A chord of a circle of radius $r$ subtends an angle of $60^{\circ}$ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to $\pi r^2 \left( \frac{1}{6} - \frac{\sqrt{3}}{4} \right)$.                                               [Page No. 148]

Minor sector area (60°) = $\pi r^2 \times \frac{60}{360} = \frac{\pi r^2}{6}$

Since the chord subtends 60° at centre, triangle formed has two sides = $r$ and included angle = 60°.

This is an equilateral triangle with side $r$.

$$\text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} r^2$$

Minor segment area = Sector area - Triangle area

$$\begin{aligned} &= \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \\ &= r^2 \times \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) & [\text{Factoring out } r^2] \\ &= r^2 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right). \end{aligned}$$

Hence, proved.