X Maths gb
AJAY PARMAR
1
GROUP TUITION
Chapter:14 Surface area and volume
Formulae for Area
Cube
2
Curved surface area of cube = 4l
Total surface area of cube = 6l2
Cuboid
Curved surface area of cuboid = 2h(l + b)
Total surface area of cuboid = 2(lb + bh + hl)
Cylinder
Curved surface area of cylinder = 2  r h
Total surface area of cylinder = 2r (r + h)
Cone
In a cone l = slant height or lateral height h = vertical height and r = radius  l2 = r2 + h2
Curved surface area of cone =  r l
Total surface area of cone =  r (l + r)
Sphere
2
Surface area of Sphere = 4  r
Hemisphere
2
Curved surface area of hemisphere = 2  r
Total surface area of hemisphere = 3  r2
Frustum
Bigger radius = r1, Smaller radius = r2, Height = h and Slant Height = l
l2 = h2 + (r1 – r2)2
Curved surface area of frustum =  (r1 + r2) l
Total surface area of frustum =  (r1 + r2) l + r12 + r22
Perimeter of a square = 4l
Perimeter of a triangle = a + b + c
Perimeter
Perimeter of a rectangle = 2(l + b)
Circumference of a circle = 2  r
Formulae for Volume
Volume of Cylinder =  r2 h
1
Volume of Cone =  r2 h
3
4
Volume of Sphere =
 r3
3
Volume of Hemisphere =
Volume of Frustum =
2
 r3
3
1
 h[r12 + r22 + r1r2]
3
 If One Object is melted and recast into another One Object then
Volume of the new object = Volume of the original object
 If One Big Object is melted and recast into Many (n) Small Objects of equal size then
n × Volume of smaller object = Volume of big object.
Volume of big object
 Number of small objects =
Volume of 1 small object
Ex. 14.1
1. A toy is made by mounting a cone on a hemisphere. The radius of the cone and the hemisphere is 5
cm. The total height of the toy is 17 cm. Find the total surface area of the toy.
Height of the toy = 12 cm
Hemisphere
Cone
Radius r = 5 cm
Radius r = 5 cm
12cm
17cm
Height h = 17 – 5 = 12 cm
Slant height l,
5cm
l2 = h2 + r2 = 122 + 52 = 144 + 25 = 169
 l = 13 cm
5cm
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AJAY PARMAR
GROUP TUITION
Chapter:14 Surface area and volume
TSA of toy
= CSA of the cone + CSA of the hemisphere
= rl + 2r2
= r (l + 2r)
22
× 5 (13 + 2 × 5)
7
2530
=
= 361.428571 = 361.43 cm2
7
=
Thus, the total surface area of the toy is 361.43 cm2.
2. A show piece shown in the figure is made of two solids – a cube and hemisphere. The base of the block
is a cube with edge 7 cm and the hemisphere fixed on the top has diameter 5.2 cm. Find the total
surface area of the piece.
Cube
Hemisphere
Length l = 7 cm
Radius r =
diameter
5.2
=
= 2.6 cm
2
2
TSA of the show piece
= TSA of the cube + CSA of the hemisphere – Area of the base of the hemisphere
7cm
= 6l2 + 2r2 – r2
2
2
= 6l + r
= 6 × (7)2 +
22
× (2.6)2
7
= 294 + 21.25 = 315.25 cm2
Thus, the total surface area of the show piece is 315.25 cm2.
3. A vessel is in the form of a hemisphere mounted on a hollow cylinder. The diameter of the hemisphere
is 21 cm and the height of the cylinder is 25 cm. If the vessel is to be painted at the rate of ` 3.5 per
cm2, then find the total cost to paint the vessel from outside.
Hemisphere
Cylinder
Radius = r =
diameter
21
=
cm = 10.5cm
2
2
Radius r =
diameter
21
=
cm = 10.5cm
2
2
