0.7  7  h  7  4.4
7
 2  22  0.7  h  30.8
 30.8  h  30.8
Divide by 30.8 on both sides, we get:
30.8  h 30.8 m

30.8
30.8
 h  1m
So, the height of cylinder is 1 m.
Q7.
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 Rs. 40 per m2 .
Sol (i) Well is of circular base and some depth.
 Well is of cylindrical shape.
Height of well, h = depth of well = 10 m
Inner diameter of well, d = 3.5 m
Inner radius of well, r 
d 3.5

m  1.75 m
2
2
Inner curved surface area of well
= Curved surface area of cylinder = 2rh
22


 2
 1.75  10 m2
7


 2  22  0.25  10 m2
 110 m2
 Inner curved surface area of well is 110 m2
(ii) Cost of plastering 1m2 area = Rs. 40
Cost of plastering110 m2 area = Rs. 40  110  Rs.4400.
Q8.
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.
Sol:
Heating system is of cylindrical shape.
Height of pipe, h  length of pipe = 28 m
Radius of pipe, r =
diameter of pipe
2

5
cm  2.5 cm
2

2. 5
m
100
[ 1 cm =
1
m]
100
 0.025 m
Total surface area of heating system
 2r r  h
 [2 
22
 0.025(0.025  28)] m2
7
22


 2
 0.025  28 .025 m2
7


 44  0.025  28 .025  2

m
7


 1.1 28 .025  2

m
7


 30 .8275  2

m
7


 4.40 m2
(Approx.)
So, the total radiating surface is 4.40 m2 .
Q9.
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 height.
(ii)
How much steel was actually used, if
1
of the steel actually used was
12
wasted in making the tank.
Sol (i) Tank is of cylindrical shape.
Height of tank, h = 4.5 m
Diameter of tank= 4.2 m
Diameter
2
Radius of tank, r =

4.2
m  2.1 m
2
Curved surface Area of cylinder = 2rh
 Curved surface Area of tank = 2rh
22


 2
 2.1 4.5 m 2
7


 44  0.3  4.5 m2
= 13 .2  4.5 m 2
 59.40 m2
So, lateral surface area of tank is 59 .40 m 2 .
(ii) As tank is closed,
Steel used in tank = Total surface area of tank
 2r r  h
 22

 2 
 2.12.1  4.5 m 2
7


22


 2
 2.1 6.6 m 2
7


 2  22  0.3  6.6 m2
 44  1.98 m2
 87.12 m2
Let `x' m2 steel be actually used in making the tank.
12
1045.44
1

2
 x1 
 87.12 
 95.04 m2 .
  87 .12 m  x 
11
11
 12 
Q10. In figure 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 height of 30
cm. A margin of 2.5 cm is to be given for folding it over the top and bottom
of the frame. Find how much cloth required for covering the lampshade.
Fig. 13.12
Sol:
Frame is of cylindrical shape. So the decorative cloth used is also in the form of
cylindrical shape.
Height of frame = (margin fold + actual height + margin)
 2.5  30  2.5cm
 35 cm
Diameter of frame, d = 20 cm
Radius of frame, r =

Diameter
2
20
cm  10 cm
2
Curved surface Area of frame = C.S.A. of cylinder = 2rh
22


 2
 10  35 cm2
7


 44  10  5 cm2
= 2200 cm2
But cloth used for covering frame = C.S.A. of frame.
Hence, 2200 cm2 cloth is used for making cover of frame.
Q11. 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 competitions with cardboard. If there
were 35 competitors, how much cardboard was required to be bought for
the competition?
Sol:
Radius of pen holder, r = 3 cm
Height of pen holder, h = 10.5 cm
Surface Area of 1 Pen holder
= curved surface area of penholder + base area of pen holder
= 2rh  r 2
22
22


 3  10 .5 
 3  3 cm2
= 2
7
7


22  3  3 

2
  2  22  3  1.5 
 cm
7


22  9  2

  44  4.5 
cm
7 

198 

2
 198 
 cm
7 


1386 198
1584
cm2 
cm2
7
7
As 1 competitor makes 1 pen holder
 35 competitors make 35 pen holders.
And surface area of 1 pen holder = 
1584
cm2
7
 Surface area of 35 pen holders =
1584
 35 cm2
7
 7920 cm2
So, 7920 cm2 cardboard is required for competition.
Exercise - 13.3
Assume  
22
, unless stated otherwise.
7
Q.1
Diameter of the base of cone is 10.5 cm and its slant height is 10 cm. Find
its curved surface area.
Sol:
Slant height of cone,  = 10 cm
Diameter of base of cone = 10.5 cm
Radius of base, r =

Diameter
2
10.5
105
cm 
cm
2
2  10
Curved surface area of cone = r
 22 105



 10 cm2
 7 2  10

 11 15 cm2
 165 cm2
Q2.
Find the total surface area of a cone, if its slant height is 21 cm and
diameter of its base is 24 m.
Sol:
Slant height of cone,  = 21 m
Diameter of base of cone, d = 24 m
Radius of base of cone, r =

Diameter
2
24
m  12m
2
Total surface Area of cone = r   r 
 22

   12 21  12 m2
7

 22


 12  33 m 2
 7


8712 2
m
7
T.S.A. of cone = 1244.57 m2
So, total surface area of cone is 1244 .57 m 2 .
Q3.
Curved surface area of a cone is 308 cm2 and its slant height is 14 cm.
Find:
(i) Radius of the base
(ii) Total surface area of cone.
Sol :(i) Let the radius of base of cone be r.
Slant height of cone,   14 cm
We know that, curved surface area of cone  r
 22

 308   
 r  14 
 7

 308   22  2  r 
 22  2  r   308 
 44 r  308
r 
308
44
 r  7 cm
 Radius of base of cone is 7 cm.
(ii)
Total surface area of cone = r   r 
 22

=   714  7 cm2
7

 22

=   7  21cm2
7

 22  21 cm2
 462 cm2
 Total surface area of cone is 462 cm2 .
Q4.
(i)
(ii)
A conical tent is 10 cm high and radius of its base is 24 m. Find
Slant height of the tent
Cost of the canvas required to make the tent, if the cost of 1 m2 canvas is
Rs. 70
Sol:
Let ABC be a conical tent.
10 m
A
B
C
O
24 m
Now, height of cone, h = 10 m
Radius of base, r = 24 m
(i) Let  be the slant height of cone.
In AOB
AB 2  AO2  OB 2
[By Pythagoras theorem]
  2  h2  r 2


 (10 m) 2  (24 m) 2  10 2 m2  24 2 m2  10   24  m2
2
2
 100  576 m2
 676 m2
 26 m 
2
   26 m
(Taking square root on both sides)
Therefore, slant height of cone is 26 m.
(ii)
Canvas required for tent = curved surface area of cone = r
 22


