In the diagram above, O is the center of the circle of radius 3.5cm, ∠POQ = 60°. What is the area of the minor sector POQ?
[Take π = 22/7].
Area of sector = \(\frac{\theta}{360°} \times \pi r^2\)
= \(\frac{60}{360} \times \frac{22}{7} \times 3.5 \times 3.5\)
= \(\frac{77}{12}\)
= \(6\frac{5}{12}\) cm\(^2\)
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