Height h = 25 cm
TSA of the vessel
= CSA of the cylinder + CSA of the hemisphere
= 2rh + 2r2
= 2r (h + r)
Cost of painting 1 cm2 = ` 3.5
10.5cm
 Total cost = ` [3.5 × 2r (h + r)]
` 3.5 × 2 ×
25cm
10.5cm
22
21
×
× (25 + 10.5)
2
7
= ` 3.5 × 66 × 35.5 = ` 8200.50
The cost of painting the vessel from outside is ` 8200.50.
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AJAY PARMAR
3
GROUP TUITION
Chapter:14 Surface area and volume
4. Chirag made a bird bath for his garden in the shape of a cylinder with a hemispherical depression at
one end, (see the figure). The height of the cylinder is 1.5 m and its radius is 50 cm. Find the total area
of the bird bath. ( = 3.14)
50cm
Cylinder
Hemisphere
Height h = 1.5 m = 150 cm
Radius r = 50 cm
Radius r = 50 cm
TSA of the bird bath
1.5m
= CSA of the cylinder + CSA of the hemisphere
= 2rh + 2r2
= 2r (h + r)
= 2 × 3.14 × 50 (150 + 50)
= 314 × 200 = 62800 cm2
Total surface area of bird bath is 62800 cm2.
5. A solid is composed of a cylinder with hemispherical ends on both the sides. The radius and the height
of the cylinder are 35 cm and 20cm respectively. Find the total surface area of the solid.
Cylinder
Hemisphere(2)
Radius r = 35 cm
Radius r = 35 cm
Height h = 20 cm
TSA of the solid
= CSA of the cylinder + 2 × CSA of the hemisphere
20cm
= 2rh + 2 × 2r2
35cm
= 2r (h + 2r)
=2×
22
× 35 (20 + 2 × 35)
7
35cm
= 220 × 90 = 19800 cm2
Total surface area of the given solid is 19800 cm2.
6. The radius of a conical tent is 4 m and slant height is 5m. How many meters of canvas of width 125
cm will be used to prepare 12 tents? If the cost of canvas is ` 20 per meter, then what is the total cost
of 12 tents? ( = 3.14)
Tent (Cone)
Radius r = 4 m
Slant height l = 5 m
Area of the canvas required to prepare 12 tents = 12 × rl
Width of the canvas = 125 cm = 1.25 m
Area of the canvas = Length × Width [as shape of cloth is rectangle]
Area of canvas = Area of 12 tents
L × 1.25 = 12 × rl
 Length =
12  3.14  4  5
= 602. 88 m
1.25
 Length = 602.88 m
Cost of 1 m canvas = ` 20
 Total cost = ` (602.88 × 20) = ` 12057.60
The total cost of 12 tents is ` 12057.60
7. If the radius of a cone is 60 cm and its curved surface area is 23.55 m 2, then find its slant height.
( = 3.14)
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GROUP TUITION
Chapter:14 Surface area and volume
Cone
Radius r = 60 cm = 0.60 m
Curved surface area of the cone = rl
 23.55 = 3.14 × 0.60 × l
l=
2355  100  100
1250
=
= 12.50 m
100  314  60
100
The slant height of the cone is 12.50 m
8. The cost of painting the surface of sphere is ` 1526.04 at the rate of ` 6 per m2. Find radius of sphere.
( = 3.14)
Area of the region painted at the cost of ` 6 = 1 m2
 Area of the region painted at the cost of ` 1526 =
1526.04
= 254.34 m2
6
 Surface area of the sphere = 254.34 m2
Surface area of a sphere = 4r2
 254.34 = 4 × 3.14 × r2
1
= 20.25
4  3.14
 r2
= 254.34 ×
r
= 20.25 = 4.5 m
Radius of the sphere is 4.5 m
Ex. 14.2
1. The curved surface area of a cone 550 cm2. If its diameter is 14 cm, find its volume.
Cone
Radius r =
diameter
14
=
= 7 cm.
2
2
CSA of the cone = 550 cm2
rl = 550
22
× 7 × l = 550
7
550
 l=
= 25 cm
22

For a cone, l2 = r2 + h2
 h2 = l2 – r2 = 625 – 49 = 576
 h = 24 cm
Volume of a cone =
1 2
1
22
r h = ×
× 7 × 7 × 24 = 1232 cm3
3
3
7
Volume of the cone is 1232 cm3.
2. A solid is in the form of cone with hemispherical base. The radius of the cone is 15 cm and the total
height of the solid is 55 cm. Find the volume of the solid.
Total height of solid = 55 cm
Hemisphere
Cone
Radius r = 15 cm
Radius r = 15 cm
Height h = Total height – Radius = 55 – 15 = 40 cm
Volume of the combined solid
= Volume of the cone + Volume of the hemisphere
2
1 2
r h + r3
3
3
πr 2
=
(h + 2r)
3
22
15  15
=
×
(40 + 2 × 15)
7
3
=
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40cm
55cm
15cm
15cm
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AJAY PARMAR
GROUP TUITION
Chapter:14 Surface area and volume
=
22
× 15 × 5 × 70 = 16500 cm3
7
Volume of the given combined solid is 16500 cm3.
3. How many liters of milk can be stored in a cylindrical tank radius 1.4 m and height 3 m?
Cylindrical tank
Radius r = 1.4 m
height h = 3 m
= r2h =
Volume of the cylindrical tank
22
× 1.4 × 1.4 × 3 = 18.48 m3
7
1 m3 = 1000 liters
 18.48 m3 = (18.48 × 1000) liters = 18480 liters
Thus, the capacity of the cylindrical tank is 18480 liters.
4. The spherical balloon with radius 21 cm is filled with air. Find the volume of air contained in it.
For the spherical balloon, radius r = 21 cm
Volume of the air contained in the spherical balloon = Volume of a sphere =
=
4 3
r
3
4
22
×
× 21 × 21 × 21 = 38808 cm3
3
7
Thus, the volume of the air contained in the balloon is 38808 cm3.
5. A solid has hemispherical base with diameter 8.5 cm and diameter of cylinder with height 8 cm and
diameter of cylinder is 2 cm. Find the volume of this solid. ( = 3.14)
Cylinder
Radius r =
2
diameter
= = 1 cm
2
2
Height h = 8 cm.
Radius of the hemisphere R =
8.5
diameter
=
= 4.25 cm
2
2
Volume of the combined solid = Volume of the cylinder + Volume of the hemisphere
= r2h +
2
R3
3
= 3.14 × 1 × 1 × 8
8cm
+
2
× 3.14 × 4.25 × 4.25 × 4.25
3
8.25cm
= 25.12 + 160.70
= 185.82 cm3
The volume of the given combined solid is 185.82 cm3.
8.25cm
6. A playing top is made up of steel. The top is shaped like a cone surmounted by a hemisphere. The
total height of top is 5 cm and the diameter of the top is 3.5 cm. Find the volume of the top.
Total height of top = 5 cm
Hemisphere
Cone
Radius r =
3.5
diameter
=
= 1.75 cm
2
2
Radius r =
3.5
diameter
=
= 1.75 cm
2
2
Height h = Total height– Radius = 5 – 1.75 = 3.25 cm.
Volume of the top
= Volume of the cone + Volume of the hemisphere
2
1 2
r h + r3
3
3
2
= πr (h + 2r)
3
1.75  1.75
22
=
×
(3.25 + 2 × 1.75)
3
7
=
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1.75cm
1.75cm
5cm
3.25 cm
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GROUP TUITION
Chapter:14 Surface area and volume
=
22  0.25  1.75
× 6.75
3
= 22 × 0.25 × 1.75 × 2.25
= 21.65625 = 21.66 cm3
Volume of the top is 21.66 cm3.
7. How many liters of petrol will be contained in a closed cylindrical tank with hemisphere at one end
having radius 4.2 cm and total height 27.5 cm?
Total height = 27.5cm
Hemisphere
Cylinder
Radius r = 4.2 cm
Radius r = 4.2 cm
Height h = Total height – Radius = 27.5 – 4.2 = 23.3 cm
Volume of the tank
= Volume of the cylinder + Volume of the hemisphere
2 3
r
3
2
= r2 (h + r)
3
2
22
=
× 4.2 × 4.2 (23.3 + × 4.2)
3
7
= r2h +
23.3cm
4.2cm
4.2cm
= 22 × 0.6 × 4.2(26.1)
= 1446.984 cm3
=
1446.984
liters ( 1 liter = 1000 cm3)
1000
= 1.45 liters
1.45 liters of petrol will be consumed in the given tank.
8. The capacity of a cylindrical tank at a petrol pump is 57,750 liters. If its diameter is 3.5 m, find the
height of cylinder.
Cylinder
Capacity = 57750 liters.
57750 liters =
57750
m3 = 57.75 m3
1000
 Volume of the cylindrical tank = 57.75 m3
Radius of the cylindrical tank r =
3.5
diameter
=
= 1.75 m and height = h m
2
2
Volume of the cylindrical tank = r2h
22
× 1.75 × 1.75 × h
7
22
175
175
 57.75 =
×
×
×h
7
100
100
5775
7
100
100
h=
×
×
×
=6m
100
22
175
175
 57.75
=
The height of the cylindrical tank is 6 m.
9. A hemispherical pond is filled with 523.908 m3 of water. Find the maximum depth of the pond.
Hemisphere
Volume = 523.908 m3 and its maximum depth (radius) = r m
Volume of a hemisphere =
2 3
r
3
2
22
×
× r3
3
7
3
523908
7
3
r =
× ×
2
1000
22
 523.908 =
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GROUP TUITION