 24  26 m 2
 7


13728 2
m
7
Cost of 1 m 2 canvas = Rs. 70
Cost of
13728 2
13728
m canvas = Rs. 70 
7
7
 Rs.10  13728
 Rs.137280
 The cost of canvas required for making tent is Rs. 137280.
Q5.
Sol:
What length of tarpaulin 3 m wide will be required to make conical tent of
height 8 m and base radius 6m? Assume that the extra length of material
that will be required for stitching margins and wastage in cutting is
approximately 20 cm (use   3.14 ).
Height h of conical tent 8 m
Radius (r) of conical tent = 6 m
Slant height   of cone =

h2  r 2
8m2  6m2
 64m2  36m2
=

100m2
10m2
=10 m
   10 m
Curved surface area of conical tent = r
 3.14  6  10 m2
 18 .84  10 m2
 188.4 m2
Now, 20 cm tarpaulin is wasted in cutting and margin.
Let x be the actual length of tarpaulin. So, length of tarpaulin used for making
20
) m  ( x  0.2) m
tent  ( x 
100
Width of tarpaulin = 3 m
 Tarpaulin used in making tent = C.S.A. of conical tent
 x  0.2m  3m  r
 x  0.2m  3m  188 .4 m2
188 .4 m 2
 x  0.2m 
3m
 x  0.2 m  62.8 m
 x  62.8 m  0.2 m
 x  63.0 m
 x  63 m
So, required length of tarpaulin is 63 m.
Q6.
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
.
cm
13
Sol:
5m
C
B
12 cm A
Slant height   of conical tomb = 25 m
Diameter of base of conical tomb = 14 m
Radius (r) of base of conical tomb =
Diameter 14

m  7m
2
2
Curved surface Area of conical tomb = r
 22


 7  25 m2
 7

 22  25 m2
 550 m2
Cost of white washing 100 m2 area = Rs. 210
Cost of white washing 1m2 area = Rs.
210
100
Cost of white washing 550 m2 area  Rs.
210
 550  Rs.21 55  Rs.1155.
100
Therefore, cost of white-washing curved surface of conical tomb is Rs. 1155.
Q7.
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.
Sol:
Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 m
Slant height (  ) of conical cap =
h2  r 2

24 cm 2  7 cm2
 576 cm2  49 cm2
 625 cm2
 252 cm2 
25 cm2
= 25 cm
   25 cm
Curved surface area of 1 conical cap = r
 22


 7  25 cm2
 7

 22  25 cm2
 550 cm2
Sheet required for 1 cap = C.S.A. of conical cap = 550 cm2
Sheet required for 1 conical cap = 550 cm2
 Sheet required for 10 conical caps = 10  550 cm2  5500 cm2
Therefore, 5500 cm2 sheet is required for making 10 conical caps.
Q8.
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 height 1 m. If the outer side of each of cone is to be painted
and the cost of painting is Rs. 12 per m2 , what will be the cost of painting
all these cones?
(Use   3.14 and
Sol:
1.04  1.02 )
Height h of hollow cone = 1m
Diameter (d) of base of hollow cone = 40 cm
Radius (r) of base of hollow cone =
Diameter
2
40 cm
20
 20 cm 
m  0.2 m
2
100
1
m]
[ 1 cm =
100

Slant height of hollow cone,   h2  r 2
=
1m2  0.2m2
 1m2  0.04m2
 1.04 m2
 1.02 m
(Given)
Outer area of 1 cone = C.S.A. of 1 cone = r
 3.14  0.2  1.02 m2
 0.64056 m 2
Outer area of 1 cone = 0.64056 m2
Outer area of 50 cones = 50  0.64056 m2
 32.02800 m2
 32.028 m2
Cost of painting 1 m2 area = Rs. 12
Cost of painting 32.028 m2 area = Rs 12  32 .028   Rs.384 .336
 Rs.384.34
So, Rs. 384.34 will be spent for painting all hollow cones.
Exercise - 13.4
Assume  
22
, unless stated otherwise.
7
Q.1
Find the surface area of a sphere of radius:
(i)
10.5 cm
Sol:
Radius (r) of sphere = 10.5 cm
Surface area of sphere = 4r 2
 4
22
2
 10.5 cm
7
22


 4
 10 .5  10 .5 cm2
7


 4  22  1.5  10 .5 cm2
 88  15 .75 cm2
 1386.00 cm2
 1386 cm2
 Surface area  1386 cm 2 .
(ii)
5.6 cm
Sol : Radius (r) of sphere = 5.6 cm
Surface area of sphere = 4r 2
 4
22
2
 5.6 cm
7
22


 4
 5.6  5.6 cm2
7


 4  22  0.8  5.6 cm2
 88  4.48 cm2
 394.24 cm2
 Surface area of sphere of radius 5.6 cm is 394 .24 cm 2 .
(iii)
14 cm
Sol:
Radius (r) of sphere = 14 cm
Surface area of sphere = 4r 2
= 4
22
2
 14 cm
7
22


 4
 14  14 cm2
7


 4  22  2  14 cm2
 2464 cm2
 Surface area of sphere with radius 14 cm is 2464 cm2 .
Q2.
Find the surface area of a sphere of diameter:
(i)
14 cm
Sol:
Diameter of sphere, d = 14 cm
Radius of sphere, r 

Diameter
2
14
cm  7 cm
2
Surface area of sphere = 4r 2
 4
22
2
 7 cm
7
22

 2
 4
 7  7 cm
7


 616 cm2
 Surface area of sphere with diameter 14 cm is 616 cm2 .
(ii)
21 cm
Sol:
Diameter (d) of sphere = 21 cm
Radius (r) of sphere =

Diameter
2
21
cm
2
Surface area of sphere = 4r 2
 4
22  21

  cm
7 2

2
22 21 21  2

 4
  cm
7
2 2

 22  3  21cm2
 1386 cm2
 Surface area of sphere with diameter 21 cm is 1386 cm2 .
(iii)
3.5 m
Sol:
Diameter (d) of sphere = 3.5 m =
Radius of sphere, r 
35
m
10
Diameter 35 / 10

m
2
2

35
7
m m
2  10
4
22  7 
Surface area of sphere = 4r  4 
  m
7 4 
2
2
22 7 7  2

 4
  m
7 4 4

7

  22  m 2
4


154 2
m  38.5 m2
4
 The surface area of sphere with diameter 3.5 m is 38 .5 m 2 .
Q3.
Find the total surface area of a hemisphere of radius 10 cm. (use   3.14 )
Sol:
Radius of hemisphere, r = 10 cm
Total surface area of hemisphere = 3r 2
 3  3.14  (10 cm)2
 3  3.14  10  10 cm2
 3  3.14  100 cm2
 9.42  100 cm2
 942 cm2
 The total surface area of given hemisphere is 942 cm2 .
Q4.
Sol:
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.
Radius r1  of spherical balloon = 7 cm
Radius r2  of spherical balloon, when air is pumped into it = 14 cm
Surface area of balloon of radius r1  = 4r1 
2
Surface area of balloon of radius r2  = 4r2 
2
Ratio of surface areas of balloon in two stages

