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| Title of object: |
area of regular polygon |
| Canonical Name: |
AreaOfRegularPolygon |
| Type: |
Theorem |
| Created on: |
2007-06-03 00:22:24 |
| Modified on: |
2007-06-03 00:22:24 |
| Classification: |
msc:51-00 |
Preamble:
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{pstricks}
\usepackage{psfrag}
\usepackage{graphicx}
\usepackage{amsthm}
\usepackage{xypic}
\newtheorem{thm*}{Theorem}
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Content:
\PMlinkescapeword{regular}
\PMlinkescapeword{divides}
\begin{thm*}
Given a \PMlinkname{regular $n$-gon}{RegularPolygon} with apothem of length $a$ and perimeter $P$, its area is
$$A=\frac{1}{2}aP.$$
\end{thm*}
\begin{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,
\begin{center}
$\begin{array}{rl}
A & \displaystyle =n\left( \frac{1}{2}as \right) \\
& \\
& \displaystyle =\frac{1}{2}a(ns) \\
& \\
& \displaystyle =\frac{1}{2}aP. \end{array}$
\end{center}
\end{proof}
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{center}
\begin{pspicture}(-2,-2)(2,2)
\psline[linecolor=blue](0,0)(0,-1.732)
\psline[linecolor=red](-2,0)(2,0)
\psline[linecolor=red](-1,-1.732)(1,1.732)
\psline[linecolor=red](1,-1.732)(-1,1.732)
\pspolygon(2,0)(1,1.732)(-1,1.732)(-2,0)(-1,-1.732)(1,-1.732)
\end{pspicture}
\end{center} |
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