Area of Sector/Proof 2

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Theorem

Let $\mathcal C = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.

Let $BAC$ be the sector of $\mathcal C$ whose angle between $AB$ and $AC$ is $\theta$.

Sector.png


Then the area $\mathcal A$ of sector $BAC$ is given by:

$\mathcal A = \dfrac {r^2 \theta} 2$

where:

$r = AB$ is the length of the radius of the circle
$\theta$ is measured in radians.


Proof

From Area of Circle, the area of $\mathcal C$ is $\pi r^2$.

From Full Angle measures $2 \pi$ Radians, the angle within $\mathcal C$ is $2 \pi$.

The fraction of the area of $\mathcal C$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$.

Hence the result.

$\blacksquare$