Chapter:14 Surface area and volume
117907  3  7
1000
63  63  63
3
r =
10  10  10
 r3 =
 r3 =  63 
3
 10 
 r = 6.3 m
Thus, the maximum depth of the hemispherical pond is 6.3 m
10. A gulab – jamun contains 40 % sugar syrup in it. Find how much syrup would be there in 50 gulab –
jamuns, each shaped like a cylinder with two hemispherical ends with total length 5 cm and diameter
2.8 cm.
Total length = 5cm
Hemisphere(2)
Cylinder
Radius r =
2.8
diameter
=
= 1.4 cm
2
2
Radius r =
2.8
diameter
=
= 1.4 cm
2
2
Height h = Total length – 2 × Radius of hemispherical ends
5cm
= 5 – 2 × 1.4 = 2.2 cm
Volume of one gulab jamun
= Volume of the cylindrical part + Volume of the hemispherical part
= r2h + 2 ×
= r2(h +
=
2 3
r
3
2.8cm
2.8cm
4
r)
3
4
22
× 1.4 × 1.4 (2.2 + × 1.4)
3
7
= 6.16 × 4.07
= 25.0712 = 25.07 cm3
 Volume of 50 gulab jamuns = 50 × 25.07
There is 40 % syrup in the volume of each gulab jamun.
 Volume of the syrup in 50 gulab jamun
= 40 % of the syrup in 50 gulab jamuns
= 40 % of (50 × 25.07)
=
40
× 50 × 25.07
100
= 501.4 cm3
=
501.4
liter
1000
= 0.5014
= 0.5 liter
There is 0.5 litre of syrup in 50 gulab jamuns.
11. The height and the slant height of a cone are 12 cm and 20 cm respectively. Find its volume. ( = 3.14)
Cone
Height h = 12cm
Slant height l = 20 cm and radius = r cm
For a cone,
 l2 = h2 + r2
 202 = 122 + r2
 r2 = 202 – 122
 r2 = 400 – 144 = 256
 r = 16 cm
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GROUP TUITION
Chapter:14 Surface area and volume
Volume of a cone =
1 2
1
r h = × 3.14 × 16 × 16 × 12 = 3215.36 cm3
3
3
The volume of the cone is 3215.36 cm3.
12. Find the total volume of a cone having a hemispherical base, if a radius of the base is 21 cm and
height of cone is 60 cm.
Hemisphere
Cone
Radius r = 21 cm
Radius r = 21 cm
Height h = 60 cm
Volume of the combined solid
= Volume of the cone + Volume of the hemisphere
60cm
2
1 2
r h + r3
3
3
1
= r2(h + 2r)
3
1
22
= ×
× 21 × 21 × (60 + 2 × 21)
3
7
=
21cm
21cm
= 462 × 102 = 47124 cm3
Volume of the given solid is 47124 cm3.
13. If the slant height of a cone is 18.7 cm and the curved surface area is 602.8 cm 2, find the volume of
cone. ( = 3.14)
Cone
Slant height l = 18.7 cm and CSA = 602.8 cm2
CSA of a cone = rl
 602.8 = 3.14 × r × 18.7
r=
6028
100
10
×
×
10
314
187
 r = 10.27 cm
For a cone,
 h2 = l2 – r2
 h2 = 18.72 + 10.272
 h2 = 349.69 – 105.4729 = 244.2171
 h = 15.63 cm
Volume of a cone =
1 2
1
r h = × 3.14 × 10.27 × 10.27 × 15.63 = 1725.47 cm3
3
3
The volume of the cone is 1725.47 cm3.
14. If the surface of spherical ball is 1256 cm2, then find the volume of the sphere. ( = 3.14)
Radius of the spherical ball = r cm
Surface area of the spherical ball = 1256 cm2
Surface area of the sphere = 4r2
 1256 = 4 × 3.14 × r2
 r2 =
1256
4  3.14
 r2 = 100
 r = 10 cm
Volume of a sphere =
4 3
4
r = × 3.14 × 10 × 10 × 10 = 4186.67 cm3
3
3
Volume of the spherical ball is 4186.67 cm3
Ex. 14.3
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AJAY PARMAR
GROUP TUITION
Chapter:14 Surface area and volume
1. A hemispherical bowl of internal radius 12 cm contains some liquid. This liquid is to be filled into
cylindrical bottles of diameter 4 cm and height 6 cm. How many bottles can be filled with this liquid?
Hemisphere
Cylinder
Radius r = 12 cm,
Radius R =
4
diameter
= = 2 cm
2
2
Height H = 6 cm
Suppose, n bottles can be filled with the liquid in the hemispherical bowl.
 Volume of n cylindrical bottles = Volume of the hemispherical bowl
2 3
r
3
2
 n × 2 × 2 × 6 = × 12 × 12 × 12
3
2 12  12  12
n= ×
3
2 2 6
 n × R2H =
 n = 48
Thus, 48 bottles can be filled with the liquid in the hemispherical bowl.
2. A cylindrical container having diameter 16 cm and height 40 cm is full of ice cream. The ice cream is
to be filled into cones of height 12 cm and diameter 4 cm, having a hemispherical shape on the top.
Find the number of such cones which can be filled with the ice cream.
Ice cream cone
Cylinder
Hemisphere
Radius = R =
4
= 2 cm
2
diameter
16
=
= 8 cm
2
2
Cone
Radius r =
Radius = R = 2 cm
Height h = 40 cm
Height H = 12 cm
Volume of the ice cream used to fill one cone
= Volume of the cone + Volume of the hemisphere
2cm
2cm
12 cm
2
1
R2H + R3
3
3
1
= R2 (H + 2R)
3
=
Number of cones that can be filled
=
3  8  8  40
Volume of ice cream in the cylinder
3  8  8  40
  r2  h
=
=
=
= 120
1
2

