What is the area of the segment formed by a chord in a circle of radius 8 cm, if the angle subtended at the center is 60°?

Correct option is A

​Given :

Radius of circle r = 8 cm
Angle subtended at the center $$\theta = 60^\circ$$​​

Formula Used :

​$$\text{Area of sector} = \frac{\theta}{360^\circ}\pi r^2$$​​
Area of triangle formed by two radii

​$$\text{Area of triangle} = \frac{1}{2} r^2 \sin \theta$$​​

Solution :

Area of sector:
=$$\frac{60}{360} \times \pi \times 8^2$$ ​

​$$= \frac{1}{6} \times \pi \times 64$$ 

​$$= \frac{32\pi}{3}$$​​

Area of triangle:
=$$\frac{1}{2} \times 8^2 \times \sin 60^\circ$$​​
=$$\frac{1}{2} \times 64 \times \frac{\sqrt{3}}{2}$$​​
= $$16\sqrt{3}$$​​

Area of segment = Area of sector - Area of triangle
​$$= \frac{32\pi}{3} - 16\sqrt{3}$$​​