Area of Regular Polygon by Circumradius

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Theorem

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

Let the circumradius of $P$ be $r$.


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

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


Proof

RegularPolygonAreaInscribed.png

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

$\blacksquare$


Sources

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