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[parent] area of regular polygon (Theorem)
Theorem   Given a regular $ n$-gon with apothem of length $ a$ and perimeter $ P$, its area is

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

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.


\begin{pspicture}(-2,-2)(2,2) \psline[linecolor=blue](0,0)(0,-1.732) \psline[lin... ...2) \pspolygon(2,0)(1,1.732)(-1,1.732)(-2,0)(-1,-1.732)(1,-1.732) \end{pspicture}



"area of regular polygon" is owned by Wkbj79.
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Cross-references: hexagon, sides, triangles, congruent, vertices, center, line segments, area, length, apothem
There is 1 reference to this entry.

This is version 3 of area of regular polygon, born on 2007-06-03, modified 2007-06-03.
Object id is 9501, canonical name is AreaOfRegularPolygon.
Accessed 1553 times total.

Classification:
AMS MSC: 51-00 (Geometry :: General reference works )
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