Question

In the below given figure, $ABC$ is an equilateral triangle of side $14$ cm. $M$ is the centre of the circumcircle. Find the area of the shaded region(ie. excluding triangle enclosed by circle)

• A. $115.27{\text{cm}}^2$
• B. $96.63{\text{cm}}^2$
• C. $120.46{\text{cm}}^2$
• D. $146.72{\text{cm}}^2$

Answer 1

The correct option is C $120.46{\text{cm}}^2$

ABC is on equilateral triangle of side 14 cm.

M is the centre of the circumcircle.

Height (H) of the equilateral triangle is given by $\frac{\sqrt{3}}{2}\times s$, where s is the side of the equilateral triangle

$H\Rightarrow \frac{\sqrt{3}}{2}\times 14$ $(\because s=14cm)$

$H\Rightarrow 7\sqrt{3}$ cm ...(i)

$\because$ in an equilateral triangle , orthocenter, centroid and circumcenter and incenter coincide, and the radius (R) of the circumcircle = $\frac{2}{3}H$

$\Rightarrow R=\frac{2}{3}\times 7\sqrt{3}$ [From (i)]

$\Rightarrow \frac{14\sqrt{3}}{3}$

Area of circle = $\pi R^2=\frac{22}{7}\times \frac{14\sqrt{3}}{3}\times \frac{14\sqrt{3}}{3}=\frac{44\times 52}{9}c{m}^2$ ..(i)

Area of $\Delta ABC=\frac{\sqrt{3}}{4}\times s^2=\frac{\sqrt{3}}{4}\times 14\times 14=\frac{196\sqrt{3}}{4}=49\sqrt{3}c{m}^2$

$=84.87c{m}^2$

Area of shaded region = Area of circle - Area of $\Delta ABC$

$=(205.33-84.87)c{m}^2$

$=120.46c{m}^2$