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- CBSE Class 9 Textbook Solutions
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- Maths
- Chapter 11 Surface Areas and Volumes
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Ex. 11.1
Ex. 11.2
Ex. 11.3
Ex. 11.4
Surface Areas and Volumes Exercise Ex. 11.1
Solution 1
Radius of base of cone = 
cm = 5.25 cm
Slant height of cone = 10 cm
CSA of cone =
= 
Thus, the curved surface area of cone is 165 
.
Solution 2
Radius of base of cone = 
m = 12 cm
Slant height of cone = 21 m
Total surface area of cone = 
(r + l)
Solution 3
(i) Slant height of cone = 14 cm
Let radius of circular end of cone be r.
CSA of cone = 
Thus, the radius of circular end of the cone is 7 cm.
(ii) Total surface area of cone = CSA of cone + Area of base
=
Thus, the total surface area of the cone is 462 
.
Solution 4
(i) Height (h) of conical tent = 10 m
Radius (r) of conical tent = 24 m
Let slant height of conical tent be l.
l = 26 m
.
Thus, the slant height of the conical tent is 26 m.
(ii) CSA of tent =
= 
Cost of 1
canvas = Rs 70
Cost of
canvas = 
= Rs 137280
Thus, the cost of canvas required to make the tent is Rs 137280.
Solution 5
Height (h) of conical tent = 8 m
Radius (r) of base of tent = 6 m
Slant height (l) of tent =
CSA of conical tent =
= (3.14
6 10)
= 188.4
Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L - 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
[(L - 0.2 m) 3] m = 188.4
L - 0.2 m = 62.8 m
L = 63 m
Thus, the length of the tarpaulin sheet will be 63 m.
Solution 6
Slant height (l) of conical tomb = 25 m
Base radius (r) of tomb =
= 7 m
CSA of conical tomb =
=
Cost of white-washing 100
area = Rs 210
Cost of white-washing 550 area =Rs
= Rs 1155
Thus, the cost of white washing the conical tomb is Rs 1155.
Solution 7
Radius (r) of conical cap = 7 cm
Height (h) of conical cap = 24 cm
Slant height (l) of conical cap =
CSA of 1 conical cap =
= 
CSA of 10 such conical caps = (10
550)
= 5500
Thus, 5500 sheet will be required to make the 10 caps.
Solution 8
Radius (r) of cone =
= 0.2 m
Height (h) of cone = 1 m
Slant height (l) of cone =
CSA of each cone =
= (3.14
0.2 1.02)
= 0.64056
CSA of 50 such cones = (50 0.64056) = 32.028
Cost of painting 1 area = Rs 12
Cost of painting 32.028 area = Rs (32.028 12) = Rs 384.336
Thus, it will cost Rs 384.34 (approximately) in painting the 50 hollow cones.
Surface Areas and Volumes Exercise Ex. 11.2
Solution 1
(i) Radius of sphere = 10.5 cm
Surface area of sphere = 
(ii) Radius of sphere = 5.6 cm
Surface area of sphere = = 
(iii) Radius of sphere = 14 cm
Surface area of sphere = = 
Solution 2
(i) Radius of sphere 
Surface area of sphere
(ii) Radius of sphere 
Surface area of sphere
(iii) Radius of sphere 
Surface area of sphere = 
Solution 3
Radius of hemisphere = 10 cm
Total surface area of hemisphere 
Solution 4
Radius
of spherical balloon = 7 cm
Radius 
of spherical balloon, when air is pumped into it = 14 cm
Solution 5
Inner radius (r) of hemispherical bowl = 
Surface area of hemispherical bowl = 
Cost of tin-plating 100
area = Rs 16
Cost of tin-plating 173.25 area 
= Rs 27.72
Thus, the cost of tin-plating the inner side of hemispherical bowl is Rs 27.72
Solution 6
Let radius of the sphere be r.
Surface area of the sphere = 154 
= 154 cm2
Thus, the radius of the sphere is 3.5 cm.
Solution 7
Let diameter of earth be d. Then, diameter of moon will be 
.
Radius of earth = 
Radius of moon = 
Surface area of moon = 
Surface area of earth = 
Required ratio =
Thus, the required ratio of the surface areas is 1:16.
Solution 8
Inner radius of hemispherical bowl = 5 cm
Thickness of the bowl = 0.25 cm
Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm
Outer CSA of hemispherical bowl = 
Thus, the outer curved surface area of the bowl is 173.25 
.
