Perimeter 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 perimeter $\PP$ of $P$ is given by:
 * $\PP = 2 n r \sin \dfrac \pi n$

Proof

 * RegularPolygonAreaInscribed.png

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

Then $\PP$ is equal to $n$ times the base of $\triangle OAB$.

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

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

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

Then:
 * $d = 2 r \sin \dfrac \pi n$

So: