Q1. [1]
Area of a segment of a circle of radius 'r' and central angle 60° is :
- A $\frac{\pi r^2}{2} - \frac{1}{2}r^2$
- B $\frac{2\pi r}{4} - \frac{\sqrt{3}}{4}r^2$
- C $\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2$
- D $\frac{2\pi r}{4} - r^2 \sin 60°$
Previously asked in CBSE board exam
2026 30/2/1 Q15
Generated by claude-sonnet-4-6 · 2026-06-15 10:38 · grounding rag
Model Answer
Answer: (C) $\dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4}r^2$
Area of segment = Area of sector − Area of triangle $= \dfrac{60}{360}\times\pi r^2 - \dfrac{\sqrt{3}}{4}r^2 = \dfrac{\pi r^2}{6} - \dfrac{\sqrt{3}}{4}r^2$
Explanation
- Area of sector (θ = 60°): $\frac{60}{360}\times\pi r^2 = \frac{\pi r^2}{6}$
- Area of equilateral triangle formed (since both radii = r and included angle = 60°, the triangle is equilateral): $\frac{\sqrt{3}}{4}r^2$
- Subtracting gives option C. Options A, B, D are incorrect — B and D involve arc length terms ($2\pi r$) mixed with area, which is dimensionally wrong.
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