Question

In fig., an equilateral triangle ABC of side $6$ cm has been inscribed in a circle. Find the area of the shaded region. (Take $\pi =3.14$)

Answer 1

Draw $OP\perp BC,\therefore AB=BC=CA=6cm$

$ar.△ABC=\frac{\sqrt{3}}{4}{\left(C\right)}^2=9\sqrt{3}c{m}^2$

$\angle BAC=60°,BP=PC=\frac{BC}{2}=\frac{6}{2}=3cm$

In $△BOC,\angle BOC+\angle OBC+\angle OCB=180°$

$\Rightarrow 120°+\angle OBC+\angle OBC=180°\Rightarrow \angle OBC=\frac{180°-120°}{2}=30°[\because OB=OC]$

$\cos 30°=\frac{BP}{OB}\Rightarrow OB=\frac{6}{\sqrt{3}}cm$

$\therefore$Radius of circle$=\frac{6}{\sqrt{3}}cm$

Area of circle$=\pi \times {\left(\frac{6}{\sqrt{3}}\right)}^2=3.14\times 12=37.68c{m}^2$

Area of shaded region$=37.68-9\sqrt{3}=22.10c{m}^2$