# 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 $\frac{1}{2}}as$, 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

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

∎

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 |