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 = 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
- 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
- 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
- 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
- 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
- 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
- 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
- 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
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
1 thought on “Surface Area and Volumes – Exercise 13.4 – Class IX”
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