On increasing the diameter of a circle by 40%, its area will be increased by

Solution:

(c) 96%
Let d be the original diameter.

Radius $=\frac{d}{2}$

Thus, we have:

Original area $=\pi \times\left(\frac{d}{2}\right)^{2}$

$=\frac{\pi d^{2}}{4}$

New diameter $=140 \%$ of $d$

$=\left(\frac{140}{100} \times d\right)$

$=\frac{7 d}{5}$

Now,

New radius $=\frac{7 d}{5 \times 2}$

$=\frac{7 d}{10}$

New area $=\pi \times\left(\frac{7 d}{10}\right)^{2}$

$=\frac{49 \pi d^{2}}{10}$

Increase in the area $=\left(\frac{49 \pi d^{2}}{10}-\frac{\pi d^{2}}{4}\right)$

$=\frac{24 \pi d^{2}}{100}$

$=\frac{6 \pi d^{2}}{25}$

We have:

Increase in the area $=\left(\frac{6 \pi d^{2}}{25} \times \frac{4}{\pi d^{2}} \times 100\right) \%$

$=96 \%$