# In the following figure, the area of segment ACB is

Question:

In the following figure, the area of segment ACB is

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

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

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

(d) None of these

Solution:

We have to find area of segment ACB.

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

We know that $\theta=120^{\circ}$.

Substituting the values we get,

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

$\therefore$ Area of the PAQ 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,

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

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

Therefore, area of the segment $A C B$ is $\left(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\right) r^{2}$.

Hence, the correct answer is option (d).