AMIT BAJAJ 13 SURFACE AREAS AND VOLUMES CHAPTER

CHAPTER
13
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SURFACE AREAS AND VOLUMES
Points to Remember :
1. Cuboid
(i) Volume = lbh
(ii) Curved surface area = 2h (l + b)
(iii) Total surface area = 2 (lb + bh + lh)
(iv) Diagonal  l 2  b 2  h 2
2. Cube
(i) Volume = a3
(ii) Curved surface area = 4a2
(iii) Total surface area = 6a2
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(iv) Diagonal  3 .a
3. Cylinder
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(i) Volume = r2h
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(ii) Curved surface area = 2rh
(iii) Total surface area = 2r (r + h)
4. Cone
(i) Volume 
1 2
r h
3
(ii) slant height, l  h 2  r 2
(iii) curved surface area   r l
(iv) Total surface area   r (l  r )
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
1
5. Sphere
(i) Volume  4 r 3
3
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(ii) Total surface area = 4  r2
6. Hemi-sphere
(i) Volume  2 r 3
3
(ii) Curved surface area = 2r2
(iii) Total surface area = 3r2
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ILLUSTRATIVE EXAMPLES
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Example 1. 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 surface area and by how much?
—NCERT
Solution.
(i) Lateral surface area (L1) of cubical box = 4 × (edge)2
= 4 × (10)2 cm2
= 400 cm2
and, lateral surface area (L2) of cuboidal box
= 2 (length + breadth)× height
= 2 (12.5 + 10) × 8 cm2 = 360 cm2
Clearly, L1 > L2.
Now, L1 – L2 = 400 cm2 – 360 cm2 = 40 cm2
 The cubical box has larger lateral surface area and is greater by 40 cm2.
(ii) Total surface area of the cubical box (S1) = 6 (edge)2
= 6 × (10)2 cm2 = 600 cm2
Total surface area of the cuboidal box (S2)
= 2 (lb + lh + bh)
= 2 (12.5 × 10 + 10 × 8 + 8 × 12.5) cm2
= 2 (125 + 80 + 100) cm2 = 610 cm2
Clearly, S2 > S1.

S2 – S1 = 610 cm2 – 600 cm2 = 10 cm2
Thus, the cuboidal box has greater surface area and is greater by 10 cm2.
Example 2. Two cubes each of 15 cm edge are joined end to end. Find the surface area of the resulting cuboid.
Solution.
here,
l = length of resulting cuboid = 15 cm + 15 cm = 30 cm
b = breadth of resulting cuboid = 15 cm
h = height of resulting cuboid = 15 cm
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SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
 Surface area of resulting cuboid
= 2 (lb + bh + lh)
= 2 (30 × 15 + 15 × 15 + 30 × 15) cm2
= 2 (450 + 225 + 450) cm2 = 2 (1125) cm2
= 2250 cm2 Ans.
Example 3. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open 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 m2 costs Rs. 20.
–NCERT
Solution.
We have, Length; l = 1.5 m, Breadth, b = 1.25 m and Depth = Height, h = 0.65 m
(i) Since the plastic box is open at the top,
 Plastic sheet required for making such box
= 2 (l + b) × h + lb
= 2 (1.5 + 1.25) × 0.65 m2 + 1.5 × 1.25 m2
= 2 × 2.75 × 0.65 m2 + 1.875 m2
= 3.575 m2 + 1.875 m2 = 5.45 m2
(ii) Cost of 1 m2 of sheet = Rs. 20
 Total cost of 5.45 m2 of sheet = Rs. 5.45 × 20 = Rs. 109 Ans.
Example 4. 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 4 m ×
3 m?
–NCERT
Solution.
here, l = 4m, b = 3m, h = 2.5 m.
Since there is no tarpaulin for the floor.
 Tarpaulin required = [2 (l + b) × h + lb]
= [2 (4 + 3) × 2.5 + 4 × 3] m2
= (2 × 7 × 2.5 + 12) m2
= (35 + 12) m2 = 47 m2 Ans.
Example 5. The sum of length, breadth and height of a cuboid is 21 cm and the length of its diagonal is 12 cm.
Find the surface area of the cuboid.
Solution.
Let the length, breadth and height of the cuboid be l cm, b cm and h cm respectively.
Then,
l + b + h = 21
...(1)
Now, diagonal = 12 cm
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
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l 2  b 2  h 2  12

l2 + b2 + h2 = 144
...(2)
Now,
l + b + h = 21
Squaring both sides, we get
(l + b + h)2 = (21)2

l2 + b2 + h2 + 2lb + 2bh + 2lh = 441

144 + 2 (lb + bh + lh) = 441

2 (lb + bh + lh) = 441 – 144 = 297
 Surface area of the cuboid is 297 cm2 Ans.
MATHEMATICS–IX
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( using (2))
SURFACE AREAS AND VOLUMES
3
Example 6. Aggarwal sweets stall was placing an order for making card board boxes for packing their sweets.
Two size of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller
of dimensions 15 cm × 12 cm × 5 cm. 5% of the total surface area is required extra, for all the
overlaps. If the cost of cardboard is Rs. 5 for 1000 cm2, find the cost of cardboard required for
supplying 300 boxes of each kind.
Solution.
Surface area of Ist box = 2 (25 × 20 + 20 × 5 + 25 × 5) cm2
= 2 (500 + 100 + 125) cm2 = 1450 cm2
Surface area of IInd box = 2 (15 × 12 + 12 × 5 + 15 × 5) cm2
= 2 (180 + 60 + 75) cm2 = 630 cm2
Total combined surface area = 1450 cm2 + 630 cm2 = 2080 cm2

