Question

The diameter of a sphere is decreased by 25 %. By what percent its curved surface area decrease ?

Answer 1

Let r be the radius and S be the curved surface area. Then,

$S=4\pi r^2$

Decrease in diameter = 25 % of 2r =$(\frac{25}{100}\times 2r)=\frac{r}{2}$

Decreased diameter = $2r-\frac{r}{2}=\frac{3r}{2}$

Decreased radius = $\frac{3r}{4}$

Let $S_1$ be the new curved surface area . Then,

$S_1=4\pi \times (\frac{3r}{4}{)}^2=\frac{9\pi r^2}{4}$

Decrease in curved surface area = $S-S_1=\frac{7\pi r^2}{4}$

So, percentage decrease in curved surface area is

$\frac{S-S_1}{S}\times 100$ % = $\frac{700}{16}=43.75$ %