Surface area of balloon with radius (r1 )
Surface area of balloon with radius r2 

4(r1 )2
4(r2 )2

(r1 )2
77
1 1 1



2
14  14 2  2 4
(r2 )
 Ratio of surface areas = 1: 4.
Q5.
A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the
cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2 .
Sol:
Diameter of hemispherical bowl = 10.5 cm
Radius (r) of hemispherical bowl =
10.5
cm
2
Inner curved surface area of hemispherical bowl
 2r 2
 2
22  10.5


cm 
7  2

2
22 10 .5 10 .5  2

 2


cm
7
2
2 

 11 1.5  10 .5 cm2
 173.25 cm2
Cost of tin-plating on 100 cm2 = Rs. 16
Cost of tin-plating on 1 cm2 area = Rs. 16/100
Cost of tin-plating on 173.25 cm2 area = Rs
 Rs.
16
 173.25
100
2772.00
100
 Rs.
2772
100
 Rs.27.72
 Cost of tin-plating on inside of hemispherical bowl is Rs. 27.72
Q6.
Find the radius of a sphere whose surface area is 154 cm2 .
Sol:
Let the radius of sphere be r.
Surface area of sphere = 154 cm2
But, surface area of sphere = 4 r 2
 4r 2  154 cm2
 4

22 2
 r  154 cm2
7
88 2
 r  154 cm2
7
Multiply by

7
on both sides, we get
88
7 88 2
7

r 
 154 cm2
88 7
88
 r2 
77
cm2
4
77
7

r 
cm2   cm 
2 2
2

2
2
Taking square root on both sides, we get:
7

r   cm 
2

2
2
r 
7
cm
2
 r  3.5 cm
So, radius of sphere is 3.5 cm.
Q7.
The diameter of the moon is approximately one fourth of the diameter of
earth. Find the ratio of their surface areas.
Sol:
Let the diameter of earth be d, and then the diameter of moon be
Radius of earth, re 
diameter of earth
d
=
2
2
Radius of moon, rm 
diameter of moon
2
d
d
d
 4 

2 2 4 8
 d
2
Surface area of earth, Se  4(re )  4 
2
Se  4   
 4 
2
d2
22
d2
4
Similarly, surface area of moon, sm  4rm 
2
2
d2
 d
 4     4  2
8
8
d2
 4 
64
 d2 
4 
64
Sm
4
1
  2

Required ratio =
Se
 d  64 16
4 
 4
 Ratio of surface areas of moon and earth is 1: 16.
d
.
4
Q8.
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 bowl.
Sol:
Inner radius of bowl = 5 cm
Thickness of bowl = 0.25 cm
Outer radius of hemispherical bowl, r = Inner radius + thickness
= 5 cm + 0.25 cm = 5.25 cm
2
Outer curved surface area of hemispherical bowl = 2r
 2
22
2
 5.25 cm
7
22


 2
 5.25  5.25 cm2
7


 2  22  0.75  5.25 cm2
 44  3.9375 cm2
 173.25 cm2
So, outer curved surface area of hemispherical bowl is 173 .25 cm 2 .
Q9.
A right circular cylinder just encloses a sphere of radius r (see figure
13.22). Find
(i)
Surface area of sphere
(ii)
curved surface area of cylinder
(iii)
Ratio of the areas obtained in (i) and (ii).
Fig. 13.22
Sol :(i) Radius of sphere = r
Surface area of sphere = 4r 2
(ii)
r
r
r+r
r
Radius of base of cylinder = radius of sphere = r
Height of cylinder h = diameter of sphere = r  r  2r
Curved surface area of cylinder = 2rh
 2r 2r 
 2  2    r  r  4r 2
Curved surface area of cylinder = 4r 2
(iii) Ratio of areas in (i) and (ii) 

Surface area of sphere
Curved surface area of cylinder
4r 2 1
  1: 1
4r 2 1
Exercise - 13.5
Q1.
A matchbox measures 4 cm  2.5 cm  1.5 cm . What will be the volume of a
packet containing 12 such boxes?
Sol:
Length of matchbox,  = 4 cm
Breadth (b) of match box = 2.5 cm
Height (h) of match box = 1.5 cm
So, matchbox is of cuboidal shape
Volume of cuboid =   b h
Volume of 1 matchbox =   b h
= 4 cm 2.5 cm1.5 cm
= 4  2.5  1.5 cm3
3
= 15 cm
Volume of 1 matchbox = 15 cm3
Volume of 12 matchboxes  180 cm3
Volume of packet containing 12 matchboxes = Volume of 12 matchboxes
= 180 cm3
 The volume of packet is 180 cm3 .
Q2.
A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many
liters of water it hold? (1 m3  1000 Lt.)
Sol:
Length (  ) of water tank = 6 m
Breadth (b) of water tank = 5 m
Height (h) of water tank = depth of tank = 4.5 m
Volume of water tank =   b  h  (6  5  4.5) m3
 135.0 m3
Volume of water tank= 135 m3
Amount of water in tank = Volume of water tank
=135 m3
 135 1000Lt.
[ 1 m3 = 1000 Lt.]
 135000Lt.
Therefore, 135000 Lt. water is contained in water tank.
Q3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to
hold 380 cubic meters of a liquid?
Sol:
Let the height of cuboidal vessel be h.
Length of cuboidal vessel,  = 10 m
Breadth of cuboidal vessel, b = 8 m
Volume of cuboidal vessel = 380 m3
But as we know,
Volume of cuboidal vessel =   b  h
   b  h  380 m3
 (10 m)  (8 m)  h  380 m3
 (80 m 2 )  h  380 m3
h
380 m3 38

m  4.75 m
8
80 m2
So, vessel should be 4.75 m high to hold the given amount of liquid.
Q4.
Sol:
Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3m deep at
the rate of Rs. 30 per m3 .
Length   of cuboidal pit = 8 m
Breadth (b) of cuboidal pit = 6 m
Depth (h) of cuboidal pit = 3 m
Volume of cuboidal pit =   b  h
 8m  6m  3m
 8  6  3 m 3
 144 m3
Cost of digging 1m3 = Rs. 30
Cost of digging 144 m3 = Rs. 30 144
 Rs.4320
Therefore, cost of digging of cuboidal pit will be Rs. 4320.
Q5.
The capacity of a cuboidal tank is 50000 liters of water. Find the breadth of
tank, if the length and depth of tank are respectively 2.5 m and 10 m.
Sol : Let the breadth of tank be b.
Length   of cuboidal tank = 2.5 m
Depth (h) of cuboidal tank = 10 m
Volume of cuboidal tank =   b  h
But given capacity of cuboidal tank = Volume of cuboidal tank
= 50000 Lt.

50000 3
m
1000
3
= 50 m
3
   b  h = 50 m
 2.5m  b  10m  50 m3
 b  2.5  10 m2  50 m3
 b  25m2  50 m3
b 
50 m3
25 m 2
 b  2m
Therefore, the breadth of given tank is 2 m.
Q6.
Sol:
A village having a population of 4000 requires 150 liters 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?
Length   of tank = 20 m
Breadth (b) of tank = 15 m
Height (h) of tank = 6 m
So, tank is of cuboidal shape.
Volume of tank =   b  h
= 20m  15m  6m
 20  15  6m3
 1800 m3
 1800 1000 Lt.
1m
3

 1000 Lt.
 1800000Lt.
Amount of water in tank = 1800000 Lt.
Water consumed by 1 person in 1 day = 150 Lt.
Water consumed by 4000 persons in 1 day = 150  4000 Lt.  600000 Lt.
Let people use water of tank in ‘n’ days.
So water consumed by 4000 persons in ‘n’ days  600000  nLt.  600000 nLt.
 600000 n Lt.  1800000 Lt.
n
1800000Lt.
n3
600000 Lt.
So, water of tank will last for 3 days.
Q7.
A godown measures 40 m  25 m  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 a
godown.
Sol:
Length   of godown = 40 m
Breadth (b) of godown = 25 m
Height (h) of godown = 10 m
So, godown is cuboidal in shape.
 Volume of godown    b  h
 40  25  10 m3
 10000 m3
Now,
Length  1  of wooden crates = 1.5 m
Breadth b1  of wooden crates = 1.25 m
Height h1  of wooden crates = 0.5 m
 Wooden crate is also cuboidal in shape.
So, Volume of wooden crates =  1  b1  h1
 1.5  1.25  0.5  m3
 0.9375 m3
Now,
Number of wooden crates that can be stored into godown




Volume of godown
Volume of 1 wooden crate
10000 m3
0.9375 m3
10000 10000
9375
100000000
 10666.66
9375
So, 10666 wooden crates of given dimensions can be stored into the godown.
Q8.
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.
Sol:
Side of cube, a = 12 cm
Volume of cube  a3
 12 cm
3
 12  cm3
3
 1728 cm3
Let the side of smaller cube be a 1 .
Volume of 1 smaller cube  a1 
3
Volume of 8 smaller cubes  8a1 
3
As the solid cube is divided into 8 small cubes.
 Volume of 8 small cubes = Volume of solid cube
3
3
 8a1   1728 cm
 a1  
3
1728 3
cm
8
3
 a1   216  cm3
3
3
 a1   6 cm
Taking cube root on both sides, we get:
a    6  
1
1
3 3
1
3 3
 a1  6 cm
So, the side of small cube is 6 cm.
Surface area of solid cube  6a2
 612  cm 2
2
 6  144 cm2
 864 cm2
Surface area of small cube  6a1 
2
 6  6 cm
2
 6  6  6  cm2
 216 cm2
Required ratio =
Surface area of solid cube 864 4

  4 :1
Surface area of small cube 216 1
So, 4 : 1 is the required ratio.
Q9.
A river 3m deep and 40m wide is flowing at the rate of 2 km per hour . How
much water will fall into the sea in a minute?
Sol:
Depth of river, h = 3 m
Width of river, b = 40 m
Distance covered by water in 1 hour = 2 km  2  1000m  2000 m
Distance covered by water in 60 minutes = 2000 m
Distance covered by water in 1 minute =
[ 1 hour = 60 minutes]
2000
100
m 
m
60
3
Volume of water flowing in 1 minute    b  h
[River is cuboidal in shape, distance = length]
 100

 3  40 m 3
=
 3

 100  40 m3
 4000 m3
 , 4000 m3 amount of water will fall into the sea in a minute.
Exercise - 13.6
Q1.
The circumference of the base of a cylinder vessel is 132 cm and its height
1000 cm3  1Lt.
is 25 cm. How many liters of water can it hold?
Sol:
Let the radius of base of cylindrical vessel be r .

Height of cylindrical vessel = 25 cm
Circumference of base cylindrical vessel = 132 cm
As base of cylindrical vessel is circular in shape.
So, circumference of base  2r
 2r = 132 cm
 2

(by given)
22
 r  132 cm
7
44
 r  132 cm
7
r 
7
 132 cm
44
 r = 21 cm
Volume of cylindrical vessel  r 2h
 22

2

 21  25 cm3
 7

 22


 21  21  25  cm3
 7

 22  3  21  25 cm3
 66  525  cm3
 34650 cm3


1


3
1 cm  1000 Lt.


34650
Lt.
1000
 34.65 Lt.
Therefore, 34.65 Lt. water can be hold in given horizontal vessel.
Q2.
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 cm3 of wood has a mass of 0.6 g.
Sol:
Inner diameter (d) of cylindrical pipe = 24 cm
Inner radius (r) of cylindrical pipe 
Inner diameter
2

24
cm  12 cm
2
Outer diameter (D) of cylindrical pipe = 8 cm
Outer radius (R) of cylindrical pipe 

Outer diameter
2
28
cm  14 cm
2
Length (h) of pipe = 35 cm
Volume of wood in cylindrical pipe = Outer volume of pipe - Inner volume of pipe
2
2
= R h  r h


  R2  r 2 h


 22
14 2  12 2  35  cm3

7

 22

   35 196  144  cm3
7

 22


 35  52 cm3
 7

 110  52 cm3
 5720 cm3
Mass of 1 cm3 of wood = 0.6 g
Mass of 5720 cm3 of wood  5720 0.6 g
 3432.0 g
 3432 g

3432
kg
1000
[ 1 g =
1
kg ]
1000
 3.432 kg
Therefore, the mass of wood is 3.432 kgs.
Q3.
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?
Sol:
For Tin can:
Length   of can = 5 cm
Width (b) of can = 4 cm
Height (h) of can = 15 cm
So, tin can is cuboidal in shape.
 Volume of tin can    b  h
 5  4  15  cm3
 300 cm3
For plastic cylinder Pack:
Diameter (d) of the plastic cylinder pack = 7 cm
Radius (r) of base of plastic cylinder pack 
Diameter 7
 cm
2
2
Height (h) of plastic cylinder pack = 10 cm
Volume of plastic cylinder pack  r 2h
 22  7  2