2

[
12

2(2)]
Volume of ice cream in one cone
2

2

16
2
  R (H  2R)
3
Thus, 120 cones can be filled with the ice cream.
3. A cylindrical tank of diameter 3 m and height 7 m is completely filled with groundnut oil. It is to be
emptied in tins each of capacity 15 litres. Find the number of such tins required.
Cylinder
Radius r =
3
diameter
= m
2
2
Height h = 7 m
Volume of the groundnut oil in the cylindrical tank = r2h m3 = 1000r2h litres
Volume of one tin = 15 litres
Number of oil tins required =
1000r 2 h
Volume of cylindrical tank in litres
1000  22  3  3  7
=
=
= 3300
Volume of one tin
15  7  2  2
15
No. of tins required = 3300 tins.
4. A cylinder of radius 2 cm and height 10 cm is melted into small spherical ball of diameter 1 cm. Find
the number of such balls.
Cylinder
Spherical balls
Radius r = 2 cm
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Radius R =
1
diameter
= cm
2
2
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AJAY PARMAR
GROUP TUITION
Chapter:14 Surface area and volume
Height h = 10 cm
For each spherical ball,
Suppose, n balls are produced by melting the cylinder.
 Volume of n balls =Volume of the cylinder
4
R3 = r2h
3
4
1
1
1
 n × × × × = 2 × 2 × 10
3
2
2
2
1
 n × = 40
6
n×
 n = 40 × 6
 n = 240
Thus, 240 balls are produced.
5. A metallic sphere of radius 15 cm is melted and a wire of diameter 1 cm is drawn from it. Find the
length of wire.
Sphere
Wire (Cylinder)
Radius r = 15 cm
radius R =
1
diameter
= cm
2
2
Height H = ?
Wire is made from metallic sphere.
 Volume of wire = Volume of the sphere
R2H =
4 3
r
3
1
1
4
× × H = × 15 × 15 × 15
2
2
3
4
 H = × 15 × 15 × 15 × 2 × 2
3

 H = 18000 cm
 H = 180 m
Thus, the length of the wire is 180 m.
6. There are 45 conical heaps of wheat, each of them having diameter 80 cm and height 30 cm. To store
the wheat in a cylindrical container of the same radius, what will be the height of cylinder?
Cone (45)
Cylinder
Radius r =
80
= 40 cm
2
Radius R = ?
Height h = 30 cm
Height H = ?
Volume of cylindrical container = Volume of 45 conical heaps of wheat
 r2h = 45 ×
1
r2h
3
 40 × 40 × H = 15 × 40 × 40 × 30
 H = 450 cm = 4.5 m
Thus, the height of the cylindrical container is 4.5 m.
7. A cylindrical bucket, 44 cm high and having radius of base 21 cm, is filled with sand. This bucket is
emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 33 cm,
find the radius and the slant height of the heap.
Cylinder
Conical heap
Radius r = 21 cm
Height H = 33 cm and
Height h = 44 cm
radius R = ?
Volume of sand in the conical heap = Volume of sand the cylindrical bucket
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GROUP TUITION
Chapter:14 Surface area and volume
1
R2H = r2h
3
1 2
 R × 33 = 21 × 21 × 44
3
21 21 44  3
 R2 =
33

