Surface Area and Volumes – Exercise 13.4 – Class IX

Assume π = ²²/₇ , unless stated otherwise.

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

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Solution:

(i) r = 10.5 cm

Surface area of the sphere  = 4πr²

= 4 x ²²/₇ x (10.5)²

= 1386 cm²

(ii) r = 5.6 cm

Surface area of the sphere  = 4πr²

= 4 x ²²/₇ x (5.6)²

= 394. 24 cm²

(iii) 14 cm

Surface area of the sphere  = 4πr²

= 4 x ²²/₇ x (14)²

= 2464 cm²

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

(i) 14 cm

(ii) 21 cm

(iii) 3.5 m

Solution:

(i) d = 14 cm

r = ^d/₂ = ¹⁴/₂ = 7  cm

Surface area of the sphere  = 4πr²

= 4 x ²²/₇  x  7²

= 616 cm²

(ii) d = 21 cm

r = ^d/₂ = ²¹/₂ = 10.5 cm

Surface area of the sphere  = 4πr²

= 4 x ²²/₇ x (10.5)²

= 1386 cm²

(iii) d =3.5 m

r = ^d/₂ =  ^3.5/₂ = 1.75 m

Surface area of the sphere  = 4πr²

= 4 x ²²/₇ x (1.75)²

=  38.5 m²

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

Solution:

Total surface area of a hemisphere = 3πr²

=  3 x ²²/₇ x (10)²

= 942.857 cm² 

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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.

Solution:

Surface area of the spherical balloon when r  = 7 cm  :

Surface area = 4πr²  = 4 x ²²/₇ x 7²  = 616 cm²

Surface area of the spherical balloon when r  = 14 cm  :

Surface area = 4πr²  = 4 x ²²/₇ x 14² = 2464 cm²

The required ratio of the surface area of the spherical balloon = 616 : 2464 = 1 : 4

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5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm².

Solution:

d =  10.5 cm ;  r = ^d/₂ = ^10.5/₂ = 5.25  cm

Surface area  = 4πr² = 4 x ²²/₇ x 5.25² = 346.5 cm²

The cost of tin-plating it on the inside for 100 cm² =  Rs 16

Then, the cost of tin plating it on inside for 346.5cm² = ^346.5×16/₁₀₀ = Rs. 55.44

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6. Find the radius of a sphere whose surface area is 154 cm²

Solution:

Total surface area = 154 cm²

We know, surface area of the sphere = 4πr²

4πr² = 154

r² = ^154×7/_4×22 = ^7×7/₄

r = ⁷/₂ = 3.5 cm

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7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution:

Let diameter of the earth = d

We know, radius of the earth, r = ^d/₂

Given, diameter of the moon = ¹/₄  diameter of the earth

Then, diameter of the moon = ¹/₄ x d

We have,  radius of the moon, r = ¹/₄ x ^d/₂ = ^d/₈

Surface area of the moon = 4πr² = 4 x ²²/₇ x (^d/₈)² = ¹¹/₅₆ d

Surface area of the earth = 4πr² = 4 x ²²/₇ x (^d/₂)² = ²²/₇ d

Required ratio = (¹¹/₅₆ d)/( ²²/₇ d) = ¹/₁₆ = 1:16

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8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Solution:

Inner radius of the bowl, r = 5 cm

Thickness of  the steel  = 0.25 cm

Outer radius of  the bowl = 5 + 0.25  = 5.25 cm

Outer curved  surface area of the bowl = 2πR² = 2 x ²²/₇ x (5.25)² = 173.25 cm²

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9. A right circular cylinder just encloses a sphere of radius r (see Fig. 13.22). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii)

Solution:

(i) Surface area of the surface = 4πr²

(ii) curved surface area of the cylinder = 2πrh = 2π r x 2r = 4πr²

(iii) required ratio = 4πr²/4πr² = 1

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