A chord of a circle of radius 14 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor segment of the circle. Also find…

Area of minor segment \(\displaystyle =\frac{22}{7} \times 14 \times 14 \times \frac{60}{360}-\frac{1}{2} \times 14 \times 14 \times \frac{\sqrt{3}}{2}\) \[=\left(\frac{308}{3}-49 \sqrt{3}\right) \mathrm{cm}^{2} \text { or } 17.9 \mathrm{~cm}^{2} \] Area of major segment \(\displaystyle =\frac{22}{7} \times 14 \times 14-\left(\frac{308}{3}-49 \sqrt{3}\right)\) \[\begin{aligned} & =616-\frac{308}{3}+49 \sqrt{3} \\ & =\left(\frac{1540}{3}+49 \sqrt{3}\right) \mathrm{cm}^{2} \text { or } 598.1 \mathrm{~cm}^{2} \end{aligned} \]