Question

If the radius of a sphere is decreased by $25\%$ by what per cent its curved surface area decreased?

Answer 1

Let initial radius be r.
so, surface area=$4\pi r^2$
Now, radius decreased by $25\%$
$=r-r\times \frac{25}{100}={\left(\frac{3r}{4}\right)}^2$
$\%change=\frac{differenceinsurfacearea}{initialsurfacearea}\times 100\%$
so, difference $=4\pi r^2-4\pi \frac{9r^2}{16}$
$=4\pi r^2(1-\frac{9}{16})=4\pi r^2\left(\frac{7}{16}\right)$
Hence $\%change=\frac{4\text{/}\pi r^2\left(\frac{7}{16}\right)}{4\pi \text{/}{r}^2}\times 100\%$
$=\frac{7}{\frac{\text{/}16}{4}}\times 1\text{/}{00}^2$
$=\frac{175}{4}\%$