# Interior Angles of a Polygon

Just like how the sum of the interior angles of a triangle equals 180º, the sum of the interior angles of every polygon has a set value. This is called the Polygon Interior Angles Theorem. To determine what that value is, we are going to use the fact that the interior angles of a triangle equal 180º to figure it out.

First let’s take a four-sided figure, called a quadrilateral, like the one below: It is important to note that it does not have to be a regular quadrilateral. The Polygon Interior Angle theorem will work for all of them. The first step in proving this theorem is to break the polygon into triangles, as below: We want to divide the quadrilateral into as few triangles as possible. This is why the quadrilateral was divided into two triangles. It could, in theory, be divided into an infinite number of triangles, but that just makes the math more complicated so we’ll just use two triangles for this example.

Labeling the interior angles of the new triangles gives us this: The Sum of the Interior Angles of a Triangle Theorem gives us:

m∠a + m∠b + m∠c = 180º                                                                                                       1

m∠d + m∠e + m∠f = 180º                                                                                                        2

The sum of the interior angles of the quadrilateral is:

∑interior angles = m∠a + m∠b + m∠c + m∠d+ m∠e + m∠f                                              3

Substituting equation 1 and 2 into equation 3 gives us:

∑interior angles = 180º + 180º

∑interior angles = 360º

This same procedure can be used to find the interior angles of a five-sided figure, called a pentagon. The Sum of the Interior Angles of a Triangle Theorem gives us:

m∠a + m∠b + m∠c = 180º                                                                                                       4

m∠d + m∠e + m∠f = 180º                                                                                                        5

m∠g + m∠h + m∠i = 180º                                                                                                        6

The sum of the interior angles of the pentagon is:

∑interior angles = m∠a + m∠b + m∠c + m∠d+ m∠e + m∠f + m∠g + m∠h + m∠i          7

Substituting equation 4, 5 and 6 into equation 7 gives us:

∑interior angles = 180º + 180º + 180º

∑interior angles = 540º

Expanding this proof to a six sided figure, called a hexagon, we get the following situation: The Sum of the Interior Angles of a Triangle Theorem gives us:

m∠a + m∠b + m∠c = 180º                                                                                                       8

m∠d + m∠e + m∠f = 180º                                                                                                        9

m∠g + m∠h + m∠i = 180º                                                                                                       10

m∠j + m∠k + m∠l = 180º                                                                                                        11

The sum of the interior angles of the hexagon is:

∑interior angles = m∠a + m∠b + m∠c + m∠d+ m∠e + m∠f + m∠g + m∠h + m∠i + m∠j + m∠k + m∠l                                                                                                                                              12

Subbing equations 8, 9, 10, and 11 into equation 12 gives us:

∑interior angles = 180º + 180º + 180º + 180º

∑interior angles = 720º

This is enough to notice a pattern is beginning to form. Each time a side is added to a polygon the sum of the interior angles for that polygon increases by 180º. Now, if we were to look at additional polygons, with additional sides, we would notice that the pattern contniues. Writing a general equation for this pattern:

∑interior angles = (n-2) * 180