Area of Regular Polygon by Circumradius

Theorem

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

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

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

$\AA = \dfrac 1 2 n r^2 \sin \dfrac {2 \pi} n$

Then $\AA$ 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:

$(1): \quad h = r \cos \dfrac \pi n$

$(2): \quad d = 2 r \sin \dfrac \pi n$

So:

$\ds \AA$

$=$

$\ds n \frac {h d} 2$

$\ds $

$=$

$\ds \frac n 2 \paren {r \cos \frac \pi n} \paren {2 r \sin \dfrac \pi n}$

substituting from $(1)$ and $(2)$ above

$\ds $

$=$

$\ds \frac 1 2 n r^2 2 \paren {\cos \frac \pi n} \paren {\sin \dfrac \pi n}$

rearranging

$\ds $

$=$

$\ds \frac 1 2 n r^2 \paren {\sin \frac {2 \pi} n}$

$\ds $

$=$

$\ds \frac 1 2 n r^2 \sin \frac {2 \pi} n$

simplifying

$\blacksquare$