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

The area of a segment of a circle can be found using the formula:

\(\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}\)

The area of a sector with central angle \(\theta\)(in radians) and radius \(r\)is given by:

\(\text{Area of sector} = \frac{\theta}{2\pi} \times \pi r^2 = \frac{\theta r^2}{2}\)

For \(\theta = 60^\circ\), we first convert this angle to radians:

\(60^\circ = \frac{\pi}{3} \text{ radians}\)

Thus, the area of this sector is:

\(\text{Area of sector} = \frac{\frac{\pi}{3} \times r^2}{2} = \frac{\pi r^2}{6}\)

The area of an equilateral triangle with side length \(r\)is:

\(\text{Area of triangle} = \frac{\sqrt{3}}{4}r^2\)

Therefore, the area of the segment is:

\(\text{Area of segment} = \frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2\)

This matches the given correct option, which is:

\(\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2\)

Thus, the correct answer is indeed:

\(\frac{\pi r^2}{6} - \frac{\sqrt{3}}{4}r^2\)