Area of Regular Polygon by Inradius

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 $\AA$ of $P$ is given by:

$\AA = n r^2 \tan \dfrac \pi n$

Proof

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

Then $\AA$ 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:

\(\ds \AA\)

\(=\)

\(\ds n \frac {r d} 2\)

Area of Triangle in Terms of Side and Altitude

\(\ds \)

\(=\)

\(\ds \frac n 2 r \paren {2 r \tan \dfrac \pi n}\)

substituting from above

\(\ds \)

\(=\)

\(\ds n r^2 \tan \dfrac \pi n\)

rearranging

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 4$: Geometric Formulas: $4.19$