Question

The radius of a sphere is increased by $p\%$. By what percent its surface area increases?

Answer 1

The surface area of a sphere is $4\pi r^2$

If radius increases by $p\%$

The new radius will be $r(1+\frac{p}{100})$

So, the new surface area will be $4\pi r^2{(1+\frac{p}{100})}^2$

$\therefore \%$ change in surface area

$=\frac{4\pi r^2{(1+\frac{p}{100})}^2-4\pi r^2}{4\pi r^2}\times 100$

$=\left[\right(1+\frac{p}{100}{)}^2-1]\times 100$

$=(1+\frac{p^2}{{100}^2}+\frac{2p}{100}-1)\times 100$

$=(\frac{p^2}{{100}^2}+\frac{2p}{100})\times 100$

$=\frac{p^2}{100}+2p$