Surface Areas and Volumes – Exercise 13.4 – (MATHEMATICS) – 9th Class

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Exercise 31.4

Assume \pi = {{22} \over 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.

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Sol.
(i) Radius of a sphere (r) = 10.5 cm
∴ Surface area of the sphere = 4πr²
 = 4 \times {{22} \over 7} \times 10.5 \times 10.5
= 1386 cm²

(ii) Radius of a sphere (r) = 5.6 cm
∴ Surface area of the sphere = 4πr2
 = 4 \times {{22} \over 7} \times 5.6 \times 5.6
= 394.24 cm²

(iii) Radius of a sphere (r) = 14 cm

Surface area of the sphere = 4πr2

 = 4 \times {{22} \over 7} \times 14 \times 14 = 2464c{m^2}


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

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

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Sol. (i) Diameter (d) = 14 cm
Radius (r)   = {{14} \over 2} = 7cm

Now, surface area of the sphere = 4πr2

 = 4 \times {{22} \over 7} \times 7 \times 7 = 4 \times 22 \times 7 = 616c{m^2}

(ii) Diameter (d) = 21 cm

∴ Radius (r) =  = {{21} \over 2}cm

Now, surface area of the sphere = 4nr2

 = 4 \times {{22} \over 7} \times {{21} \over 2} \times {{21} \over 2} = 1386c{m^2}

(iii) Diameter (d) = 3.5 m

∴ Radius (r) =   = {{3.5} \over 2}m = {7 \over 4}m
Now, surface area of the sphere = 4nr2

 = 4 \times {{22} \over 7} \times {7 \over 4} \times {7 \over 4} = {{22 \times 7} \over 4} = 38.5{m^2}


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

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Sol. Total surface area of hemisphere = 3πr²
= 3 × 3.14 × 10 × 10 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.

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Sol. S1 = Surface area with radius 7 cm = 4π(7)² cm².
S2 = Surface area with radius 14 cm = 4π(14)2 cm2.

∴  {{{S_1}} \over {{S_2}}} = {{4\pi {{(7)}^2}} \over {4\pi {{(14)}^2}}} \times {1 \over 4} \Rightarrow {S_1}:{S_2} = 1:4


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.

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Sol. Inner diameter = 10.5 cm   ⇒  Inner radius = 5.25 cm.

Inner surface area =  = 2 \times {{22} \over 7} \times {(5.25)^2}c{m^2} = 173.25c{m^2}
Cost of tin-plating = Rs.\left( {16 \times {{173.25} \over {100}}} \right) = Rs.27.72


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

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Sol. Let r be the radius of the sphere.

Surface area of the sphere = 154 cm²

4 \times {{22} \over 7} \times {r^2} = 154 \Rightarrow {r^2} = {{154 \times 7} \over {88}} = {{49} \over 4}

⇒                             r = {7 \over 2}cm \Rightarrow r = 3.5cm


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

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Sol. Let diameter of the earth = d units

Radius of the earth = {d \over 2} units

∴ Diameter of the moon = {d \over 4} units

⇒  Radius of the moon = {d \over 8} units.

∴   {{SurfaceAreaOfTheMoon} \over {SurfaceAreaOfTheEarth}} = {{4\pi {{\left( {{d \over 8}} \right)}^2}} \over {4\pi {{\left( {{d \over 2}} \right)}^2}}} = {4 \over {64}} = {1 \over {16}}

∴ Surface area of the moon : surface area of the earth

= 1 : 16.


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.

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Sol. Inner radius = 5 cm.
∴  Outer radius = (5 + 0.25) cm = 5.25 cm.

∴  Outer curved surface area = 2 \times {{22} \over 7} \times {(5.25)^2}c{m^2}

= 173.25 cm2


9. A right circular cylinder just encloses a sphere of radius r (see figure). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder;
(iii) ratio of the areas obtained in (i) and (ii)

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Sol. (i) Surface area of the sphere = 4πr²

(ii) For cylinder: radius of base = r, height = 2r.

∴   Curved surface area of the cylinder = 2π(r)(2r)
= 4πr2

(iii) Ratio of the required areas is 1 : 1.


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