Perimeter of Regular Polygon by Circumradius

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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

RegularPolygonAreaInscribed.png

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

$\blacksquare$


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