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## Chapter 13: Surface Areas and Volumes

TEXTBOOK’S EXERCISE – 13.4

1. Find the surface area of a sphere of radius:

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Sol. (i) r = 10.5 cm

Surface area of the sphere

= 4πr2 = 4 × 22/7 × (10.5)2 cm2

= 4 × 22/7 × 10.5 × 10.5 cm2 = 1386 cm2

(ii) r = 5.6 cm

Surface area of the sphere

= 4πr2 = 4 × 22/7 × (5.6)2 cm2

= 4 × 22 7 × 5.6 × 5.6 cm2

= 394.24 cm2

(iii) r = 14 cm

Surface area of the sphere = 4πr2

= 4 × 22/7 × 14 × 14 cm2 = 88 × 28 cm2

= 2464 cm2

2. Find the surface area of a sphere of diameter :

(i) 14 cm (ii) 21 cm (iii) 3.5 m

Sol. (i) r = 14/2 cm = 7 cm

Surface area of the sphere

= 4πr2 = 4 × 22/7 × 72 cm2

= 4 × 22/7 × 7 × 7 cm2

= 88 × 7 cm2 = 616 cm2

(ii) r = 21/2 cm = 10.5 cm

Surface area of the sphere

= 4πr2 = 4 × 22/7 × (10.5)2 cm2

= 4 × 22 7 × 10.5 × 10.5 cm2 = 1386 cm2

(iii) r = 3.5/2 m = 1.75 m

Surface area of the sphere

= 4πr2 = 4 × 22/7 × (1.75)2 m2

= 4 × 22/7 × 1.75 × 1.75 m2 = 38.5 m2

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Sol. r = 10 cm

Total surface area of the hemisphere = 3πr2

= 3 × 3.14 × (10)2 cm2

= 3 × 3.14 × 100 cm2 = 942 cm2

4. The radius of a spherical balloon 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.

Sol. When r = 7 cm

The surface area of the balloon

= 4πr2 = 4 × π × 7 × 7 cm2

When r = 14 cm

The surface area of the balloon

= 4πr2 = 4 × π × 14 × 14 cm2

The required ratio of the surface areas of the balloon

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