    10 cm3
 7 2



 22 7 7


   10 cm3
 7 2 2

 11  7  5 cm3
 385 cm3
So, it is clear plastic cylinder pack has greater capacity than tin can.
Volume of plastic cylinder pack – Volume of tin can
 385 cm3  300 cm3
 85 cm 3
 Plastic cylinder pack has more capacity by 85 cm3.
Q4.
If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then
find:
(i) Radius of its base
(ii) its volume. (Use   3.14 )
Sol:(i) Let r be the radius of base of cylinder.
Height (h) of cylinder = 5 cm
The lateral surface area of cylinder = 94.2 cm2
But, Lateral surface area of cylinder  2rh
 2rh  94.2 cm2
 2  3.14  r  5  94.2
 31.4  r  94.2
Divide by 31.4 on both sides, we get:
r 
94.2
31.4
942
 r  10
314
10
r 
942 10

cm
10 314
 r  3 cm
So, the radius of base of cylinder is 3 cm.
(ii) Volume of cylinder  r 2h


 3.14  3   5 cm3
2
 3.14  9  5 cm3
 3.14  45 cm3
 141.30 cm3
Therefore, the volume of cylinder is 141.3 cm3.
Q5.
It costs Rs. 2200 to paint the inner curved surface of a cylindrical vessel
10m deep. If the cost of painting is at the rate of Rs. 20 per m2, find
(i)
Inner curved surface area of vessel,
(ii)
Radius of base,
(iii)
Capacity of vessel.
Sol: (i) Depth (h) of cylindrical vessel = 10 m
Let r be the radius of base of vessel.
Cost for painting inner curved surface of cylindrical vessel = Rs. 2200
Area painted in Rs. 20 = 1 m2
Area painted in Rs. 1 
1 2
m
20
 1

 2200  m 2
Area painted in Rs. 2200  
 20

 110 m2
Therefore, the inner curved surface area of cylindrical vessel is 110 m 2.
(ii) Curved surface area of cylindrical vessel = 110 m2
And as we know curved surface area of cylindrical vessel  2rh
 2rh  110 m2
 2
22
 r  10 m  110 m2
7
 2  22  r  10m  110  7  m2
 440  r m  770 m 2
770 m 2
r 
440 m
r 
7
m  1.75 m
4
Therefore, the radius of base of vessel is 1.75 m.
2
(iii) Capacity of vessel  r h
 22  7  2


    10 m 3
 7 4



 22 7 7


   10 m 3
 7 4 4


11 7  5 m3

385 3
m
4
4
 96.25 m3
[1 m3  1 k]
 96.25 k
Therefore, capacity of vessel is 96.25 k.
Q6.
The capacity of closed cylindrical vessel of height 1 m is 15.4 liters. How
many square meters of metal sheet would be needed to make it?
Sol:
Let r be the radius of base of cylindrical vessel.
Height (h) of vessel = 1 m
Volume of vessel  15.4 Lt. 

15.4 3
m
1000
154
m3
10000
But vessel is cylindrical in shape.
 Volume of vessel  r 2h
So,
r 2h 
154
m3
10000

22 2
154
 r  1m 
m3
7
10000

22 2
154
 r  1m 
m3
7
10000
1

3
 1 Lt.  1000 m 


7  m3
 154


 10000 22  1  m
 r2  
 r2 
77 2
m
10000
2
 7  2
r  
 m
 100 
2
 7

r  
m
 100 
2
2
r 
7
m
100
[Taking square root on both sides]
Sheet required to make vessel = Total surface area of cylindrical vessel
 2r r  h
 22 7  7

 2 

 1m2

7 100  100 

 44  1  7  100  2


m
 100  100 
 44 107  2

m
=
 100 100 

4708 2
m  0.4708 m2
10000
Therefore, 0.4708 m2 sheet required to make given vessel.
Q7.
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 graphite is 1mm. If the length of the pencil is 14 cm, find the
volume of the wood and that of the graphite.
Sol:
For graphite:
Diameter (d) of base of graphite used in pencil = 1 mm 
Radius r of base of solid cylinder of graphite =
1
cm
10
Diameter
2
1
cm
10

2

1
cm
2  10

1
cm
20
Height (h) of solid cylinder of graphite= Height of pencil = 14 cm
2
Volume of solid cylinder of graphite  r h
 22  1  2

      14 cm3
 7  20 

1
 22 1

 

 14 cm3
 7 20 20

 11  1 1 1 3

cm
 10  10 

11
cm3
100
 0.11cm3
Therefore, the volume of graphite is 0.11 cm3.
For pencil:
Diameter of base of pencil = 7 mm 
7
cm
10
7
Diameter 10

cm
Radius (R) of base of pencil 
2
2

7
7
cm 
cm
2  10
20
Height (h) of pencil = 14 cm
2
Volume of pencil = R h
 22  7  2

      14  cm3
 7  20 

7
 22 7

 

 14 cm3
 7 20 20

 11  7  7  1  3

cm
 1 10  10 

539
cm3
100
= 5.39 cm3
Volume of wood used in pencil = volume of pencil – volume of graphite
 5.39 cm3  0.11cm3  5.28 cm3
Therefore, the volume of wood is 5.28 cm3.
Q8.
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 the
hospital has to prepare daily to serve 250 patients?
Sol:
Diameter of base of cylindrical bowl = 7 cm
Radius (r) of base of cylindrical bowl =
Height (h) of soup in bowl = 4 cm
2
Volume of soup in bowl = r h
Diameter 7 cm

2
2
 22  7 7

      4 cm3

 7 2 2
 22  1 7  1  3

cm
 1 1 1 

154
cm3
1
 154 cm3
Amount of soup for 1 patient = 154 cm3
Amount of soup for 250 patients  250  154  cm3
 38500 cm3

38500
Lt.
1000
1


3
1 cm  1000 Lt.


 38.5 Lt.
38.5 Lt. 38.5  soup should be prepared daily in hospital.
Exercise - 13.7
Q1.
Find the volume of the right circular cone with
(i)
radius 6 cm, height 7 cm
(ii)
radius 3.5 cm, height 12 cm
Sol:(i) Radius (r) of base of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone =
1 2
r h
3
 1 22

2
 
 6   7 cm3
3 7

 1 22

 
 6  6  7 cm3
3 7

3
= 1 22  2  6  1cm
 264 cm3
So, volume of cone with radius 6 cm and height 7 cm is 264 cm 3.
(ii) Radius (r) of base of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone =
1 2
r h
3
 1 22

 
 3.5  3.5  12 cm3
3 7

 1 22  0.5  3.5  4 cm3
 154.00 cm3
 154 cm3
So, volume of cone with radius 3.5 cm and height 12 cm is 154 cm 3.
Q2.
Find the capacity in liters of a conical vessel with
(i)
Radius 7 cm, slant height 25 cm.
Sol:
Radius (r) of base of conical vessel = 7 cm
Slant height   of vessel = 25 cm
Height (h) of vessel   2  r 2

252  72 cm
=
625  49 cm
 
2
 h2  r 2

 576 cm
 24 cm
Volume of conical vessel =
1 2
r h
3
 1 22

2
 
 7   24 cm3
3 7

 1 22

 
 7  7  24 cm3
3 7

 1 22  1 7  8 cm3
 1232 cm3

1232
Lt.
1000
1


3
 1 cm  1000 Lt.


