Question

Find the area of the shaded region in the figure, where a circular arc of radius $6cm$ has been drawn with vertex $O$ of an equiateral triangle OAB of side $12cm$ as center.

Answer 1

Here we can see that

Area of required figure $=$ Area of circle$+$Area of triangle$-$ Area of the sector

$\Rightarrow A\left(circle\right)=\pi r^2$

$\Rightarrow A\left(circle\right)=\pi 6^2$

$\Rightarrow A\left(circle\right)=36\pi$

$\Rightarrow A\left(triangle\right)=\frac{\sqrt{3}}{4}{a}^2$

$\Rightarrow A\left(triangle\right)=\frac{\sqrt{3}}{4}(12{)}^2$

$\Rightarrow A\left(triangle\right)=36\sqrt{3}$

$\Rightarrow A\left(sector\right)=\frac{1}{2}{r}^2\theta$

$\Rightarrow A\left(sector\right)=\frac{1}{2}(6{)}^2\frac{\pi }{3}$

$\Rightarrow A\left(sector\right)=6\pi$

Therefore, Required $Area=36\pi +36\sqrt{3}-6\pi$

$\Rightarrow$ Required $Area=30\pi +36\sqrt{3}$