# What is the sum, in degrees, of the measures of the interior angles of a pentagon?

Jan 3, 2017

${540}^{o}$ or $5 \times {108}^{o}$

#### Explanation:

In any $n$-gon, the sum of the interior angles is $\left(n - 2\right) \cdot {180}^{o}$.
This can be proven by taking a regular n-gon (why not?), and drawing a (isosceles) triangle from the centre to one of the sides.

The centre angle will be ${360}^{o} / n$ and the other angles of that triangle will be $\frac{1}{2} \left(180 - \frac{360}{n}\right)$.
The inner angle of the n-gon will be just twice that, or $180 - \frac{360}{n}$

The sum of all inner angles will then be:
$n \cdot \left(180 - \frac{360}{n}\right) = n \cdot \left(180 \times \frac{n}{n} - \frac{360}{n}\right) =$

$\cancel{n} \cdot \left(\frac{180 n - 360}{\cancel{n}}\right) = \left(n - 2\right) \cdot {180}^{o}$

In case of a regular n-gon, each angle will be $\frac{n - 2}{n} \cdot {180}^{o}$