Area of Segment of Circle

Theorem

Let $C$ be a circle of radius $r$.

Let $S$ be a segment of $C$ such that its base subtends an angle of $\theta$ at the center of the circle.

Then the area $\AA$ of $S$ is given by:

$\AA = \dfrac 1 2 r^2 \paren {\theta - \sin \theta}$

where $\theta$ is measured in radians.

Let $BDCE$ be the segment $S$.

Let $b$ be the length of the base of $S$.

Let $BACE$ be the sector of $C$ whose angle is $\theta$.

The $\AA$ is equal to the area of $BACE$ minus the area of the isosceles triangle $\triangle ABC$ whose base is $b$.

Let $h$ be the altitude of $\triangle ABC$.

From Area of Sector, the area of sector $BACE$ is $\dfrac 1 2 r^2 \theta$.

From Area of Isosceles Triangle, the area of $\triangle ABC$ is $\dfrac 1 2 r^2 \sin \theta$.

Thus:

$\ds \AA$

$=$

$\ds \frac 1 2 r^2 \theta - \frac 1 2 r^2 \sin \theta$

$\ds $

$=$

$\ds \frac 1 2 r^2 \paren {\theta - \sin \theta}$

$\blacksquare$