# It is about finding area want help in part b?

May 4, 2018

Given triangle ABC is a equilateral triangle, one interior angle is $\frac{\pi}{3}$.

Since the 3 arcs are equal,
$\angle C A B = \frac{\pi}{3}$
$2 \alpha = \frac{\pi}{3}$

Since $\angle A O B = 2 \frac{\pi}{3}$ and $A O = O B = r$,
Triangle AOB is a equilateral triangle.
$\therefore r = 4$

Substitute $2 \alpha = \frac{\pi}{3}$ and $r = 4$ into the answer for (a)(ii),
Area of segment AB
=$\frac{1}{2} {r}^{2} \left(\frac{\pi}{3} - \sin \left(\frac{\pi}{3}\right)\right)$
= $\frac{1}{2} {\left(4\right)}^{2} \left(\frac{\pi}{3} - \frac{\sqrt{3}}{2}\right)$
= $\frac{8 \pi}{3} - 4 \sqrt{3}$

Area of triangle ABC
=$\frac{1}{2} a b \sin C$
=$\frac{1}{2} \left(4\right) \left(4\right) \left(\sin \left(\frac{\pi}{3}\right)\right)$
=$4 \sqrt{3}$

=$4 \sqrt{3} - 3 \left(\frac{8 \pi}{3} - 4 \sqrt{3}\right)$
=$4 \sqrt{3} - 8 \pi + 12 \sqrt{3}$
=$16 \sqrt{3} - 8 \pi$