Question

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

• A. $(36\sqrt{3}+\frac{660}{7})c{m}^2$
• B. $(32\sqrt{3}+\frac{660}{7})c{m}^2$
• C. $(36\sqrt{3}+\frac{550}{7})c{m}^2$
• D. $(36\sqrt{3}+\frac{660}{9})c{m}^2$

Answer 1

The correct option is A

$(36\sqrt{3}+\frac{660}{7})c{m}^2$

OAB is an equilateral triangle with each angle equal to ${60}^{\circ }$.
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
$=\frac{\sqrt{3}}{4}\times (OA{)}^2=\frac{\sqrt{3}}{4}\times 144=36\sqrt{3}c{m}^2$
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 ${60}^{\circ }$
$=\left(\frac{{60}^{\circ }}{{360}^{\circ }}\right)\times \pi r^{2}c{m}^2$
$=\frac{1}{6}\times \frac{22}{7}\times 6^{2}c{m}^2=\frac{132}{7}c{m}^2$
Area of the shaded region= Area of the equilateral triangle + Area of the circle - Area of the sector
$=36\sqrt{3}c{m}^2+\frac{792}{7}c{m}^2-\frac{132}{7}c{m}^2$
$=(36\sqrt{3}+\frac{660}{7})c{m}^2$