Surface Area and Volumes – Exercise 13.4 – Class IX

For previous exercise – Surface Area and Volumes – Exercise 13.3 – 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


  1. 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/23.5/2 = 1.75 m

Surface area of the sphere  = 4πr2

= 4 x 22/7 x (1.75)2

=  38.5 m2


  1. 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 


  1. 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


  1. 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


  1. 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


  1. 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


  1. 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


  1. Surface Area and Volumes – Exercise 13.4 – Class IXA 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


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


 

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