Question

In the figure given below find the area of the shaded region where a circular arc of radius $6cm$ has been drawn with vertex O of an equilateral triangle $OAB$ of side $12cm$ as centre.

Answer 1

Given:-

Radius of circle $\left(r\right)=6cm$

Side of equilateral triangle $\left(a\right)=12cm$

To find:- Area of shaded region $=?$

Solution:-

Area of circle :

$=\pi r^2=\frac{22}{7}\times {\left(6\right)}^2=\frac{792}{7}{cm}^2$

Area of equilateral triangle:

$=\frac{\sqrt{3}}{4}{\left(\text{side}\right)}^2=\frac{\sqrt{3}}{4}\times {\left(12\right)}^2=36\sqrt{3}{cm}^2$

Area of small sector

$=\frac{\theta }{360°}\times \pi r^2=\frac{60°}{360°}\times \frac{22}{7}\times {\left(6\right)}^2=\frac{132}{7}{cm}^2$

Therefore,

Area of shaded region $=$ Area of circle $+$ Area of triangle $-$ Area of small sector

$=\frac{792}{7}+36\sqrt{3}-\frac{132}{7}=\frac{660}{7}+36\sqrt{3}{cm}^2$

Hence area of shaded region is $(\frac{660}{7}+36\sqrt{3}{cm}^2)$.