Cost of cardboard for 1 cm 3  Rs.
5
1000
 Cost of cardboard for 655200 cm2  Rs.
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Area of overlaps = 5% of 2080 cm2  5  2080 cm 2  104 cm 2
100
 Total surface area of 2 boxes (one of each kind)
= (2080 + 104) cm2 = 2184 cm2
 Surface area of 300 boxes of each kind
= 300 × 2184 cm2 = 655200 cm2
Now, cost of cardboard for 1000 cm2 = Rs. 5
5
 655200
1000
= Rs. 3276 Ans.
Example 7. 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.
–NCERT
Solution.
Diameter of cylindrical pillar = 50 cm
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50
cm  25 cm  0.25 m
2
also, height (h) = 3.5 m
 radius (r) 
Now, curved surface = 2rh  2  22  0.25  3.5 m 2 = 5.5 m2
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Cost of painting 1 m2 = Rs. 12.50
 Cost of painting 5.5 m2 = Rs. 12.50 × 5.5 = Rs. 68.75 Ans.
Example 8. In the given figure, 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 is required for covering the
lampshade.
–NCERT
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SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
Solution.
20
cm  10 cm
2
and, height (h) = 30 cm + 2 × 2.5 cm = 35 cm
(for margin)
 Cloth required for covering the lampshade = Its curved surface area = 2  rh
Here, radius (r ) 
 2
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22
 10  35 cm 2 = 2200 cm2 Ans.
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Example 9. It is required to make a closed cylindrical tank of height 120 cm and base diameter 140 cm from a
metal sheet. How many square metres of the sheet is required for the same?
Solution.
We have, diameter of base = 140 cm.
140
 radius of base 
cm  70 cm
2
and, height of cylinder = 120 cm.
 Total surface area of required tank
= 2  r (r + h)
 2
22
 70 (70  120) cm 2
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83600 2
m  8.36 m 2 Ans.
10000
Example 10. 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?
–NCERT
Solution.
Cardboard required by each competitor
= curved surface area of one penholder + base area
= 2rh + r2, where r = 3 cm, h = 10.5 cm
 440  190 cm 2  83600 cm 2 
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

22
 22
  2   3 10.5   (3) 2  cm 2
7
 7


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= (198 + 28.28) cm2 = 226.28 cm2 (approx)
 Cardboard required for 35 competitors
= 35 × 226.28 cm2 = 7920 cm2 (approx) Ans.
Example 11. The diameter of a roller is 84 cm and its length is 100 cm. It takes 300 complete revolution to move
once over to level a playground. Find the area of the playground in m2.
Solution.
Radius of roller (r ) 
84
cm  42 cm
2
Length of the cylindrical roller (h) = 100 cm.
Area moved by the roller in one revolution
 2  rh  2 
MATHEMATICS–IX
22
 42  100 cm 2
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SURFACE AREAS AND VOLUMES
5

Area moved in 300 revolutions
22
 42  100  300 cm 2
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= 44 × 180000 cm2 = 44 × 18 m2 = 792 m2 Ans.
Example 12. Find (i) the curved surface area of a cylindrical petrol storage tank that is 4.2 m in diameter and
 2
4.5 m high. (ii) how much steel was actually used if
making the closed tank?
Solution.
4.2
m  2.1 m , h  4.5 m
2
(i) Curved surface area = 2  r h
Here, r 
22
 2.1 4.5 m 2
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= 59.4 m2
(ii) Total surface area of closed tank
 2
 2 r h  2 r 2  2r (r  h)
B
22
 2.1 ( 2.1  4.5) m 2
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= 87.12 m2
 2
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1
of the steel actually used was wasted in
12
—NCERT
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Let the total sheet used for making the cylindrical tank be x m2. Given, wastage 
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according to given question, x 
x
 87.12
12
x 2
m .
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87.12  12
x  87.12  x 
 95.04 m 2
12
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 Steel used for making closed tank including wastage = 95.04 m2 Ans.
Example 13. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base
and (ii) total surface area of the cone.
—NCERT
Solution.
(i) Here,  r l = 308 cm2, l = 14 cm.

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22
308

 14  r  308  r 
 7 cm.
7
22  2
and, (ii) Total surface area = r (r + l)
22
 7 (7  14) cm 2
7
= 22 × 21 cm2 = 462 cm2
Example 14. How many meters of cloth, 5 m wide, will be required to make a conical tent, the radius of whose
base is 7 m and height is 24 m?
Solution.
Radius of the tent, r = 7 m
height of the tent, h = 24 m

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
slant height, l  r 2  h 2  7 2  24 2 m  625 m  25 m
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX

curved surface area =  r l
22
 7  25 m 2  550 m 2
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i.e., area of the cloth = 550 m2

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Now, length of cloth required  area  550 m  110 m
width
5
 length of cloth required = 110 m.
Example 15. 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 m2 canvas is Rs. 70.
Solution.
(i) here, r = 24 m, h = 10 m
Let l be the slant height of the cone. then,
l2 = h2 + r2  l  h 2  r 2  24 2  10 2
 576  100  676  26 m.
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(ii) Canvas required to make the conical tent = curved surface of the cone
B
22
 rl 
 24  26 cm 2
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Now, Rate of canvas for 1 m2 = Rs. 70
–NCERT
 Total cost of canvas  Rs. 22  24  26  70 = Rs. 137280 Ans.
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Example 16. 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.
–NCERT
Solution.
here, radius of cap (r) = 7 cm
height of cap (h) = 24 cm
Let l be the slant height. then,
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l  h 2  r 2  24 2  7 2  576  49  625  25 cm
Sheet required for one cap = curved surface of the cone = rl
22
 7  25 cm 2  550 cm 2
7
 Sheet required for 10 such caps = 10 × 550 cm2 = 5500 cm2. Ans.
Example 17. A bus stop in 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 the cones 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 1.04  1.02 ).
Solution.
here, radius (r ) 
–NCERT
40
cm  20 cm  0.2 m and height (h) = 1 m
2
slant height (l )  r 2  h 2  0.04  1  1.04  1.02 m
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
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Now, curved surface of 1 cone =  rl 
2
 0.2  1.02 m 2
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 Curved surface of 50 such cones  50  22  0.2  1.02 m 2
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Now, cost of painting 1 m2 = Rs. 12
 Total cost of painting  Rs. 12  50 
22
 0.2  1.02 = Rs. 384.68 (approx) Ans.
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Example 18. A corn cob (see figure), shaped somewhat like a cone, has the radius of its broadest end as 2.1 cm
and length as 20 cm. If each 1 cm2 of the surface of the cob carries an average of four grains, find
how many grains you would find on entire cob?
—(NCERT)
Solution.
We have, r = 2.1 cm, h = 20 cm
let, slant height be l cm.
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then, l  r 2  h 2  (2.1) 2  (20) 2 cm  4.41  400 cm

