# Area of Regular Polygon by Circumradius

## Theorem

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

Let $C$ be a circumcircle of $P$.

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

Then the area $\AA$ of $P$ is given by:

$\AA = \dfrac 1 2 n r^2 \sin \dfrac {2 \pi} n$

## 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 $\triangle OAB$.

Let $d$ be the length of one side of $P$.

Then $d$ is the length of the base of $\triangle OAB$.

Let $h$ be the altitude of $\triangle OAB$.

The angle $\angle AOB$ is equal to $\dfrac {2 \pi} n$.

Then:

$(1): \quad h = r \cos \dfrac \pi n$
$(2): \quad d = 2 r \sin \dfrac \pi n$

So:

 $\ds \AA$ $=$ $\ds n \frac {h d} 2$ Area of Triangle in Terms of Side and Altitude $\ds$ $=$ $\ds \frac n 2 \paren {r \cos \frac \pi n} \paren {2 r \sin \dfrac \pi n}$ substituting from $(1)$ and $(2)$ above $\ds$ $=$ $\ds \frac 1 2 n r^2 2 \paren {\cos \frac \pi n} \paren {\sin \dfrac \pi n}$ rearranging $\ds$ $=$ $\ds \frac 1 2 n r^2 \paren {\sin \frac {2 \pi} n}$ Double Angle Formula for Sine $\ds$ $=$ $\ds \frac 1 2 n r^2 \sin \frac {2 \pi} n$ simplifying

$\blacksquare$