- CBSE
- Class 9
- 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 cm^{3}

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 m^{3}

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 cm^{3}

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 cm^{2}

### 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 cm^{3}

Thus, the volume of cone so formed by the triangle is 100 cm^{3}.

### 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 =

m^{3}(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 m^{3}.

### 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 cm^{3}

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 (r_{1}) of hemispherical tank = 1 m

Thickness of hemispherical tank = 1 cm = 0.01 m

Outer radius (r_{2}) 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 cm^{2}

r^{2} = 154 cm^{2}

Volume of sphere

### Solution 8

(i) Cost of white washing the dome from inside = Rs 498.96

Cost of white washing 1 m^{2} area = Rs 2

CSA of inner side of dome = = 249.48 m^{2}

(ii) Let inner radius of hemispherical dome be r.

CSA of inner side of dome = 249.48 m^{2}

2r^{2} = 249.48 m^{2}

Volume of air inside the dome = Volume of the hemispherical dome

= 523.908 m^{3}

Thus, the volume of air inside the dome is approximately 523.9 m^{3}.

### 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 = 4r^{2}

Surface area of iron sphere of radius r' = 4 (r')^{2} = 4 (3r)^{2} = 36 r^{2}

### Solution 10

Radius (r) of capsule

Volume of spherical capsule

Thus, approximately 22.46 mm^{3} of medicine is required to fill the capsule.