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 404.41 cm  20.11 cm
curved surface area of corn cob =  r l
22
 2.1  20.11 cm 2
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= 132.726 cm2
Now, number of grains on 1 cm2 = 4
 number of grains on 132.726 cm2 = 4 × 132.726
= 530.904
 531
Hence, total number of grains on the corn cob = 531 Ans.
Example 19. The surface area of a sphere is 154 cm2. Find its radius.
Solution.
Let the radius of the sphere be r cm.
then, 4  r2 = 154
(given)

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
r2 
 r

8
—NCERT
154 154  7 49


4
4  22
4
49
7
cm  cm
4
2
radius of the sphere 
7
cm.
2
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
Example 20. The radius of a spherical baloon 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.
—NCERT
Solution.
Let S1 and S2 be the total surface area in two cases of r = 7 cm and R = 14 cm.

S1 = 4 r2 = 4 (7)2 cm2
and
S2 = 4 R2 = 4 (14)2 cm2

Required ratio 
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S1
4π  7  7
1

 i.e. 1 : 4 Ans.
S 2 4π  14 14 4
Example 21. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio
of their surface areas.
–NCERT
Solution.
Let the diameter of earth be R and that of the moon will be
The radii of moon and earth are
R
R
and
respectively..
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2
2
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R
4
R
1
4 
8
Ratio of their surface area     64  1  4  1 i .e. 1 : 16 Ans.
2
1
64 1 16
R
4 
4
2
Example 22. 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.
–NCERT
Solution.
Inner radius, r = 5 cm
Thickness of Steel = 0.25 cm
 Outer radius, R = (r + 0.25) cm = (5 + 0.25) cm = 5.25 cm
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 Outer curved surface  2 r 2   2   5.25  5.25  cm 2
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

= 173.25 cm2
Example 23. The internal and external diameters of a hollow hemispherical vessel are 20 cm and 28 cm respectively. Find the cost of painting the vessel all over at 15 paisa per cm2.
Solution.
Outer radius of vessel, R = 14 cm
Inner radius of vessel, r = 10 cm
Area of the outer surface = 2  R2
= 2  × (14)2 cm2 = 392  cm2
Area of the inner surface = 2  r2
= 2  × (10)2 cm2 = 200  cm2
Area of the ring at the top =  (R2 – r2)
=  (142 – 102) cm2
=  (14 – 10) (14 + 10) cm2
= 96  cm2
 Total area to be painted
=3 92  cm2 + 200  cm2 + 96  cm2
= 688 cm2
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Now, cost of painting 1 cm2 = 15 paisa  15 Rs.
100
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
9
15
 688 π Rs
100
15
22

 688  Rs
100
7
= 324.34 Rs. Ans.
Example 24. A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold?
(1 m3 = 1000 l).
–NCERT
Solution.
Here, l = 6m, b = 5m and h = 4.5 m
 Volume of the tank = lbh = (6 × 5 × 4.5) m3 = 135 m3
 The tank can hold = 135 × 1000 litres = 135000 litres of water. ( 1 m3 = 1000 litres)
Example 25. The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length
and depth are respectively 2.5 m and 10 m.
—NCERT
Solution.
Given capacity of a cuboidal tank
50000 3
 50000 l 
m  50 m 3
1000
Let the breadth of cuboidal tank be b m.
according to given question, we have
2.5 × b × 10 = 50
50
 25 b  50  b 
b2
25
 breadth of the tank is 2m. Ans.
Example 26. 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?
—NCERT
Solution.
Volume of water that flows in 1 hour (60 minutes)
= volume of water of a cuboid whose dimensions are 3 m, 40 m and 2000 m.
( 2 km = 2000 m)
= 3 × 40 × 2000 m3
 Volume of water that flows in 1 minute
 Cost of painting 688  cm 2 
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3  40  2000 3
m  4000 m 3 Ans.
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Example 27. Three cubes whose edges are 3 cm, 4 cm and 5 cm respectively are melted and recast into a single
cube. find the surface area of the new cube.
Solution.
Let x cm be the edge of new cube. Then, volume of the new cube = sum of the volumes of three
cubes.
 x3 = 33 + 43 + 53 = 27 + 64 + 125 = 216 = (6)3
 x = 6 cm.
 Edge of the new cube is 6 cm.
and, surface area of the new cube = 6 (6)2 cm2 = 216 cm2 Ans.
Example 28. 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?
–NCERT
Solution.
Here l = 20 m, b = 15 m, and h = 6 m
 Capacity of the tank = lbh = (20 × 15 × 6) m3 = 1800 m3
Water requirement per person per day = 150 litres
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SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
Water required for 4000 person per day = (4000 × 150) l
 4000 150  3
3

 m  600 m
 1000 
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Number of days the water will last

Capacity of tank
Total water required per day

1800
 30
600
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Thus, the water will last for 30 days Ans.
Example 29. A godown measures 60 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 the godown.
–NCERT
3
3
Solution.
Volume of the godown = (60 × 25 × 10) m = 15000 m
Volume of 1 crate = (1.5 × 1.25 × 0.5) m3 = 0.9375 m3
Number of crate that can be stored in the godown

