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\(^2\)
The correct answer is: 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 : \(\frac{240}{360} \times \pi \times (2\sqrt{3})^{2}\)
= \(8\pi cm^{2}\)
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