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# Class 9 NCERT Solutions Maths Chapter 11 - Surface Areas and Volumes

## 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  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

Surface area of sphere

Surface area 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 .
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
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
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

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

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
2r2 = 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 = 4r2
Surface area of iron sphere of radius r' = 4 (r')2    = 4 (3r)2 = 36 r2