area of regular polygon

Theorem 1.

Given a regular $n$ -gon (http://planetmath.org/RegularPolygon) with apothem of length $a$ and perimeter (http://planetmath.org/Perimeter2) $P$ , its area is

A=\frac{1}{2}aP.

Proof.

Given a regular $n$ -gon $R$ , line segments can be drawn from its center to each of its vertices. This divides $R$ into $n$ congruent triangles. The area of each of these triangles is $\displaystyle\frac{1}{2}as$ , where $s$ is the length of one of the sides of the triangle. Also note that the perimeter of $R$ is $P=ns$ . Thus, the area $A$ of $R$ is

$\begin{array}[]{rl}A&\displaystyle=n\left(\frac{1}{2}as\right)\\ &\\ &\displaystyle=\frac{1}{2}a(ns)\\ &\\ &\displaystyle=\frac{1}{2}aP.\end{array}$

∎

To illustrate what is going on in the proof, a regular hexagon appears below with each line segment from its center to one of its vertices drawn in red and one of its apothems drawn in blue.

Title	area of regular polygon
Canonical name	AreaOfRegularPolygon
Date of creation	2013-03-22 17:11:06
Last modified on	2013-03-22 17:11:06
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	6
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 51-00