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.
A. 32\(\pi\) cm\(^2\) B. 4\(\pi\) cm\(^2\) C. 8\(\pi\) cm\(^2\) D. 16\(\pi\) cm\(^2\)
Correct Answer: C
Explanation
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 : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\) = \(8\pi cm^{2}\)