Solution 9
(i) Surface area of sphere = 
(ii) Height of cylinder = r + r = 2r
Radius of cylinder = r
CSA of cylinder =
(iii) Required ratio =
Surface Areas and Volumes Exercise Ex. 11.3
Solution 1
(i) Radius (r) of cone = 6 cm
Height (h) of cone = 7 cm
Volume of cone
(ii) Radius (r) of cone = 3.5 cm
Height (h) of cone = 12 cm
Volume of cone 
Solution 2
(i) Radius (r) of cone = 7 cm
Slant height (l) of cone = 25 cm
Height (h) of cone 
Volume of cone 
Capacity of the conical vessel =
litres= 1.232 litres
(ii) Height (h) of cone = 12 cm
Slant height (l) of cone = 13 cm
Radius (r) of cone 
Volume of cone 
Capacity of the conical vessel =
litres =
litres.
Solution 3
Height (h) of cone = 15 cm
Let radius of cone be r.
Volume of cone = 1570 cm3
r = 10 cm
Thus, the radius of the base of the cone is 10 cm.
Solution 4
Height (h) of cone = 9 cm
Let radius of cone be r.
Volume of cone = 48
cm3
Thus, the diameter of the base of the cone is 2r = 8 cm.
Solution 5
Radius (r) of pit =
Depth (h) of pit = 12 m
Volume of pit =
= 38.5 m3
Capacity of the pit = (38.5
1) kilolitres = 38.5 kilolitres
Solution 6
(i) Radius of cone = 
=14 cm
Let height of cone be h.
Volume of cone = 9856 cm3
h = 48 cm
Thus, the height of the cone is 48 cm.
(ii) Slant height (l) of cone 
Thus, the slant height of the cone is 50 cm.
(iii) CSA of cone = 
rl= 
= 2200 cm2
Solution 7
When the right angled 
ABC is revolved about its side 12 cm, a cone of height (h) 12 cm, radius (r) 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone
= 100
cm3
Thus, the volume of cone so formed by the triangle is 100 cm3.
Solution 8
When the right angled 
ABC is revolved about its side 5 cm, a cone of radius (r) 12 cm, height (h) 5 cm, and slant height (l) 13 cm will be formed.
Volume of cone =
Required ratio 
Solution 9
Radius (r) of heap
Height (h) of heap = 3 m
Volume of heap= 
Slant height (l) =
Area of canvas required = CSA of cone
Surface Areas and Volumes Exercise Ex. 11.4
Solution 1
(i) Radius of sphere = 7 cm
Volume of sphere = 
(ii) Radius of sphere = 0.63 m
Volume of sphere =
m3(approximately)
Solution 2
(i) Radius (r) of ball = 
Volume of ball = 
Thus, the amount of water displaced is 
.
(ii) Radius (r) of ball =
= 0.105 m
Volume of ball = 
Thus, the amount of water displaced is 0.004851 m3.
Solution 3
Radius (r) of metallic ball = 
Volume of metallic ball = 
Mass = Density
Volume = (8.9 * 38.808) g = 345.3912 g
Thus, the mass of the ball is approximately 345.39 g.
Solution 4
Let diameter of earth be d. So, radius earth will be
.
Then, diameter of moon will be
. So, radius of moon will be
.
Volume of moon = 
Volume of earth = 
Thus, the volume of moon is
of volume of earth.
Solution 5
Radius (r) of hemispherical bowl =
= 5.25 cm
Volume of hemispherical bowl
= 303.1875 cm3
Capacity of the bowl
= 0.3031875 litre = 0.303 litre (approximately)
Thus, the hemispherical bowl can hold 0.303 litre of milk.
Solution 6
Inner radius (r1) of hemispherical tank = 1 m
Thickness of hemispherical tank = 1 cm = 0.01 m
Outer radius (r2) of hemispherical tank = (1 + 0.01) m = 1.01 m
Volume of iron used to make the tank = 
Solution 7
Let the radius of the sphere be r.
Surface area of sphere = 154 cm2
r2 = 154 cm2
Volume of sphere
Solution 8
(i) Cost of white washing the dome from inside = Rs 498.96
Cost of white washing 1 m2 area = Rs 2
CSA of inner side of dome =
= 249.48 m2
(ii) Let inner radius of hemispherical dome be r.
CSA of inner side of dome = 249.48 m2
2
r2 = 249.48 m2
Volume of air inside the dome = Volume of the hemispherical dome
= 523.908 m3
Thus, the volume of air inside the dome is approximately 523.9 m3.
Solution 9
(i) Radius of 1 solid iron sphere = r
Volume of 1 solid iron sphere
Volume of 27 solid iron spheres 
It is given that 27 sold iron spheres are melted to from 1 iron sphere. So, volume of
this iron sphere will be equal to volume of 27 solid iron spheres.
Radius of the new sphere = r'.
Volume of new sphere 
(ii) Surface area of 1 solid iron sphere of radius r = 4
r2
Surface area of iron sphere of radius r' = 4 (r')2 = 4 (3r)2 = 36 r2
Solution 10
Radius (r) of capsule
Volume of spherical capsule
Thus, approximately 22.46 mm3 of medicine is required to fill the capsule.
