If the diameter of a sphere is decreased by 25%, then by what percent its curved area would decrease?

Share with your friends

Share 34

Question

If the diameter of a sphere is decreased by 25%,
then by what percent its curved area would decrease?

Lalit Mehra · Accepted Answer

Let d be the diameter of the original sphere and d' be the diameter of the new sphere.

$Given , d' = d - 25 % of d = d - \frac{25}{100} d = d - \frac{d}{4} = \frac{3 d}{4}$

Surface area of original sphere, $S_{1} = 4 π r^{2} = 4 π (\frac{d}{2})^{2}$

Surface area of new sphere, $S_{2} = 4 π (\frac{d'}{2})^{2} = 4 π (\frac{\frac{3 d}{4}}{2})^{2} = 4 π (\frac{3}{4} \times \frac{d}{2})^{2} = \frac{9}{16} \times 4 π (\frac{d}{2})^{2}$

Decrease in surface area of the sphere

$S_{1} - S_{2} = 4 π (\frac{d}{2})^{2} - \frac{9}{16} \times 4 π (\frac{d}{2})^{2} = (\frac{16 - 9}{19}) \times 4 π (\frac{d}{2})^{2} = \frac{7}{16} \times 4 π (\frac{d}{2})^{2}$

% decrease in surface area

$= \frac{Decrease in surface area of he sphere}{Surface area of original sphere} \times 100 % = \frac{\frac{7}{16} \times 4 π (\frac{d}{2})^{2}}{4 π (\frac{d}{2})^{2}} \times 100 % = \frac{700}{16} % = 43.75 %$

Thus, the percent decrease in surface area of the sphere is 43.75.