Volume of the godown
Volume of 1 crate

15000
 16000 Ans.
0.9375
B
Example 30. If the lateral surface of cylinder is 94.2 cm2 and its height is 5 cm, then find (i) radius of its base (ii)
its volume (use  = 3.14)
.
–NCERT
Solution.
(i) Let r be the radius of the base and h be the height of the cylinder. Then,
Lateral surface = 94.2 cm2

2rh = 94.2

2 × 3.14 × r × 5 = 94.2

r
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94.2
3
2  3.14  5
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Thus, the radius of its base = 3 cm.
(ii) Volume of the cylinder = r2h = (3.14 × 32 × 5) cm3
= 141.3 cm3 Ans.
Example 31. 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 m2, Find :
(i) inner curved surface area of the vessel
(ii) radius of the base
(iii) capacity of the vessel
–NCERT
Solution.
(i) inner curved surface area of the vessel 
Total cost of painting
Rate of painting
 2200  2
2

 m  110 m
20


MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
11
(ii) Let r be the radius of the base and h be the height of the cylindrical vessel.

2rh = 110
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 r  10  110
7

2

r
110  7
7
  1.75
2  22  10 4
Thus, the radius of the base = 1.75 m
 22 7 7

(iii) Capacity of the vessel = r2h      10  m 3
7
4
4


J
A
= 96.25 m3
Example 32. 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?
–NCERT
Solution.
Capacity of a closed cylindrical vessel = 15.4 litres
1  3

3
 15.4 
 m  0.0154 m
1000 

B
Let r be the radius of the base and h be the height of the vessel. Then,
Volume = r2h = r2 × 1 = r2
( h = 1m)

r2 = 0.0154

22 2
 r  0.0154
7
 r2 

T
I
0.0154  7
 0.0049
22
r  0.0049  0.07
Thus, the radius of the base of vessel = 0.07 m.
Metal sheet needed to make the vessel
= Total surface area of the vessel
= 2rh + 2r2 = 2r (h + r)
M
A
 2
22
 0.07  (1  0.07) m 2
7
= 44 × 0.01 × 1.07 m2 = 0.4708 m2
Example 33. A patient in a hospital in 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?
–NCERT
Solution.
Diameter of the cylindrical bowl = 7 cm
7
cm
2
Height of serving bowl = 4 cm.
 Radius 
12
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
 Soup served to 1 patient = Volume of the bowl = r2h
 22 7 7

     4  cm 3 = 154 cm3
7
2
2


Soup served to 250 patients = (250 × 154) cm3 = 38500 cm3 = 38.5 l. Ans.
J
A
Example 34. 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 gm.
—NCERT
Solution.
here, inner radius (r ) 
24
cm  12 cm
2
28
cm  14 cm
2
h = length of the pipe = 35 cm.
 volume of wood used in making the pipe
and outer radius (R) =
  R 2h   r 2h
  h (R 2  r 2 )

22
 35  [14 2  12 2 ] cm 3
7

22
 35  (14  12) (14  12) cm 3
7
B
J
A
22
 35  2  26 cm 3  5720 cm 3
7
Now, 1 cm3 of wood = 0.6 g
 5720 cm3 of wood = 0.6 × 5720 g = 3432.0 g
= 3.432 kg. Ans.
Example 35. 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 10 cm, find the weight of the whole pencil if the specific gravity of the wood is 0.7 gm/cm3
and that of the graphite is 2.1 gm/cm3.
—NCERT
Solution.
For graphite cylindrical rod :
T
I

M
A
radius (r) of graphite cylinder  1  1 cm  1 cm
2 10
20
and, length of graphite rod (h) = 10 cm.
 volume of graphite cylindrical rod =  r2 h
2
22  1 

   10 cm 3
7  20 
 Weight of graphite used for pencil = volume × specific gravity

22 1
1
   10  2.1 gm
7 20 20
( 1 cm3  2.1 gm)
= 0.165 gm.
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
13
Again, for pencil including graphite rod, we have,
7
7
mm 
cm
2
20
and, length of pencil (h) = 10 cm
radius of pencil (R) 

J
A
volume of pencil =  R 2 h
2
22  7 

   10 cm 3
7  20 
 volume of wood used for pencil
  R2 h   r 2 h
J
A

22 7
7
22 1
1 1
   10 cm 3 
   cm 3
7 20 20
7 20 20 10

11 1
  48 cm 3
7 20
11 1
  48  0.7 gm ( 1 cm3  0.7 gm)
7 20
= 2.64 gm.
 Total weight = 2.64 gm + 0.165 gm
= 2.805 gm Ans.
Example 36. A well of diameter 3m is dug 14 m deep. The earth taken out of it has been spread evenly all around
it to a width of 4 m to form an embankment. Find the height of the embankment.
 weight of wood 
Solution.
T
I
B
3
m , height (h) = 14 m
2
 Volume of the earth taken out of the well
Radius of the well (r ) 
22 3 3
   14 m 3  99 m 3
7 2 2
Outer radius of the embankment
  r2 h 
M
A
3
11
m4 m  m
2
2
 Area of embankment = outer area – inner area
R
  R 2  r 2

2
2
22  11   3   22  11 3   11 3 
       
     
7  2   2   7  2 2   2 2 


22
 7  4  88 m 2
7
 Height of the embankment


14
Volume 99
9

m  m  1.125 m Ans.
Area
88
8
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
Example 37. The height of a cone is 15 cm. If its volume is 1570 cm3. Find the radius of the base (Use  = 3.14).
–NCERT
Solution.
Here h = 15 cm and volume = 1570 cm3
Let the radius of the base of the cone be r cm.
Now, Volume = 1570 cm3



