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.