sum of interior angles of a polygon Archives

Sum of the interior angles of a polygon

October 21, 2009 GB High School Geometry, High School Mathematics, Questions and Quandaries

We were taught that if we let $m$ be the angle sum (the total measure of the interior angles) and $n$ be the number of vertices (corners) of a polygon, then $m = 180(n-2)$ . For example, a quadrilateral has $4$ vertices, so its angle sum is $180(4-2) = 360$ degrees. Similarly, the angle sum of a hexagon (a polygon with $6$ sides) is $180(6-2) = 720$ degrees.

But where did this formula come from? Does this formula work for all polygons? (Note that in this discussion, when we say polygon, we only refer to convex polygons).

Before we answer these questions, let us first have a brief review of some elementary concepts.

Polygons and Interior Angles

A polygon is a closed figure with finite number of sides. In the figures below, $ABCDE$ is a polygon with $5$ sides and ( $5$ vertices). It is clear that the number of sides of a polygon is always equal to the number of its vertices.

A polygon has interior angles. In the first figure below, angle $B$ measuring $91$ degrees is an interior angle of polygon $ABCDE$ . The angle sum $m$ of $ABCDE$ (not drawn to scale) is given by the equation

$m = 126 + 91 + 113 + 102 + 108 = 540$ degrees.

twopentagons

In the second figure, if we let $a_1, a_2$ and $a_3$ be the measure of the interior angles of triangle $ABE$ , then the angle sum m of triangle $ABE$ is given by the equation $m = a_1 + a_2 + a_3$ .

Angle Sum

To generalize our calculation of angle sum, we use the fact that the angle sum of a triangle is $180$ degrees. Notice that any polygon maybe divided into triangles by drawing diagonals from one vertex to all of the non-adjacent vertices. In the second figure above, the pentagon was divided into three triangles by drawing diagonals from vertex $E$ to the non-adjacent vertices $B$ and $C$ forming $BE$ and $CE$ . Now let $a_k, b_k$ and $c_k$ , where $k = 1, 2, 3$ be measures of the interior angles of the three triangles as shown on the second figure.

Calculating the angle sum of pentagon $ABCDE$ we have

pentagonanglesum

Notice that the angle measures in the first line of our equation is just a rearrangement of the measures of the interior angles of the three triangles. Hence, the angle sum of the pentagon is equal to the angle sum of the three triangles. Therefore, we can conclude that the sum of the interior angles of a polygon is equal to the angle sum of the number of triangles that can be formed by dividing it using the method described above. Using this conclusion, we will now relate the number of sides of a polygon, the number of triangles that can be formed by drawing diagonals and the polygon’s angle sum.

table

From the table above, we observe that the number of triangles formed is $2$ less than the number of sides of the polygon. This is true, because $n - 2$ triangles can be formed by drawing diagonals from one of the vertices to $n - 3$ non-adjacent vertices. Therefore, there the angle sum $m$ of a polygon with $n$ sides is given by the formula

$m = 180(n - 2)$

A More Formal Proof

Theorem: The sum of the interior angles of a polygon with $n$ sides is $180(n-2)$ degrees.

Proof:

Assume a polygon has $n$ sides. Choose an arbitrary vertex, say vertex $V$ . Then there are $n - 3$ non-adjacent vertices to vertex $V$ . If diagonals are drawn from vertex $V$ to all non-adjacent vertices, then $n - 2$ triangles will be formed. The sum the interior angles of $n -2$ triangles is $180(n - 2)$ . Since the angle sum of the polygon with $n$ sides is equal to the sum the interior angles of $n - 2$ triangles, the angle sum of a polygon with $n$ sides is $180(n-2)$ . $\blacksquare$