1 2
r h  1570
3
1
 3.14  r 2  15  1570
3
1570
r2 
 100
3.14  5
 r  100  10
J
A
J
A
Thus, the radius of the base of cone is 10 cm Ans.
Example 38. The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is 2512 cm3, find
the slant height and radius of the base of the cone. (use  = 3.14)
Solution.
Let the radius of cone be 5x and height be 12 x.
 Volume of the cone  1 r 2 h
3
according to question, we get
1
 3.14  (5 x ) 2  (12 x)  2512
3

T
I
1 314

 25 x 2  12 x  2512
3 100
2512
8 x  2
 x 
314
 radius of base = 5x = 5 × 2 cm = 10 cm
and, height = 12 x = 12 × 2 cm = 24 cm
3
M
A
B
 slant height  r 2  h 2  10 2  24 2  100  576  676  26
 radius of cone = 10 cm and slant height = 26 cm Ans.
Example 39. 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
—NCERT
Solution.
Given, volume of cone = 9856 cm3
and, radius of base (r) 
28
cm  14 cm
2
(i) we know, volume of cone  1 r 2 h
3
1 22
  14  14  h  9856

3 7
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
15

h
3  9856
cm  48 cm.
22  2 14
(ii) Slant height of a cone (l)
J
A
 h2  r 2
 (48) 2  (14) 2 cm
 2304  196 cm
 2500 cm  50 cm
(iii) Curved surface area of a cone
=rl
J
A
22
 14  50 cm 2  2200 cm 2 .
7
Example 40. 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.
–NCERT

Solution.
Diameter of the base of the cone = 10.5 m
10.5 
radius  r  
 m  5.25 m
 2 
Height of the cone = 3m
B
 1 22

 Volume of the cone  1 r 2 h     5.25  5.25  3  m 3 = 86.625 m3
3
7
3


To find the slant height l
T
I
We have, l 2  h 2  r 2  32  (5.25)2 = 9 + 27.5625 = 36.5625
 l  36.5625  6.0467 m (approx.)
Canvas required to protect wheat from rain = Curved surface area
M
A
 22

 rl    5.25  6.0467  m 2 = 99.77 m2 (approx)
 7

Example 41. 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.
The solid obtained is a cone with r = 5 cm and h = 12 cm.

Volume  1 r 2 h
3

1
 3.14  5  5  12 cm 3
3
 100 3.14 cm3
= 314 cm3 Ans.
16
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
Example 42. If the surface area of a sphere is 616 cm2, find its volume.
Solution.
Let r be the radius of sphere.
then, 4  r2 = 616
J
A
616 616  7

 49
4
4  22

r2 

r  49 cm  7 cm
 volume of sphere  4  r 3  4  22  7  7  7 cm 3
3
3 7
J
A
= 1437.3 cm3 Ans.
Example 43. 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?
–NCERT
Solution.
Let the radius of earth be R.
then, radius of moon 
R
.
4
 volume of earth  4  R 3
3
3
and, volume of moon 
4 R
  .
3 4
T
I
4
 R3
volume of earth
64
3



volume of moon 4  R  3
1
 
3 4
B
i.e., volume of earth is 64 times the volume of the moon.
1
times that of earth.
64
Example 44. 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 cm3?
–NCERT
Solution.
Diameter of the ball = 4.2 cm
M
A
i.e., volume of moon is
4.2 
Radius  
 cm  2.1 cm
 2 
Volume of the ball  4 r 3
3

4 22
  2.1  2.1  2.1 cm 3  38.808 cm 3
3 7
Now, Density of metal = 8.9 gm per cm3
 Mass of the ball = 38.808 × 8.9 g = 345.3912 g = 345.4 g (approx) Ans.
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
17
Example 45. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1m, then find the
volume of the iron used to make the tank.
–NCERT
Solution.
Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel, then
R = 1.01 m and r = 1 m ( as thickness = 1cm = 0.01 m)
Volume of iron used
= External volume – Internal volume

2 3 2 3 2
R  r   ( R 3  r 3 )
3
3
3

2 22
  [(1.01)3 – (1) 3 ] m 3
3 7
J
A
J
A
44
44
 (1.030301  1) m 3 
 0.030301 m 3
21
21
= 0.06348 m3 (approx)
Example 46. A dome of a building is in the form of hemi-sphere. From inside, it was white-washed at the cost
of Rs. 498.96. If the cost of white-washing is Rs. 2 per square meter, find:
(i) the inside surface area of the dome
(ii) the volume of the air inside the dome.
–NCERT
Solution.
Let r be the inner radius of the hemispherical dome. Then, inside surface area of the hemisphere
= 2  r2.
Since, at the rate of Rs. 2 per square metre, the total cost of white-wash is Rs. 498.96,

according to question, 2r 2  249.48

r2 

249.48  7
 39.69
2  22
r  39.69 m  6.3 m
(i) Inside surface are of the dome
= 2  r2 = 249.48 m2
(ii) Inside volume of the dome
M
A
B
498.96 2
m  249.48 m 2
2
T
I
 surface area of hemisphere 
(calculated above)
2 3 2 22
r    (6.3) 3 m 3
3
3 7
= 523.908 m3 Ans.
Example 47. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere
with surface are S. Find the:
(i) radius r of the new sphere.
(ii) ratio of S and S.
–NCERT

Solution.
18
4
(i) Volume of 27 solid spheres of radius r  27  r 3
3
4 3
Volume of the new sphere of radius r  r 
3
SURFACE AREAS AND VOLUMES
...(1)
...(2)
MATHEMATICS–IX
According to the problem, we have,
4 3
4
r   27  r 3
3
3

