Question

A chord or the circle with radius 5 cm subtends an angle of ${60}^{\circ }$ at the centre. Find the area of the minor segment by this chord.

Answer 1

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

Here,$\theta ={60}^o,r=5cm$

Therefore,area of the minor segment$=(\frac{60}{360}\times 3.14-sin{30}^{0}cos{30}^o)25$

$=(0.5234-\frac{\sqrt{3}}{4})25$

$=2.26$$c{m}^2$