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 a perpendicular bisector OP from point O to BC.

AB = BC = CA = 6 cm

Area of ΔABC = $\frac{\sqrt{3}}{4}$ $\times$ $6^2$ = 9$\sqrt{3}$ $c{m}^2$

∠BAC = 60° (equilateral triangle)

BP = PC = $\frac{BC}{2}$ = $\frac{6}{2}$ = 3 cm

In ΔBOC,

∠BOC + ∠OBC + ∠OCB = 180°

⇒ 120° + ∠OBC + ∠OBC = 180°

⇒∠OBC = $\frac{180°-120°}{2}$ ( $\because$ OB= OC)

⇒∠OBC = 30°

cos 30° = $\frac{BP}{OB}$

$\frac{\sqrt{3}}{2}$ = $\frac{3}{OB}$

OB= $\frac{6}{\sqrt{3}}$

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

Area of the circle = $\pi$ ($\frac{6}{\sqrt{3}}{)}^2$

= 12 $\pi$ $c{m}^2$

∴ Area of the remaining (shaded) part = Area of the circle – Area of the equilateral triangle

= 12 $\pi$ - 9$\sqrt{3}$

= 37.68 - 15.57

=22.11 $c{m}^2$