J
A
r3  27r 3  (3r )3  r   3r
S 4r 2
r2
r2
1



  1: 9
2
2
2
S  4r
9
(3r )
9r
(ii) Required ratio 
Example 48. A wooden bookshelf has external dimensions as follows : Height = 110 cm, Depth = 25 cm, Breadth
= 85 cm (see 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 paisa per cm2 and the
rate of painting is 10 paisa per cm2, find the total expenses required for polishing and painting the
surface of the bookshelf.
–NCERT
Solution.
T
I
B
J
A
Area to be polished
= (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5) cm2
= (9350 + 4250 + 5500 + 1500 + 1100) cm2 = 21700 cm2
M
A
 cost of polishing @ 20 paisa per cm2 = 21700 
20
 Rs. 4340
100
Also, Area to be painted = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90) cm2
= (9000 + 3600 + 6750) cm2 = 19350 cm2
 Cost of painting @ 10 paisa per cm2
 19350 
10
Rs.  Rs. 1935
100
 Total expense = Rs. 4340 + Rs. 1935 = Rs. 6275 Ans.
Example 49. The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on
small supports as shown in 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 paisa per cm2 and black paint costs 5
paisa per cm2.
–NCERT
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
19
Solution.
J
A
J
A
Clearly, we have to subtract the area of the circle on which sphere is resting while calculating the
cost of silver paint.
Surface area to be painted silver
= 8 (curved surface area of the sphere – area of circle on which sphere is resting)
= 8 (4 R2 – r2) where R 
21
cm , r  1.5 cm
2
B
 441

 8π  4 
 2.25  cm 2  8π (441  2.25) cm 2  8π (438.75) cm 2
4


 Cost of silver paint @ 25 paisa per cm2
25 
 22
 Rs.  8   438.75 
  Rs. 2757.86 (approx)
7
100


T
I
Again, surface area to be painted black
= 8 × curved surface area of cylinder
= 8 × 2rh = 8  2 
M
A
22
 1.5  7 cm 2  528 cm 2
7
 Cost of black paint @ 5 paisa per cm2
5
 Rs. 528 
 Rs. 26.40
100
 Total cost of painting = Rs. 2757.86 + Rs. 26.40
= Rs. 2784.26 (approx) Ans.
Example 50. The diameter of a sphere is decreased by 25%. By what percent does its curved surface decrease?
–NCERT
Solution.
2
d
Let d be the diameter of the sphere. Then, its surface area  4   d 2
2
On decreasing its diameter by 25%,
New diameter  d1  d  25% of d  d  25 d  75 d  3 d
100
100
4
20
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
2
d 
 1 3d 
 New surface area  4 1   4  
2 4 
 2
 4.
2
J
A
9d 2
9

. d 2
64 16
 Decrease in surface area
 d 2 
9
9
7

d 2  1  d 2  d 2
16
16
 16 
 Percentage decrease in surface area

decrease in surface area
 100%
original surface area
J
A
7
d 2
7
 16 2  100%   100% = 43.75% Ans.
16
d
PRACTICE EXERCISE
Questions based on Surface area of cuboid and cube
B
1. Find the surface area of a cuboid 16 m long, 14 m broad and 7 m high.
2. Find the length of the longest pole that can be placed in a room 12 m long, 8 m broad and 9 m high.
3. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of Rs. 8.50 per m2.
4. Find the percentage increase in the surface area of a cube when each side is doubled.
5. The paint in a certain container is sufficient to paint an area equal to 9.375 m 2. How many bricks of
dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
—NCERT
6. A small indoor green house (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?
—NCERT
7. Three cubes each of side 6 cm are joined end to end. Find the surface of the resulting cuboid.
8. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at top. Ignoring the
thickness of the plastic sheet, find:
(i) area of the sheet required to make the box.
(ii) the cost of the sheet for it, if a sheet measuring 1 m2 cost Rs. 22.
9. If the surface area of the cube is 96 cm2, find its edge and length of its diagonal.
10. The dimensions of a rectangular box are in the ratio of 2 : 3 : 4 and the difference between the cost of
covering it with sheet of paper at the rate of Rs. 4 and Rs. 4.50 per m2 is Rs. 650. Find the dimensions of
the box.
11. Mary wants to decorate her christmas tree. She wants to place the tree on a wooden box covered with
coloured paper with picture of Santaclaus on it (see figure). She must know the exact quantity of paper
to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively,
how many square sheets of paper of side 40 cm would she require?
—NCERT
T
I
M
A
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
21
J
A
12. The length and breadth of a hall are in the ratio 4 : 3 and its height is 5.5 meters. The cost of decorating
its walls (including doors and windows) at Rs. 6.60 per m2 is Rs. 5082. Find the length and breadth of the
room.
Questions based on Surface area of a cylinder
J
A
13. The curved surface area of a right circular cylinder of height 14 cm is 176 cm2. Find the diameter of the
base of the cylinder.
14. 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). Find its :
—NCERT
M
A
T
I
15.
16.
17.
18.
19.
22
B
(i) inner curved surface area
(ii) outer curved surface area
(iii) Total surface area
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.
An iron pipe 20 cm long has exterior diameter 25 cm. If the thickness of the pipe is 1 cm, find the total
surface area of the pipe.
A rectangular sheet of paper 88 cm × 50 cm is rolled along its length and a cylinder is formed. Find curved
surface area of the cylinder formed.
A solid cylinder has total surface area of 462 cm2. Its curved surface area is one-third of its total surface
area. Find the radius and height of the cylinder.
The total surface area of a hollow metal cylinder, open at both the ends of external radius 8 cm and height
10 cm is 338  cm2. Find thickness of the metal in the cylinder.
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
20. In the given figure, you see the frame of a lamp shade. 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 is required for covering the lampshade.
—NCERT
Question based on surface area of a Cone
B
J
A
J
A
21. Find the curved surface area of a cone, if its slant height is 50 cm and the diameter of its base is 28 cm.
22. Find the total surface area of a cone, if its slant height is 21 cm and diameter of its base is 24 cm.
—NCERT
T
I
22 

