# Area and Perimeter of a Pentagon

A pentagon is a polygon that has five vertices and five sides. The sum of the angles in the pentagon is always $540$°. There are two types of pentagons—regular pentagons and irregular pentagons.

A regular pentagon is a pentagon where all the sides are the same length, and all the angles are the same size.

An irregular pentagon is a pentagon where the sides don’t have the same length, and the angles aren’t the same size.

Note! It’s not possible to have one but not the other—a pentagon will always either have both the same size of angles and same side lengths, making it regular, or have neither, making it irregular.

## Perimeter of a Pentagon

You find the perimeter of a pentagon by adding the lengths of its five sides together.

Formula

### PerimeterofaPentagon

 $P={\text{s}}_{1}+{\text{s}}_{2}+{\text{s}}_{3}+{\text{s}}_{4}+{\text{s}}_{5}$

Example 1

You find the perimeter of the pentagon above by adding the length of all the sides together:

$\begin{array}{llll}\hfill P& =4\phantom{\rule{0.17em}{0ex}}\text{cm}+5\phantom{\rule{0.17em}{0ex}}\text{cm}+4\phantom{\rule{0.17em}{0ex}}\text{cm}+3.5\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{=}+3\phantom{\rule{0.17em}{0ex}}\text{cm}=19.5\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill P& =4\phantom{\rule{0.17em}{0ex}}\text{cm}+5\phantom{\rule{0.17em}{0ex}}\text{cm}+4\phantom{\rule{0.17em}{0ex}}\text{cm}+3.5\phantom{\rule{0.17em}{0ex}}\text{cm}+3\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =19.5\phantom{\rule{0.17em}{0ex}}\text{cm}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

## Area of a Pentagon

You can find the area of a regular pentagon by dividing the pentagon into five equal triangles. Then you can find the area of just one of the triangles and multiply the answer by 5.

Formula

### AreaofaPentagon

Example 2

You can find the area of the pentagon above by finding the area of a triangle and multiplying it by 5: $\begin{array}{llll}\hfill {A}_{\text{triangle}}& =\frac{2\phantom{\rule{0.17em}{0ex}}\text{cm}\cdot 3\phantom{\rule{0.17em}{0ex}}\text{cm}}{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =3\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {A}_{\text{pentagon}}& =5\cdot {A}_{\text{triangle}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =5\cdot 3\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =15\phantom{\rule{0.17em}{0ex}}{\text{cm}}^{2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

The area of the pentagon is $15$ cm2.