# Area of Regular Polygon by Inradius

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## Theorem

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 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:

\(\displaystyle \AA\) | \(=\) | \(\displaystyle n \frac {r d} 2\) | 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

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