23. The curved surface area of a cone is 4070 cm2 and its radius is 35 cm. What is its slant height?  use π 

7 

24. The radius and slant height of a cone are in the ratio 4 : 7. If its curved surface area is 792 cm2, find its
22 

radius.  use π 

7 

25. The circumference of the base of a 10 m high conical tent is 44 m. Calculate the length of canvas used in
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making the tent if width of canvas is 2 m.  use π  22 
7 

26. What length of tarpaulin 3 m wide will be required to make 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  = 3.14)
—NCERT
27. 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 m2?
—NCERT
2
28. The curved surface area of a cone of radius 6 cm is 188.4 cm . Find its height.
Questions based on surface area of sphere and hemi-sphere
29. The surface area of a sphere is 5544 cm2. Find its radius.
30. A hemi-spherical 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.
—NCERT
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
23
31. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their
surface areas.
—NCERT
32. In the given figure, a right cylinder just encloses a sphere of radius r. Find
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(i) Surface area of the sphere
(ii) curved surface area of the cylinder
(iii) ratio of the areas obtained in (i) and (ii).
33. The internal and external diameters of a hollow hemi-spherical vessel are 20 cm and 28 cm respectively.
Find the cost of painting the vessel all over at 15 paisa per cm2.
34. A toy is in the form of a cone mounted on a hemi-sphere. The diameter of the base and the height of the
cone are 6 cm and 4 cm respectively. Find the surface area of the toy. (use  = 3.14)
Question based on Volume of Cuboid and Cube
B
35. The total surface area of a cube is 1350 cm2. Find its volume.
36. 500 persons took dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level
of water in the tank, if the average displacement of water by a person is 4 m3?
37. Three cubes of a metal with edges 6 cm, 8 cm and 10 cm respectively are melted and formed into a single
cube. Find the edge of the new cube formed. Also, find its volume.
38. The volume of a cuboid is 1536 m3. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find
surface area of the cuboid.
39. A field is 70 m long and 40 m broad. In one corner of the field, a pit which is 10 m long, 8 m broad and 5
m deep, has been dug out. The earth taken out of it is evenly spread over the remaining part of the field.
Find the rise in the level of the field.
40. The areas of three adjacent faces of a cuboid are 15 cm2, 40 cm2 and 24 cm2. Find the volume of the
cuboid.
41. How many bricks, each measuring 25 cm × 15 cm × 8 cm will be required to build a wall 10 m × 4 dm × 5 m
when one-tenth of its volume is occupied by mortar?
42. A rectangular reservoir is 120 m long and 75 m wide. At what speed per hour must water flow into it
through a square pipe of 20 cm wide so that the water rises by 2.4 m in 18 hours.
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Question based on Volume of a Cylinder
43. The radius of a cylinder is 14 cm and its height is 40 cm. Find (i) curved surface area (ii) the total surface
area (iii) volume of the cylinder.
44. The total surface area of a cylinder is 462 cm2. Its curved surface is one-third of its total surface area. Find
the volume of the cylinder.
45. The curved surface area and the volume of a pillar are 264 m2 and 396 m3 respectively. Find the diameter
and the height of the pillar.
24
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
46. The sum of the height and radius of the base of a solid cylinder is 37 m. If the total surface area of the
cylinder is 1628 m2, find its volume.
47. A cylindrical tube, open at both ends, is made up of metal. The internal radius of the tube is 5.2 cm and
its length is 25 cm. The thickness of the metal is 8 mm. Calculate the volume of the metal.
48. 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?
—NCERT
49. If the diameter of the cross-section of a wire is decreased by 5%, how much percent will the length be
increased so that the volume remains the same?
—NCERT
50. Water flows out through a circular pipe, whose internal diameter is 2 cm, at the rate of 70 cm per second
into a cylindrical tank, the radius of whose base is 40 cm. By how much time will the level of water rise in
half an hour?
51. A rectangular piece of paper is 22 cm long and 12 cm wide. A cylinder is formed by rolling the paper along
its length. Find the volume of the cylinder.
52. A well, with inner radius 4m, is dug 14 m deep. The earth taken out of it has been spread evenly all round
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22 

it to a width of 3m to form an embankement. Find the height of this embankement.  use π 

7 

Question based on Volume of a Cone
B
53. The base radii of two cones of the same height are in the ratio 3 : 4. Find the ratio of their volumes.
22 

54. A cone of height 24 cm has curved surface area 550 cm2. Find its volume.  use π 

7 

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55. The radius and height of a right circular cone are in the ratio of 5 : 12. If its volume is 314 cm3, find the slant
height and radius of the base of the cone. (use  = 31.4)
56. Find the slant height and curved surface area of a cone whose volume is 12935 cm3 and the radius of the
base is 21 cm.
57. A semi-circular thin sheet of metal of diameter 28 cm is bent and an open conical cup is made. Find the
capacity of the cup.
58. Monica has a piece of canvas whose area is 551 m2. She uses it to have a conical tent made with a base
radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts
to approximately 1 m2, find the volume of the tent that can be made with it.
—NCERT
59. Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.
60. A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is
made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved
surface area.
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Question based on Volume of a Sphere and Hemi-Sphere
61. Find the surface area of a sphere whose volume is 606.375 m3.
62. A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameter 0.6 cm.
Find the number of small balls thus obtained.
63. The surface areas of two spheres are in the ratio 1 : 4. Find the ratio of their volumes.
64. The diameter of a metallic sphere is 6 cm. It is melted and drawn into a wire having diameter of the crosssection as 2 mm. Find the length of the wire.
MATHEMATICS–IX
SURFACE AREAS AND VOLUMES
25
65. The radii of the internal and external surfaces of a metallic spherical shell are 3 cm and 5 cm respectively.
It is melted and recast into a solid cylinder of height 10
2
cm. Find the diameter of the base of the
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cylinder.
66. The largest sphere is carved out of a cube of side 7 cm. Find the volume of the sphere. (use  = 3.14)
67. Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.
68. A hemi-spherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume
of the steel used in making it.
69. A hemi-spherical bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical
shaped small bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?
70. A hemi-sphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Find the height
of the cone.
Miscellaneous Questions
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71. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume
be
1
of the volume of the given cone, at what height above the base is the section made?
27
B
72. A circus tent consists of a cylindrical base surmounted by a conical roof. The radius of the cylinder is 20
m. The height of the tent is 63 m and that of the cylindrical base is 42 m. Find the volume of air contained
in the tent and the area of canvas used for making it.
73. How many litres of water flows out of pipe having an area of cross-section of 5 cm2 in one minute, if the
speed of water in the pipe is 30 cm/sec ?
74. A sphere of diameter 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the
vessel is 6 cm. If the sphere is completely submerged in water, find by how much will the surface level of
water be raised.
75. An ice-cream cone has a hemispherical top. If the height of the conical portion is 9 cm and base radius
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22 