 1.232 Lt.
 Capacity of conical vessel with radius 7 cm and slant height 25 cm is 1.232 Lt.
(ii)
Height 12 cm, slant height 13 cm.
Sol:
Height (h) of vessel = 12 cm
Slant height   of vessel = 13 cm
Radius (r) of base of vessel =

 2  h2
 
2
 h2  r 2

132  122 cm
 169  144 cm
 25 cm
 5 cm
Volume of vessel =
1 2
r h
3
 1 22

2
 
 5   12 cm3
3 7

 1 22

 
 5  5  12 cm3
3 7


2200
cm3
7

2200
1

Lt.
7
1000

11
Lt.
35
1


3
 1 cm  1000 Lt.


Therefore, capacity of vessel with height 12 cm and slant height 13 cm is
11
Lt.
35
Q.3
The height of a cone is 15 cm. If its volume is 1570 cm3 , find the radius of
the base. (Use   3.14 ).
Sol:
Let r be the radius of base of cone.
Height (h) of cone = 15 cm
Volume of cone = 1570 cm3
As we know, volume of cone =

1 2
r h
3
1 2
r h  1570 cm3
3

1
 3.14  r 2  15 cm  1570 cm3
3
 15.70 r 2 cm  1570 cm3
 r2 
1570 cm3
15 .70 cm
 r2 
1570  100
cm2
1570
 r 2  100 cm2
2
 r 2  10 cm
 r  10 cm
Therefore, the radius of base is 10 cm.
Q4.
If the volume of a right circular cone of height 9 cm is 48 cm3 , find the
diameter of its base.
Sol:
Let r be the radius of base of cone.
Height (h) of cone = 9 cm
Volume of cone  48 cm3
As we know, volume of cone =

1 2
r h  48  cm3
3
1 2
r h
3
 r2 
 r2 
 r2 
48
cm3
1
h
3
48  3
cm3
h
48  3  cm3
9 cm
 r 2  16 cm2
2
 r 2  4 cm
 r  4 cm
Diameter of base  2  radius of base  2  4 cm  8 cm
Therefore, the diameter of base of cone is 8 cm.
Q5.
A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilo
-liters?
Sol:
Diameter of top of conical pit = 3.5 m
Radius (r) of top of conical pit =
Diameter
2

3 .5
m
2

35
35
m
m
2  10
20
Depth (h) of conical pit = 12 m
Capacity of conical pit = Volume of conical pit 
1 2
r h
3
 1 22  35  2

 
    12 m3
 3 7  20 

 1 22 35 35

 


 12 m3
3
7
20
20



11 35 3
m
10

385 3
m
10
 38.5 m3
 38.5 K
 1 m
3
 1 k

 Capacity of conical pit is 38.5 k .
Q6.
The volume of a right circular cone is 9856 cm3. 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.
Sol:(i) Let h be the height of cone.
Diameter of base of cone = 28 cm
Radius of base of cone =
Diameter 28

cm  14 cm
2
2
Volume of cone = 9856 cm3
As we know, volume of cone =

1 2
r h
3
1 2
r h  9856 cm3
3
 1 22

2
 14   hcm2  9856 cm3
3 7

 
 1 22

 14  14  h cm2  9856 cm3
3 7

 
 4312
 21


h cm2  9856 cm3

3
 9856  21  cm

2
 4312  cm
h  
h 
206976
cm
4312
 48 cm
So, height of cone is 48 cm.
(ii) Slant height of cone  h 2  r 2

482  142 cm
 2304  196 cm
 2500 cm
 50 cm
So, slant height of cone is 50 cm.
(iii) Curved surface area of cone  r
 22


 14  50 cm2
 7

 22  2  50 cm2
 2200 cm2
So, curved surface area of cone is 2200 cm2.
Q7.
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.
Sol:
Consider the triangle ABC,
13
12 cm
cm
B
C
5 cm
A
When we revolve the right triangle ABC about side 12 cm, we obtain a cone
(right circular cone) with height (h) 12 cm, radius (r) 5 cm and slant height 13 cm.
Volume of obtained cone =
1 2
r h
3
1

2
     5   12 cm3
3

1

   25  12   cm3
3

 25  4  cm3
 100  cm3
So, volume of obtained cone is 100  cm3 .
Q8.
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 two solids obtained in Questions 7 and 8.
Sol:
Consider the right triangle ABC,
cm
13
B
12 cm
5 cm
C
A
When the right triangle ABC is revolved about side 5 cm, we obtain a right
circular cone with height h1  as 5 cm, radius r1  as 12 cm and slant height  1 
as 13 cm.
Volume of cone =
1 2
r1 h1
3
1

2
     12   5 cm3
3

1

     144  5 cm3
3

 240  cm3
Required Ratio =
Volume of cone in Question 7
Volume of cone in Question 8

100  cm3
240  cm3

10 5

24 12
So, the volume of cone obtained is 240  cm3 and required ratio is 5: 12.
Q9.
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.
Sol:
Diameter of base of heap of wheat = 10.5 m
Radius (r) of base of heap of wheat =
Diameter
2

10.5
m
2

105
21
m
m
20
4
Height (h) of heap of wheat = 3m
As heap of wheat is of conical form.
Volume of heap of wheat =
1 2
r h
3
 1 22  21  2

 
    3 m 3
 3 7  4 

 1 22 21 21

 
   3 m 3
4 4
3 7

 1 11  3  21  3

m
 1 1 2  4 

11 3  21 3
m
2 4

693 3
m  86.625 m3
8
So, volume of heap of wheat is 86.625 m3
Area of canvas required = curved surface area of heap of wheat
 r
 r h 2  r 2
2 
 22 21
 21  2
2



3    cm
 7
4
4 


 22 21
441  2
  
9
cm
7
4
16


 11 3 144  441  2
cm
 

2
16


 33 585  2
cm
 

2
16


 33


36 .5625 cm2
 2

[  2  r 2  h2 ]
 33

   6.05 cm2
 2


199.65
cm2 = 99.825 cm2
2
Therefore, the area of canvas required is 99.825 cm2.
Exercise 13.8
Q1.
Find the volume of a sphere whose radius is
(i)
7 cm
Sol:
Radius (r) of sphere = 7 cm
Volume of sphere 
4 3
r
3
 4 22
3
 
 7   cm3
3 7

 4 22

 7  7  7 cm3
= 
3 7

4

   22  7  7 cm3
3


4312 3
cm
3
1
 1437 cm3
3
1
So, volume of sphere with radius 7 cm is 1437 cm3 .
3
(ii)
0.63 m
Sol:
Radius (r) of sphere = 0.63 m
Volume of sphere 
4 3
r
3
 4 22
3
 