 R2 = 1764  R = 42 cm
For the conical heap,
l2 = R2 + H2 = 422 + 332 = 1764 + 1089 = 2853
 l = 53.41 cm
Thus, the radius of the conical heap is 42 cm and its slant height is 53.42 cm.
Ex. 14.4
1. An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base
made of the same metallic sheet. The total vertical height of the bucket is 40 cm and that of cylindrical
base is 10 cm, radii of two circular ends are 60 cm and 20 cm. Find the area of the metallic sheet used.
Also find the volume of water the bucket can hold? ( = 3.14)
Frustum
Cylinder
60cm
bigger radius r1 = 60 cm; smaller radius r2 = 20 cm
Radius R = 20 cm
Height h = total height– height of the cylindrical
Height H = 10 cm
30cm
= 40 – 10 = 30 cm
40cm
Slant height l
20cm
l2 = h2 + (r1 – r2)2 = 302 + 402 = 2500
10cm
l = 50 cm
Area of the metal sheet used to make the bucket
= CSA of the cylinder + CSA of the frustum of a cone + Area of the base of the frustum
= 2RH + l(r1 + r2) + r22
= [2RH + l(r1 + r2) + r22]
= 3.14[2 × 20 × 10 + 50(60 + 20) + 202]
= 3.14[400 + 4000 + 400]
= 3.14 × 4800
= 15072 cm2
Volume of the water that the bucket can hold
= Volume of the frustum of a cone
1
h (r12 + r22 + r1r2)
3
1
= × 3.14 × 30(602 + 202 + 60 × 20)
3
=
= 3.14 × 5200 = 163280 cm3 = 163.280 liters
Thus, the area of the metal sheet used is 15072 cm2 and the bucket can hold 163.280 liters of water.
2. A container, open from the top is made up of a metal sheet in the form of frustum of a cone of height
30 cm with radii 50 cm and 10 cm. Find the cost of the milk which can completely fill container at the
rate of ` 30 per litrf. Also find the cost of metal sheet used to make the container, if it cost ` 50 per 100
cm2. ( = 3.14)
60cm
Frustum
Bigger radius r1 = 50 cm; Smaller radius r2 = 10 cm and Height h = 30 cm
30cm
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AJAY PARMAR
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GROUP TUITION
Chapter:14 Surface area and volume
Slant height l
l2 = h2 + (r1 – r2)2 = 302 + 402 = 2500
l = 50 cm
Volume of the container in the shape of the frustum of a cone
1
h (r12 + r22 + r1r2)
3
1
= × 3.14 × 30(502 + 102 + 50 × 10)
3
1
= × 3.14 × 30(2500 + 100 + 500)
3
97340
= 3.14 × 3100 = 97340 cm3 =
liters = 9734 liters
1000
=
Thus, 97.34 liters of milk will fill the container completely.
Cost of 1 liter of milk = ` 30
 Cost of 97.34 liter of milk = ` (97.34 × 30) = ` 2920.20
Thus, the cost of milk that can fill the container completely is ` 2920.20
Area of the metal sheet used to make the container
= CSA of the frustum of a cone + area of the smaller circle in the base
= l(r1 + r2) + r22
= 3.14 × 50(50 + 10) + 3.14 10 × 10
= 9420 + 314
= 9734 cm2
Cost of 100 cm2 metal sheet = ` 50
9734  50 
 Cost of 9734 cm2 metal sheet = ` 
 = ` 4867

