**For previous exercise – Surface Area and Volumes – Exercise 13.3 – Class IX**

Assume π = ^{22}/_{7} , unless stated otherwise.

**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 = 4πr^{2}

= 4 x ^{22}/_{7} x (10.5)^{2}

= 1386 cm^{2}

(ii) r = 5.6 cm

Surface area of the sphere = 4πr^{2}

= 4 x ^{22}/_{7} x (5.6)^{2}

= 394. 24 cm^{2}

(iii) 14 cm

Surface area of the sphere = 4πr^{2}

= 4 x ^{22}/_{7} x (14)^{2}

= 2464 cm^{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 = 4πr^{2}

= 4 x ^{22}/_{7} x 7^{2}

= 616 cm^{2}

(ii) d = 21 cm

r = ^{d}/_{2} = ^{21}/_{2} = 10.5 cm

Surface area of the sphere = 4πr^{2}

= 4 x ^{22}/_{7} x (10.5)^{2}

= 1386 cm^{2}

(iii) d =3.5 m

r = ^{d}/_{2} = ^{3.5}/_{2} = 1.75 m

Surface area of the sphere = 4πr^{2}

= 4 x ^{22}/_{7} x (1.75)^{2}

= 38.5 m^{2}

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

Solution:

Total surface area of a hemisphere = 3πr^{2}

= 3 x ^{22}/_{7} x (10)^{2}

= 942.857 cm^{2}^{ }

**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 = 4πr^{2} = 4 x ^{22}/_{7} x 7^{2 } = 616 cm^{2}

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

Surface area = 4πr^{2} = 4 x ^{22}/_{7} x 14^{2} = 2464 cm^{2}

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

**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 cm**^{2}.

Solution:

d = 10.5 cm ; r = ^{d}/_{2} = ^{10.5}/_{2} = 5.25 cm

Surface area = 4πr^{2} = 4 x ^{22}/_{7} x 5.25^{2} = 346.5 cm^{2}

The cost of tin-plating it on the inside for 100 cm^{2} = Rs 16

Then, the cost of tin plating it on inside for 346.5cm^{2} = ^{346.5×16}/_{100} = Rs. 55.44

**Find the radius of a sphere whose surface area is 154 cm**^{2}

Solution:

Total surface area = 154 cm^{2}

We know, surface area of the sphere = 4πr^{2}

4πr^{2} = 154

r^{2} = ^{154×7}/_{4×22} = ^{7×7}/_{4}

r = ^{7}/_{2} = 3.5 cm

**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 = 4πr^{2} = 4 x ^{22}/_{7} x (^{d}/_{8})^{2} = ^{11}/_{56} d

Surface area of the earth = 4πr^{2} = 4 x ^{22}/_{7} x (^{d}/_{2})^{2} = ^{22}/_{7} d

Required ratio = (^{11}/_{56} d)/(^{ 22}/_{7} d) = ^{1}/_{16} = 1:16

**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 = 2πR^{2} = 2 x ^{22}/_{7} x (5.25)^{2} = 173.25 cm^{2}

**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 = 4πr^{2}

(ii) curved surface area of the cylinder = 2πrh = 2π r x 2r = 4πr^{2}

(iii) required ratio = 4πr^{2}/4πr^{2} = 1

**For next exercise – Surface Area and Volumes – Exercise 13.5 – Class IX**

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