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Find the surface area of a sphere of diameter
(i) \( 14 \mathrm{~cm} \)
(ii) \( 21 \mathrm{~cm} \)
(iii) \( 3.5 \mathrm{~m} \).
To do:
We have to find the surface area of the given spheres.
Solution:
(i) Diameter of the sphere $= 14\ cm$
This implies,
Radius of the sphere $(r) = \frac{14}{2}$
$= 7\ cm$
Therefore,
Surface area of the sphere $=4 \pi r^{2}$
$=4 \times \frac{22}{7} \times 7 \times 7$
$=616 \mathrm{~cm}^{2}$
(ii) Diameter of the sphere $=21 \mathrm{~cm}$
This implies,
Radius of the sphere $(r)=\frac{21}{2} \mathrm{~cm}$
Therefore,
Surface area of the sphere $=4 \pi r^{2}$
$=4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}$
$=1386 \mathrm{~cm}^{2}$
(iii) Diameter of the sphere $=3.5 \mathrm{~cm}$
This implies,
Radius of the sphere $(r)=\frac{3.5}{2}$
$=\frac{7}{2 \times 2}$
$=\frac{7}{4} \mathrm{~cm}$
Therefore,
Surface area of the sphere $=4 \pi r^{2}$
$=4 \times \frac{22}{7} \times \frac{7}{4} \times \frac{7}{4}$
$=\frac{77}{2}$
$=38.5 \mathrm{~cm}^{2}$