100

Thus, the cost of metal sheet used to make the container is ` 4867
Ex. 14
1. A tent is in the shape of cylinder surmounted by a conical top. If the height and the radius of the
cylindrical part are 3.5 m and 2m respectively and the slant height of the top is 3.5 m, find the area of
the canvas used for making the tent. Also find the cost of canvas of the tent at the rate of ` 1000 per
m2.
Cylinder
Cone
Radius r = 2 m
Radius r = 2 m
Height h = 3.5 m
Slant height l = 3.5 m
Area of the canvas used for making the tent
= CSA of cylindrical part + CSA of conical part
3.5cm
= 2rh + rl
= r(h + l)
2cm
22
× 2 (2 × 3.5 + 3.5)
7
44
105
=
×
= 66 m2
7
10
=
3.5m
Cost pf 1 m2 canvas = ` 1000
 Cost pf 66 m2 canvas = ` (66 × 1000) = ` 66000
Thus, the area of the canvas used for making the tent is 66 m2 and its cost is ` 66000.
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AJAY PARMAR
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GROUP TUITION
Chapter:14 Surface area and volume
2. A metallic sphere of radius 5.6 cm is melted and recast into the shape of a cylinder of radius 6 cm.
Find the height of the cylinder.
Sphere
Cylinder
Radius r = 5.6 cm
Radius R = 6 cm
Height H = ?
Sphere is melted and recast into the cylinder.
 Volume of the cylinder = Volume of the sphere
4 3
r
3
4
 6 × 6 × H = × 5.6 × 5.6 × 5.6
3
4  5.6  5.6  5.6
H=
= 6.5 cm
3 6  6
 R2H =
 H = 6.5 cm
Thus, the height of the cylinder is 6.5 cm
3. How many spherical balls of radius 2 cm can be made out of a solid cube of lead whose side measures
44 cm?
Cube
Sphere
Length of the cube l = 44 cm
Radius r = 2 cm
Suppose, n spherical ball can be made.
 Volume of n balls = Volume of cube
4 3
r = l3
3
4
22
n× ×
× 2 × 2 × 2 = 44 × 44 × 44
3
7
44  44  44  3  7
n=
= 2541
4  22  2  2  2
n×
 n = 2541
Thus, 2541 spherical balls can be made out of the solid cube.
4. A hemispherical bowl of internal radius 18 cm contains an edible oil to be filled in cylindrical bottles
of radius 3 cm and height 9 cm. How many bottles are required to empty the bowl?
Hemisphere
Cylindrical bottles
Radius R = 18 cm
Radius r = 3 cm
Height h = 9 cm
Suppose, n bottles are required to empty the bowl
 Volume of n bottles = Volume of bowl
 n × r2h =
 n × r2h =
2
R3
3
2 3
R
3
2
× 18 × 18 × 18
3
2  18  18  18
n=
= 48
3 3 3 9
n×3×3×9=
 n = 48
Thus, 48 bottles are required to empty the bowl.
5. A hemispherical tank of radius 2.4 m is full of water. It is connected with a pipe which empties it at
the rate of 7 liters per second. How much time will it take to empty the tank completely?
Hemispherical tank
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14
GROUP TUITION
Chapter:14 Surface area and volume
Radius r = 2.4 m
2 3 2
22
24
24
24
44  8  576 3
r = ×
×
×
×
=
m
3
3
7  1000
7
10
10
10
44  8  576

= 
 1000  liters
 7  1000

Volume =
=
44  8  576
liters
7
Time required to empty 7 liters of water = 1 second
 Time required to empty
=
44  8  576
44  8  576
liters of water =
seconds
7
77
44  8  576
minutes
49  60
= 68.96 minutes
It will take 68.96 minutes to empty the tank.
6. A shuttle cock used for playing badminton has the shape of a frustum of a cone mounted on a
hemisphere. The external radii of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock
is 7 cm. Find its external surface area.
Frustum
Hemisphere
5cm
bigger radius r1 = 5 cm
Radius r = 2 cm
smaller radius r2 = 2 cm
height h = total height– radius of the hemisphere
= 7 – 2 = 5 cm
2
2
5cm
l = h + (r1 – r2)2 = 52 + (5 – 2)2 = 25 + 9 = 34
7cm
l = 34 = 5.83 cm
External surface area of the shuttle cock
2cm
= CSA of the frustum of a cone + CSA of the hemisphere
2cm
= l(r1 + r2) + r2
2
= [l(r1 + r2) + r ]
=
22
22
22
[5.83(5 + 2) + 2 × 4]=
[5.83 × 7 + 8] =
× 48.81= 153.40 cm2
7
7
7
External surface area of the shuttle cock is 153.40 cm2.
7. A fez, the headgear cap used by the Turks is shaped like the frustum of a cone. If its radius on the
open side is 12 cm, and radius at the upper base is 5 cm and its slant height is 15 cm, find the area of
material used for making it. ( = 3.14)
Frustum
bigger radius r1 = 12 cm smaller radius r2 = 5 cm
5cm
slant height l = 15 cm
Let us consider that the top of the fez is made of some material other than cloth.
 Area of the cloth used for making the fez
= CSA of the frustum of a cone
= l(r1 + r2)
= 3.14 × 15 (12 + 5)
= 3.14 × 15 × 17 = 800.7 cm2
Thus, the area of the cloth used to make the fez is 800.7 cm2.
15cm
12cm
8. A bucket is in the form of a frustum of a cone with capacity of 12308.8 cm 3 of water. The radii of the
top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of bucket and the cost
of making it at the rate of ` 10 per cm2.
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AJAY PARMAR
15
GROUP TUITION
Chapter:14 Surface area and volume
bigger radius r1 = 20 cm smaller radius r2 = 12 cm
Volume = 12308.8 cm3
Volume of the frustum of a cone =
1
h (r12 + r22 + r1r2)
3
1
h (r12 + r22 + r1r2) = 12308.8
3
1
22
 ×
× h (202 + 122 + 20 × 12) = 12308.8
3
7
22