 0.63   m3
3 7

 4 22

 
 0.63  0.63  0.63 m3
3 7

63
63  3
 4 22 63
 



m
 3 7 100 100 100 
 4  22  3  63  63  3
m
=
1000000



1047816 3
m
1000000
 1.047816m3
 Volume of sphere with radius 0.63 m is 1.047816 m3 .
Q2.
Find the amount of water displaced by a solid spherical ball of diameter:
(i)
28 cm
Sol:
Diameter of spherical ball = 28 cm
Radius (r) of spherical ball 

Diameter
2
28
cm  14 cm
2
Amount of water displaced by spherical ball = Volume of spherical ball

4 3
r
3
 4 22
3
 
 14   cm3
3 7

 4 22

 
 14  14  14 cm3
3 7

4

   22  2  14  14 cm3
3


34496 3
cm
3
2
 11498 cm3
3
2
So, water displaced by a solid spherical ball with diameter 28 cm is 11498 cm3 .
3
(ii)
0.21 cm
Sol:
Diameter of spherical ball = 0.21 cm 
Radius (r) of spherical ball =
21
cm
100
Diameter
2
21
21
 100 cm 
cm
2
200
Amount of water displaced by spherical ball = Volume of spherical ball

4 3
r
3
 4 22  21  3  3
 

 cm
 3 7  200  
21
21  3
 4 22 21
 



cm
 3 7 200 200 200 

11 21 21
cm3
100  100  100

4851
cm3
1000000
 0.004851cm3
So, water displaced by spherical ball with diameter 0.21 cm is 0.004851 cm3 .
Q3.
The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the
density of metal is 8.9 g per cm3 ?
Sol:
Diameter of metallic ball = 4.2 cm =
42
cm
10
42
Diameter
42
21
cm 
cm
 10 cm 
Radius (r) of metallic ball =
2
2  10
10
2
Volume of metallic ball =
4 3
r
3
 4 22  21  3 
 
    cm3
 3 7  10  
 4 22 21 21 21  3
 


 cm
 3 7 10 10 10 
 4  22  21  21  3

cm
 10  10  10 

38808 3
cm
1000
Density of metal  8.9 g per cm3
But, Density =
Mass
Volume
 Mass  Desnity  Volume
38808 

 89 38808 
  8 .9 

g  
g
1000 

 10 1000 
 3453912 

g  345 .3912 g
 10000 
Therefore, the mass of the ball is 345.3912 g.
Q4.
The diameter of the moon is approximately one-fourth of the diameter of
the earth. What fraction of the volume of the earth is the volume of the
moon?
Sol:
Let the diameter of earth be d, and then the diameter of moon be d/4.
Radius (R) of earth 
Diameter of earth d

2
2
Radius (r) of moon 
d
d
Diameter of moon d / 4



2
2
2 4 8
Volume of earth, v e 
4
4  d
3
R   
3
3 2
Volume of moon, v m 
4
4  d
3
r    
3
3 8
4  d
 
3 8
3
3
3
d3
Volume of moon
8


 3
3
Volume of earth 4  d 
d
 
23
3 2


d3
8
1
 3 
512 d
64
Volume of moon
1

Volume of earth 64
 Volume of moon =
Q5.
3
1
of volume of earth.
64
How many liters of milk can a hemispherical bowl of diameter 10.5 cm
hold?
Sol : Diameter of hemispherical bowl = 10.5 cm

Radius (r) of hemispherical bowl =
21
105
cm 
cm
10
2
Diameter
2
21
 2 cm
2

Volume of hemispherical bowl =
21
21
cm 
cm
22
4
2 3
r
3
 2 22  21  3  3
 
   cm
3
7
 4  

 2 22 21 21 21  3
 
 
  cm
4 4 4 
3 7
 11  21  21  3
cm
=
44 


4851 3
cm  303.1875 cm3
16
Capacity of bowl  303 .1875 cm3
 303 .1875 

Lt.
 1000 
1


3
 1 cm  1000 Lt.


 0.3031875Lt.
 0.303 Lt. approximat ely 
 The volume of the hemispherical bowl is 0.303 Lt.
Q6.
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.
Sol:
Inner radius (r) of hemispherical tank = 1 m
Thickness of hemispherical tank = 1 cm =
1
m
100
 0.01m
1


1 cm  100 m


Outer radius (R) of hemispherical tank = Inner radius + thickness
 1m  0.01m  1.01m
Volume of iron used to make tank = Outer volume of tank – Inner volume of tank

2
2
R 3  r 3
3
3

2
 R3  r 3
3




 2 22
1.013  13  m3
 
3 7

 2 22
1.030301  1m3
 
3 7

 2 22

 
 0.030301 m 3
3 7


1.333244 3
m
21
 0.06348781 m3
 0.06348 m3 (Approximately)
Hence, volume of the iron used in making tank is 0.06348 m3 .
Q7.
Find the volume of a sphere whose surface area is 154 cm2.
Sol:
Let r be the radius of sphere.
Surface area of sphere = 154 cm2
But, surface area of sphere  4r 2
 4r 2  154 cm2
 4

22 2
 r  154 cm2
7
88 2
r  154 cm2
7
 r2 
7
 154 cm2
88
 r2 
77
cm2
2 2
7

 r   cm 
2

2
2
r 
7
cm
2
Volume of sphere 
4 3
r
3
 4 22  7 3 
 
    cm3
3
7
 2  

 4 22 7 7 7  3
 
   cm
3 7 2 2 2
 11  7  7  3

cm
3



539
2
cm3  179 cm3
3
3
Hence, the volume of given sphere is  179
2
cm3 .
3
Q8.
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 meter, find the
(i)
Inside surface area of the dome,
(ii)
Volume of the air inside the dome.
Sol:(i) Cost of white-washing from inside of dome = Rs. 498.96
Given that, cost of white-washing 1m2 area = Rs. 2
 Inside surface area of dome 
498.96 2
m
2
 249.48 m2
Curved surface area of inner side of dome = 249.48 m 2.
(ii)
Let r be the inner radius of hemispherical dome.
Inner curved surface area of hemispherical dome  2r 2
 2r 2  249 .48 m2
 2

22 2
 r  249.48 m2
7
44 2
 r  249.48 m2
7
 7

 249 .48  m 2
 44

 r2  
 1746 .36  2
m
 44 
 r2  
 r 2  39.69 m2
 r2 
63  63 2
m
10  10
 63 
 r   m
 10 
2
2
r 
63
m  6.3 m
10
Volume of the air inside the dome = inner volume of dome

2 3
r
3
 2 22  63  3  3
 
   m
3
7
 10  

 2 22 63 63 63  3
 



m
 3 7 10 10 10 
 2  22  3  63  63  3

m
10  10  10



523908 3
m
1000
 523.908 m3
So, volume of air inside the dome is 523 .908 m3 .
Q9.
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 new sphere,
(ii)
ratio of S and S' .
Sol:(i) Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere =
4 3
r
3
4
Volume of 27 solid iron spheres  27  r 3
3
Now, 27 solid spheres are melted to form 1 new iron sphere of radius r' .
 Volume of new iron sphere = volume of 27 solid iron spheres.

