Area of Regular Polygon

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Theorem

Let $P$ be a regular $n$-sided polygon whose side length is $b$.

Then the area of $P$ is given by:

$\Box P = \dfrac 1 4 n b^2 \cot \dfrac \pi n$

where $\cot$ denotes cotangent.


Proof

RegularPolygonArea.png

Let $H$ be the center of the regular $n$-sided polygon $P$.

Let one of its sides be $AB$.

Consider the triangle $\triangle ABH$.

As $P$ is regular and $H$ is the center, $AH = BH$ and so $\triangle ABH$ is isosceles.

Thus $AB$ is the base of $\triangle ABH$.

Let $h = GH$ be its altitude.

See the diagram.


Then:

\(\displaystyle \triangle ABH\) \(=\) \(\displaystyle \frac {b h} 2\) Area of Triangle in Terms of Side and Altitude
\(\displaystyle \) \(=\) \(\displaystyle \frac b 2 \frac b 2 \cot \alpha\) Definition of Cotangent of Angle
\(\displaystyle \) \(=\) \(\displaystyle \frac {b^2} 4 \cot \frac \pi n\) $\alpha$ is half the apex of $\triangle ABH$, and $n$ of such apices fit into the full circle of $2 \pi$


The full polygon $P$ is made up of $n$ of triangles, each of which has the same area as $\triangle ABH$.

Hence:

$\Box P = \dfrac 1 4 n b^2 \cot \dfrac \pi n$

$\blacksquare$


Sources