# 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 AreaOfRegularPolygon 2013-03-22 17:11:06 2013-03-22 17:11:06 Wkbj79 (1863) Wkbj79 (1863) 6 Wkbj79 (1863) Theorem msc 51-00