Area of Regular Polygon by Circumradius

Theorem
Let $P$ be a regular $n$-gon.

Let $C$ be a circumcircle of $P$.

Let the radius of $C$ be $r$.

Then the area $\mathcal A$ of $P$ is given by:
 * $\mathcal A = \dfrac 1 2 n r^2 \sin \dfrac {2 \pi} n$

Proof

 * RegularPolygonAreaInscribed.png

From Regular Polygon composed of Isosceles Triangles, let $\triangle OAB$ be one of the $n$ isosceles triangles that compose $P$.

Then $\mathcal A$ is equal to $n$ times the area of $\triangle OAB$.

Let $d$ be the length of one side of $P$.

Then $d$ is the length of the base of $\triangle OAB$.

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

The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.

Then:
 * $(1): \quad h = r \cos \dfrac \pi n$
 * $(2): \quad d = \dfrac 1 2 r \sin \dfrac \pi n$

So: