# How many sides are in a regular polygon that has exterior angles of 40°?

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Jan 22, 2016

A regular polygon with exterior angles of ${40}^{o}$ would have 9 side and be a nonagon.

#### Explanation:

The exterior angles of any regular polygon must add up to ${360}^{o}$.

Since the angle measure given iin the questions s ${40}^{o}$, take ${360}^{o} / {40}^{o}$ = 9. Meaning there are 9 exterior angles and therefore 9 sides to the polygon.

A regular polygon refers to a multi-sided convex figure where all sides are equal in length and all angles have equal degree measures.

The regular triangle has 3 interior angles of ${60}^{o}$ and 3 exterior angles of ${120}^{o}$. The exterior angle have a sum of ${360}^{o}$ $= \left(3\right) {120}^{o}$

The square has 4 interior angles of ${90}^{o}$ and 4 exterior angles of ${90}^{o}$. The exterior angle have a sum of ${360}^{o}$ $= \left(4\right) {90}^{o}$.

The square has 5 interior angles of ${108}^{o}$ and 5 exterior angles of ${72}^{o}$. The exterior angle have a sum of ${360}^{o}$ $= \left(5\right) {72}^{o}$.

In order to find the value of the interior angle of a regular polygon the equation is $\frac{\left(n - 2\right) 180}{n}$ where n is the number of sides of the regular polygon.

Triangle $\frac{\left(3 - 2\right) 180}{3} = {60}^{o}$
Square $\frac{\left(4 - 2\right) 180}{4} = {90}^{o}$
Pentagon $\frac{\left(5 - 2\right) 180}{5} = {72}^{o}$

Finally

The interior and exterior angles of a regular polygon form a linear pair and therefore are supplementary and must add up to ${180}^{o}$.

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