2.5 cm, find the volume of ice-cream in the ice-cream cone.  use π 

7 

76. In the given figure, a solid is made of a cylinder with hemispherical ends. If the entire length of the solid
is 108 cm and the diameter of the hemispherical ends is 36 cm, find the cost of polishing the surface of the
solid at the rate of 7 paisa per cm2.
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77. A spherical copper ball of diameter 9 cm is melted and drawn into a wire, the diameter of whose thickness
is 2 mm. Find the length of the wire in meters.
78. The difference between the inside and outside surfaces of a cylindrical water pipe 14 m long is 88 m2. If
the volume of pipe be 176 m3. Find the inner and outer radii of the water pipe.
26
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX
79. Water is flowing at the rate of 2.5 km/hr through a circular pipe 20 cm internal diameter, into a circular
cistern of diameter 20 m and depth 2.5 m. In how much time will the cistern be filled?
80. A conical vessel of radius 6 cm and height 8 cm is filled with water. A sphere is lowered into the water (see
figure), and its size is such that when it touches the sides of the conical vessel, it is just immersed. How
much water will remain in the cone after the overflow?
PRACTICE TEST
M.M. : 30
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Time : 1 hour
General Instructions :
Q. 1-4 carry 2 marks, Q. 5-8 carry 3 marks and Q. 9-10 carry 5 marks each.
B
1. The floor of a rectangular hall has a 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 the hall.
2. 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. 1250 per m2.
3. A right triangle PQR with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume
of the solid so obtained.
4. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
5. A joker’s cap is in the form of a right circular cone of base 7 cm and height 24 cm. Find the area of the
sheet required to make 10 such caps.
6. 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.
7. A village, having a population of 4000, requires 150 l 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 ?
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 the hospital has to prepare daily to serve 250 patients?
9. The radius and height of a cone are in the ratio 4 : 3. The area of the base is 154 cm2. Find the area of the
curved surface.
10. The diameter of a sphere is decreased by 25%. By what percent its curved surface area decrease?
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ANSWERS OF PRACTICE EXERCISE
1. 868 cm2
2. 17 m
5. 100
6. (i) 4250 cm2 (ii) 320 cm 7. 360 cm2
8. (i) 5.45 m2 (ii) 119.90 Rs.
9. 4 cm, 4 3 cm
10. 10 m, 15 m, 20 m
12. 40 m, 30 m
13. 4 cm
MATHEMATICS–IX
3. Rs. 629
11. 7
14. (i) 968 cm2 (ii) 1064.80 cm2 (iii) 2038.08 cm2
SURFACE AREAS AND VOLUMES
4. 300%
15. (i) 110 m2 (ii) Rs. 4400
27
16. 3168 cm2
17. 440 cm2
18. r = 7 cm, h = 14 cm
19. 3 cm
20. 2200 cm2
21. 2200 cm2
22. 1244.57 cm2
23. 37 cm
24. 12 cm
25. 134.2 m
26. 63 m
27. Rs. 1155
28. 8 cm
29. 21 cm
30. Rs. 27.72
31. 1 : 16
32. (i) 4  r2 (ii) 4r2 (iii) 1 : 1
33. 324.34 Rs
34. 103.62 cm2
35. 3375 cm3
36. 50 cm
37. 12 cm, 1728 cm3
38. 832 cm2
39. 14.7 cm
40. 120 cm3
41. 6000
42. 30 km/hr
44. 539 cm3
46. 4620 m3
47. 704 cm3
48. cylindrical tin, 85 cm2 49. 10.8%
50. 78.75 cm
51. 462 m3
52. 6.8 m (approx)
53. 9 : 16
54. 1232 cm3
55. 13 cm, 5 cm
56. 35 cm, 2310 cm2
57. 622.3 cm2
58. 1232 m3
59. 190.93 cm3
60. 415.8 cm3, 233.9 cm2
61. 346.5 m2
62. 1000
63. 1 : 8
64. 36 m
65. 7 cm
66. 179.5 cm3
67. 6 : 
68. 56.83 cm3
70. 28.44 cm
71. 20 cm
72. 61600 m3, 7102.85 m2
73. 9 l
74. 1 cm
75. 91.66 cm3
76. Rs. 855.36
77. 121.5 m
78. 1.5 m, 2.5 m
79. 10 hours
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80. 188.57 cm3
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43. (i) 3520 cm2 (ii) 4752 cm2 (iii) 24640 cm3
45. 6 m, 14 m
69. 54
ANSWERS OF PRACTICE TEST
1. 6 m
2. Rs. 68.75
3. 100  cm3
4. 0.303 l (approx.)
5. 5500 cm2
6. 173.25 cm2
7. 3 days
8. 38.5 l
9. 192.5 cm2
10. 43.75%
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28
SURFACE AREAS AND VOLUMES
MATHEMATICS–IX