Question

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

Answer 1

Let, Diameter = r

As we know that : Surface Area of sphere

$4\pi {r}^{2}$

Diameter is decreased by 25%

$r \times \frac{75}{100} \\ \\ = > \frac{3r}{4}$

Radius :

$\frac{3r}{4 \times 2} = \frac{3r}{8}$

Now, new surface area :

$4\pi { (\frac{3r}{8} )}^{2} \\ \\ = > \frac{9}{16} \pi {r}^{2}$

Then, percentage increased :

$\frac{4\pi {r}^{2} - \frac{9}{16} \pi {r}^{2}}{4\pi {r}^{2} } \times 100 \\ \\ = > \frac{ \frac{55}{16} \pi {r}^{2} }{4\pi {r}^{2} } \times 100 \\ \\ = > \frac{55}{16} \times 25 \\ \\ = > 85.93$

So, the answer will be 85.93%