# 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 $\mathcal P$ of $P$ is given by:

$\mathcal P = 2 n r \sin \dfrac \pi n$

## Proof

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

Then $\mathcal P$ 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:

 $\displaystyle \mathcal P$ $=$ $\displaystyle n d$ $\displaystyle$ $=$ $\displaystyle n \paren {2 r \sin \dfrac \pi n}$ substituting from above $\displaystyle$ $=$ $\displaystyle 2 n r \sin \dfrac \pi n$ simplifying

$\blacksquare$