JAMB - Mathematics (2013 - No. 40)

A chord of a circle subtends an angle of 120° at the centre of a circle of diameter \(4\sqrt{3} cm\). Calculate the area of the major sector.

32\(\pi\) cm\(^2\)

4\(\pi\) cm\(^2\)

8\(\pi\) cm\(^2\)

16\(\pi\) cm\(^2\)

Angle of major sector = 360° - 120° = 240°

Area of major sector : \(\frac{\theta}{360} \times \pi r^{2}\)

r = \(\frac{4\sqrt{3}}{2} = 2\sqrt{3} cm\)

Area : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\)

= \(8\pi cm^{2}\)