4
4
3
r '  27  r 3
3
3
 r ' 
3
3
4
 27  r 3
4
3
3
 r '  27 r 3  3  3  3r 3
3
3
 r '  3r 
Taking cube root on both sides ,we get:
 r'  3r
Hence, radius r' of new sphere is 3r.
(ii) Surface area of new sphere,
S'  4r '
2
 43r   4  3  r 2
2
2
 4  9  r 2
 36r 2
 S'  36 r 2
Surface area of solid iron sphere with radius r  4r 2
 S  4r 2
S
4r 2
 
S' 36 r 2
 S : S'  1: 9
Q10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How
much medicine (in mm3 ) is needed to fill this capsule?
Sol:
Diameter of capsule = 3.5 mm
Radius of capsule =
Diameter 3.5

mm
2
2
As capsule is in the shape of a sphere.
Medicine is needed to fill in capsule  Volume of capsule

4 3
r
3
 4 22

 
 1.75  1.75  1.75 mm 3
3 7


471.62500
mm 3
21
 22.45833 mm 3
 22.46 mm 3
(Approx.)
Hence, 22.46 mm3 medicine is needed to fill the capsule.
Exercise - 13.9
(OPTIONAL)
Q.1)
A wooden bookshelf has external dimensions as follows: height = 110 cm, breadth
= 25 cm, length =85 cm. 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 cm2 and the rate of painting is 10 paise per cm2. Find
the total expences required for polishing and painting the surface of the
bookshelf.
110 cm
85 cm
25
cm
Fig. 13.31
Sol:
External height of bookshelf = 110 cm
External breadth of bookshelf = 25 cm
External length of bookshelf = 85 cm
External surface area of shelf while leaving out the front face of the shelf
 h  2b  bh 
 110  85  285  25  25  110 cm 2
 9350  22125  2750  cm2
 9350  2  4875  cm2
 9350  9750  cm2
 19100 cm2
85 cm = length
A
75 cm
30 cm
F
110 cm = height
E
B
100 cm
cm
20
I
J
K
L
M
N
O
P
G
5 cm
H
D
C
d th
ea
r
b
5
=2
Area of front external face of bookshelf
= area of ABCD – area of EFGH + area of IJLK + area of MNPO
 AB  BC  EF  FG  IJ  JL  MN  NP
 85  110  75  100  75  5  75  5  cm2
 9350  7500  375  375  cm2 `
 9350  375  375  7500  cm2
 10100  7500  cm2
 2600 cm2
So, Area to be polished  19100  2600  cm2
 21700 cm2
cm
2
Cost for polishing 1 cm area = Paise 20
Cost for polishing 21700 cm2 area = Paise 20 21700
 Paise 434000
 Rs
434000
100
1


 IPaise  100 Rs 


 Rs.4340
We can observe from the figure that there are 3 rows, each with length    75 cm,
breadth (b) = 20 cm and height (h) = 30 cm.
Area of 1 row    h  2b  bh 
 75  30  275  20  30  20 cm 2
 2250  21500  600  cm2
 2250  22100  cm2
 2250  4200  cm2
 6450 cm2
So, area of 3 rows  3  area of 1 row
 3  6450  cm2
 19350 cm2
Cost of painting 1 cm2 area = Rs. 10/100 = Rs. 1/10
2
Cost of painting 19350 cm area = Rs.
1
 19350  Rs.1935
10
Total expenses = Rs. 4340  1935   Rs.6275
So, Rs. 6275 is the cost for polishing and painting the bookshelf.
Q2.
The front compound wall of a house is decorated by wooden spheres of
diameter 21 cm, placed on small supports as shown in figure 13.32. 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 cm2.
Fig. 13.32
Sol:
Diameter of wooden sphere = 21 cm
Radius (R) of wooden sphere =
Diameter 21

cm
2
2
Area of 1 wooden sphere  4R2
2

22  21   2
 4 
   cm
7  2  

22 21 21  2

 4 
  cm
7
2 2

 22  3  21cm2  1386 cm2
Radius (r) of base of cylindrical support = 1.5 cm
Height (h) of cylindrical support = 7 cm
Curved surface area of support  2rh
22


 2 
 1.5  7 cm2
7


 2  22  1.5 cm2
 66 cm2
Area of circular part of cylindrical support  r 2
 22
2
   1.5   cm2
7

 22  2.25  2

cm
7


 7.07 cm2 (Approx)
Area to be painted silver of 1 sphere
= surface area of sphere – area of circular base of support
 1386  7.07  cm2
 1378.93 cm2
Area to be painted silver of 8 spheres
 8  1378 .93 cm2  11031 .44 cm2
Cost of painting silver color at 1 cm2 area = 25 paise
 Rs.
1
4
Cost of painting silver color at 11031.44 cm2 area  Rs.
1
 11031.44
4
 Rs.2757.86
Area to be painted black  8  CSA of sup port
 8  66  cm 2
 528 cm2
Cost of painting black color at 1 cm2 area = 5 paise
 Rs
5
100
 Rs.
1


1P  100 Rs 


1
20
Cost of painting black color at 528 cm2 area  Rs.
1
 528
20
 Rs.26.4
 Rs.26.40
Total cost  Rs.2757.86  Rs.26.40  Rs.2784.26
Therefore, it will cost Rs. 2784.26 for painting it with black and silver.
Q3.
The diameter of a sphere is decreased by 25%. By what per cent does its
curved surface area decrease?
Sol:
Let the diameter of sphere be d.
Then the radius of sphere be d/2.
Curved surface area of sphere,
S1  4r 2
 d
 4 
2
 4
2
d2
4
 d2
As new sphere has diameter 25% less than the diameter of previous sphere
So, diameter of new sphere  d -
25
d
100
25 

 100 - 25 
 d 1  d

 100 
 100 
 75  3
d
 d
 100  4
Radius r ' of new sphere 
diameter of new sphere
2
3
d
 4
2

3 d 3
  d
4 2 8
C.S.A. of new sphere,
S 2  4 r '
2
2
9d2 9d2
9
3 
 4     d   4 


d2
64
16
16
8 
Decrease in area = C.S.A. of previous sphere - C.S.A. of new sphere
 S1  S2
 d2 
9
d2
16
16 d2  9d2 7d2


16
16
Percentage decrease in surface area =
Decrease in area
 100
S1
7d2
 162  100
d

7
175
 100 
%  43.75%
16
4
So, curved surface area is decreased by 43.75% .