Question

Adjoining figure two arcs $PAQ$ and $PBQ$ is a part of a circle with center $O$ and radius $OP$ while arcs $PBQ$ is a semicircle drawn on $PQ$ as diameter, If $OP=PQ=10cm$. Show that the area of the shaded region is $25(\sqrt{3}-\frac{\pi }{6})c{m}^2$

Answer 1

Since $OP=PQ=QO$

$\Rightarrow △POQ$ is an equilateral triangle.

$\therefore \angle POQ={60}^{\circ }$

Area of segment $PAQM$

$=\frac{\theta }{360}\pi r^2-\frac{\sqrt{3}}{4}{a}^2$

$=\frac{60}{360}\pi \times {10}^2-\frac{\sqrt{3}}{4}{10}^2$

$=(\frac{100\pi }{6}-\frac{100\sqrt{3}}{4}){cm}^2$

Area of semicircle with $M$ as centre is $\frac{\pi }{2}\times 5^2=\frac{25\pi }{2}{cm}^2$

Area of shaded region$=\frac{25\pi }{2}-(\frac{50\pi }{3}-25\sqrt{3})$

$=\frac{25\pi }{2}-\frac{50\pi }{3}+25\sqrt{3}$

$=\frac{-25\pi }{6}+25\sqrt{3}$

$=25(\sqrt{3}-\frac{\pi }{6}){cm}^2$