Perimeter of Regular Polygon by Circumradius
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Theorem
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
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:
\(\ds \PP\) | \(=\) | \(\ds n d\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {2 r \sin \dfrac \pi n}\) | substituting from above | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 n r \sin \dfrac \pi n\) | simplifying |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 4$: Geometric Formulas: $4.18$