Area of Regular Polygon by Inradius

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Theorem

Let $P$ be a regular $n$-gon.

Let $C$ be an incircle of $P$.

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


Then the area $\mathcal A$ of $P$ is given by:

$\mathcal A = n r^2 \tan \dfrac \pi n$


Proof

RegularPolygonAreaCircumscribed.png

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

Then $\mathcal A$ is equal to $n$ times the area of $\triangle OAB$.

Also, $r$ is the length of the altitude 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 $d = 2 r \tan \dfrac \pi n$

So:

\(\displaystyle \mathcal A\) \(=\) \(\displaystyle n \frac {r d} 2\) Area of Triangle in Terms of Side and Altitude
\(\displaystyle \) \(=\) \(\displaystyle \frac n 2 r \paren {2 r \tan \dfrac \pi n}\) substituting from above
\(\displaystyle \) \(=\) \(\displaystyle n r^2 \tan \dfrac \pi n\) rearranging

$\blacksquare$


Sources