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. [ Use $\sqrt{3}=1.73and\pi =3.14$.]

Answer 1

OAB is an equilateral triangle with each angle equal to ${60}^o$.

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle $=\frac{\sqrt{3}}{4}\times (OA{)}^2=\frac{\sqrt{3}}{4}\times {12}^2=36\sqrt{3}c{m}^2$

Area of the circle $=\pi R^2=\frac{22}{7}\times 6^2=\frac{792}{7}c{m}^2$

Area of the sector making angle ${60}^o=\frac{{60}^o}{{360}^o}\times \pi r^{2}c{m}^2=\frac{1}{6}\times \frac{22}{7}\times 6^{2}c{m}^2=\frac{132}{7}c{m}^2$

Area of the shaded region = Area of the equilateral triangle + Area of the circle - Area of the sector

$=36\sqrt{3}c{m}^2+\frac{792}{7}c{m}^2-\frac{132}{7}c{m}^2=(36\sqrt{3}+\frac{660}{7})c{m}^2=62.28+94.2857143=156.56c{m}^2$