Question

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made.(Take 𝛑=𝟑.𝟏𝟒 and √𝟑= 1.73).

Answer 1

$OAB$ is an equilateral triangle with each angle equal to $60°$.
Radius of the circle $=6cm$.
Side of the triangle $=12cm$.
Area of the equilateral triangle$=\frac{\surd 3}{4}\times \left(sid{e}^2\right)=\frac{\surd 3}{4}\times \left(O{A}^2\right)=\frac{\surd 3}{4}\times \left({12}^2\right)=36\surd 3c{m}^2=62.28c{m}^2$.
Area of circle $=3.14\times (6\times 6)=113.04{cm}^2$.

Area of the sector $OCDE=\frac{\theta }{360°}\times \pi \times r^2$ $=\frac{60}{360}\times 3.14\times (6\times 6)$ $=18.84c{m}^2$.
As the area of the sector is common in both the circle as well as the triangle.

So,
Area of the shaded region = Area of the equilateral triangle + Area of the circle - 2×Area of the sector.

Area of the shaded region
= 62.28 + 113.04 - 2×18.84
$=137.64c{m}^2$