What is the area of a segment of a circle of radius r subtending an angle \theta at the centre?
- A. \frac{1}{2}r^{2}\theta
- B. \frac{1}{2}r^{2}(\theta-2 \sin\frac{\theta}{2}\cos\frac{\theta}{2}) ✓
- C. \frac{1}{2}r^{2}(\theta-\sin\frac{\theta}{2}\cos\frac{\theta}{2})
- D. \frac{1}{2}r^{2}\sin\frac{\theta}{2}\cos\frac{\theta}{2}
Correct Answer: B. \frac{1}{2}r^{2}(\theta-2 \sin\frac{\theta}{2}\cos\frac{\theta}{2})
Explanation
The area of a segment is the area of the sector minus the area of the triangle: \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta. Applying the double angle identity \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, the expression becomes \frac{1}{2}r^2(\theta - 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}).
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