Question

$ABCDEF$ is a regular hexagon. With vertices $A,B,C,D,E$ and $F$ as the centre , circle of same radius $r$ are drawn. Find the area of the shaded portion shown in the figure

Answer 1

The intercept will be of ${120}^o$ every circle.

Hence,

${360}^o\to \pi r^2$

$1^o\to \frac{\pi r^2}{{360}^o}$

${120}^o\to 120\times \frac{\pi r^2}{{360}^o}=\frac{\pi r^2}{3}$

Total, $6$ circles present.

$\therefore$ Total shaded area $=6\times \frac{\pi r^2}{3}$

$\therefore$ Area $=2\pi r^2$.