Question

Area of a segment of a circle of radius 'r' and central angle \( 60^\circ \) 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. \( \frac{2\pi r}{4} - r^2 \sin 60^\circ \)

Hint:

For \( \theta = 60^\circ \), the triangle is equilateral. For \( \theta = 90^\circ \), the triangle area is \( \frac{1}{2}r^2 \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Area of Segment = Area of Sector - Area of Triangle.
Step 2: Detailed Explanation:
Central angle \( \theta = 60^\circ \).
Area of sector = \( \frac{60}{360} \pi r^2 = \frac{\pi r^2}{6} \).
Triangle with \( 60^\circ \) and two equal radii is an equilateral triangle.
Area of equilateral triangle = \( \frac{\sqrt{3}}{4} r^2 \).
Area of segment = \( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4} r^2 \).
Step 3: Final Answer:
The area is \( \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2 \).