32\(\pi\) cm\(^2\)
4\(\pi\) cm\(^2\)
8\(\pi\) cm\(^2\)
16\(\pi\) cm\(^2\)
Correct answer is C
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 : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\)
= \(8\pi cm^{2}\)