Question

Find the area of minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60⁰ .

Answer 1

Area of the minor segment of the circle = \[\sqrt{2}\]\[\frac{\theta}{360} \times \pi r^2 - r^2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}\]

\[= \frac{60}{360} \times \pi \left( 14 \right)^2 - \left( 14 \right)^2 \sin\frac{60}{2}\cos\frac{60}{2}\]
\[ = \left( 14 \right)^2 \left[ \frac{1}{6}\pi - \frac{1}{2} \times \frac{\sqrt{3}}{2} \right]\]
\[ = \left( \frac{308}{3} - 49\sqrt{3} \right) {cm}^2\]