× h × 784 = 12308.8
21
12308.8  21
h=
= 15 (approx.)
22  784

 h = 15 cm
l2 = h2 + (r1 – r2)2 = 152 + (20 – 12)2 = 225 + 64 = 289
l = 17 cm
To obtain the answer of T.B. do not consider the area of the base.
CSA of the bucket = l(r1 + r2) =
22
22
× 17 (20 + 12) =
× 17 × 32 = 1709.7 cm2
7
7
Cost of making for 1 cm2 = ` 10
 Cost of making for 1709.7 cm2 = ` (1709.7 × 10) = ` 17097
Thus, the height of the bucket is 15 cm and the cost of its making ` 17097.
Select as proper option (a), (b), (c) or (d) from given options and write in the box given on the right so
that the statement becomes correct:
1. The volume of the sphere with diameter 1 cm is … cm3.
1
1

(C)

6
24
4
4
1
1
1
1
Volume of a sphere = r3 = ×  × × × =  cm3
3
3
6
2
2
2
(A)
2

3
(B)
2. The volume of hemisphere with radius 1.2 cm is … cm3.
(A) 1.152
(B) 0.96
(C) 2.152
Volume of a hemisphere =
3. The volume of sphere is
(A) 0.5
Volume of a sphere =

4
4
 = ×  × r3
3
3
(D)
4

3
(D) 3.456
2 3
2
r = ×  × 1.2 × 1.2 × 1.2 = 1.152 cm3
3
3
4
 cm3. Then its diameter is … cm.
3
(B) 1
(C) 2
(D) 2.5
4 3
r
3
 Diameter = 2 × r = 2 × 1 = 2 cm
 r3 = 1
 r = 1 cm
4. The volume of cone with radius 2 cm and height 6 cm is … cm3.
(A) 8
(B) 12
(C) 14
(D) 16
1
1
Volume of cone = r2h = ×  × 2 × 2 × 6 = 8 cm3
3
3
5. The diameter of the base of the cone is 10 cm and its slant height is 17 cm. Then the curved surface
area of the cone is … cm2.
(A) 85
(B) 170
(C) 95
(D) 88
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GROUP TUITION
Chapter:14 Surface area and volume
Radius of the cone =
10
diameter
=
= 5 cm
2
2
CSA of cone = rl =  × 5 × 17 = 85 cm2
6. The diameter and the height of the cylinder are 14 cm and 10 cm respectively. Then the total surface
is … cm2.
(A) 44
(B) 308
(C) 748
(D) 1010
14
diameter
=
= 7 cm
2
2
22
TSA of cone = 2r(h + r) = 2 ×
× 7 (10 + 7) = 44 × 17 = 748 cm2
7
Radius of the cone =
7. The ratio of the radii of two cones having equal height is 2 : 3. Then, the ratio of their volumes is … .
(A) 4 : 6
(B) 8 : 27
(C) 3 : 2
(D) 4 : 9
Height of both the cones = h and let radii of two cones respectively be r1 = 2x and r2 = 3x
1 2
r1 h
2
4
4x2
Volume of the first cone
Ratio of volume of the cones =
= 3
= (2 x )2 = 2 = = 4 : 9
1 2
Volume of the second cone
9
(3 x)
9x
r2 h
3
8. If the radii of a frustum of a cone are 7 cm and 3 cm and the height is 3 cm, then the curved surface
area is … cm2.
(A) 50
(B) 25
(C) 35
(D) 63
For the frustum of a cone, r1 = 7 cm and r2 = 3 cm and h = 3 cm
l = h 2  (r1  r2 ) 2 = 32  (7  3) 2 = 9  16 = 25 = 5 cm
 l = 5 cm
CSA od the frustum of a cone = l(r1 + r2) =  × 5(7 + 3) = 50 cm2
9. The radii of a frustum of a cone are 5 cm and 9 cm and height is 6 cm, then the volume is … cm 3.
(A) 320
(B) 151
(C) 302
(D) 98
Here, r1 = 9 cm and r2 = 5 cm and h = 6 cm
Volume of the frustum of a cone =
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1
h (r12 + r22 + r1r2) =  × 6 (81 + 25 + 45) = 2(151) = 302 cm3
3
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