Question

Area of a segment of a circle of radius ‘r’ and central angle 60° is ______.

Options

• `(πr^2)/2 - 1/2 r^2`

• `(2πr)/4 - sqrt(3)/4 r^2`

• `(πr^2)/6 - sqrt(3)/4 r^2`

• `(2πr)/4 - r^2 sin 60^circ`

Answer 1

Area of a segment of a circle of radius ‘r’ and central angle 60° is `underlinebb((πr^2)/6 - sqrt(3)/4 r^2)`.

Explanation:

Given:

Radius = r

Central angle (θ) = 60°

Area of segment = Area of sector – Area of triangle

Area of sector = `θ/360^circ xx πr^2`

Area of sector = `60^circ/360^circ xx πr^2 = (πr^2)/6`

Since the central angle is 60° and the two sides are radii, the triangle is equilateral.

Area of equilateral triangle = `sqrt(3)/4 r^2`

Area of segment = `(πr^2)/6 - sqrt(3)/4 r^2`