Question

A circle of 'a' radius is divided into 6 equal sectors An equilateral triangle is drawn on the chord of each sectors to lie outside the circle. The area of the resulting figure is

Answer 1

In circle $\angle GOI=\angle GOH=\angle HOJ=\angle JOL=\angle LOK=\angle KOI=\frac{{360}^o}{{60}^o}={60}^o$

Nw consider triangle GOI4

In $\Delta GOI$,

GO=OI (Radius)

$\angle GOI={60}^o$

Therefore triangle GOI ids equilateral .

Hence, GI=GO=a

Hence, GI=GH=HJ=JL=LK=KI=a

Now area of the resulting figure is -

area of circle with radius $a+$( area of the equilateral triangle with side $a)\times 6$

$=\pi a^2+\frac{\sqrt{3}}{4}{a}^2\times 6$

$=\pi a^2+\frac{3\sqrt{3}{a}^2}{2}$

$=\frac{a^2(2\pi +3\sqrt{3})}{2}$