# In the following figure, If ABC is an equilateral triangle,

Question:

In the following figure, If ABC is an equilateral triangle, then shaded area is equal to

(a) $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right) r^{2}$

(b) $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right) r^{2}$

(c) $\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right) r^{2}$

(d) $\left(\frac{\pi}{3}+\sqrt{3}\right) r^{2}$

Solution:

We have given that ABC is an equilateral triangle.

$\therefore \angle A=60^{\circ}$

As we know that, $\angle B C A=\frac{1}{2} m(\angle B O C)$.

$\therefore 60^{\circ}=\frac{1}{2} m(\angle B O C)$

$m(\angle B O C)=120^{\circ}$

Area of the shaded region = area of the segment BC.

Let $\angle B O C=\theta$

$\therefore$ Area of the segment $=\left(\frac{\pi \theta}{360}-\sin \frac{\theta}{2} \cos \frac{\theta}{2}\right) r^{2}$

Substituting the values we get,

Area of the segment $=\left(\frac{\pi \times 120}{360}-\sin 60 \cos 60\right) r^{2}$

$\therefore$ Area of the segment $=\left(\frac{\pi}{3}-\sin 60 \cos 60\right) r^{2}$

Substituting $\sin 60=\frac{\sqrt{3}}{2}$ and $\cos 60=\frac{1}{2}$ we get,

$\therefore$ Area of the segment $=\left(\frac{\pi}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right) r^{2}$

$\therefore$ Area of the segment $=\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right) r^{2}$

Therefore, area of the shaded region is $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right) r^{2}$. Hence, the correct answer is option (a).