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
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).