How many degrees in a polygon?

I nearly always learn new mathematics when I work with teachers. Recently I led a daylong session on the Connected Mathematics unit Shapes and Designs with a group of sixth-grade teachers. In that unit, there is a problem that asks students to figure out the total number of degrees in the interior angles of an n-sided polygon.

Two strategies are presented in the text by fictional students. Tia’s strategy begins with this diagram:

Tia’s strategy

Cody’s strategy begins with a different diagram:

Cody's strategy

Most of the teachers in my session knew the formula T=(n-2)*180 where T is the total number of degrees in all interior angles of the polygon and n is the number of sides of the polygon. But it quickly became clear that they were struggling to use the diagrams to justify the formula. As a group, they liked Tia’s diagram because it shows (n-2) triangles, but they could not answer my questions about why Tia gets (n-2) triangles when she cuts up an n-sided polygon. Furthermore, they tended to see Cody’s strategy as incorrect because he gets n triangles, not (n-2).
We spent 45 minutes sorting this out as a large group. I pushed them to explain to me why Tia will get (n-2) triangles. And at the end I had four ways of understanding this. Three of them were new to me.

(1) Sides

This is the way I have always thought about the formula. In this way of thinking, each side of the polygon corresponds to one triangle in the dissected polygon, as in the shaded triangle below. The remaining sides of the shaded triangle are formed by the diagonals that Tia draws.

Relating sides of the polygon to the number of triangles.

But there are two sides that do not get used this way. The sides adjacent to the vertex from which Tia draws her diagonals are incorporated into the first and last triangles. Therefore, we always get two triangles fewer than the number of sides of the polygon. So (n-2) refers to the number of sides of the polygon.

(2) Vertices I

The remaining strategies were new to me. I have fleshed out the details of the arguments the teachers made.
Tia draws diagonals from one vertex (call it A) in order to form her triangles. In doing so, she draws (n-3) diagonals. Those diagonals cut the angle at A into (n-2) parts. At the other end of each diagonal (at C, D and E in the diagram below) the angles of the polygon are cut into 2 parts. There are two angles that remain intact. In forming her triangles, Tia get (n-2)+2(n-3)+2 angles in all of the triangles combined. This expression simplifies to 3n-6. Each triangle has three angles so the polygon has been cut into (3n-6)/3=(n-2) triangles.

Counting vertices in Tia's triangles

In this argument, n stands for the number of vertices in the original polygon. Of course this is the same as the number of sides, but the argument focuses on the vertices rather than on the sides.

(3) Vertices II

A simpler way to use vertices is to notice that from vertex A we can draw (n-3) diagonals. This is because we draw a diagonal to each vertex except A and the two vertices adjacent to A. Just as cutting a cake 1 gives 2 pieces, cutting our polygon n-3 times gives one piece more: (n-3)+1=n-2. So (n-3) diagonals make (n-2) triangles.

(4) Connecting Cody’s strategy to Tia’s

The extra 360 degrees in Cody's diagram

These vertex arguments were interesting for me to consider. I had always thought about n in the formula as the number of sides. More profound for me was a teacher’s observation that Cody’s strategy can be connected to Tia’s. In this case, n refers to triangles, not to sides or vertices.

Cody’s strategy involves putting a new point in the interior of the polygon. Connecting that point to each of the vertices gives n triangles. But when we total the measures of the angles of these triangles, we are including the angles in the center of the polygon and these angles really have nothing to do with the angles of the polygon. So we need to subtract their combined measure from the total. These extra angles completely surround the center point so they total 360 degrees. Therefore Cody’s formula is n*180-360, which is equivalent to (n-2)*180.

A teacher in my session observed that this new point could be anywhere in the polygon. Furthermore, if we pull that point towards a vertex of the polygon, two of Cody’s triangles get smaller and smaller in area. When the point is pulled all the way to a vertex, those triangles collapse and disappear, and the extra 360 degrees around the point disappears at the same time. So Tia’s strategy is the same as Cody’s strategy except that she has chosen her point to coincide with a vertex of the polygon. (n-2) refers to n triangles that Cody would draw minus the collapsed triangles. Click here to view a brief animation demonstrating this strategy.

(5) A new strategy

Following this teacher’s logic, I wondered what would happen if we dragged the point to the side of the polygon instead of to a vertex. I turns out that only one of the triangles collapses, and the extra point now has 180 degrees surrounding it. This leads to a new formula: (n-1)*180-180. Click here to view a brief animation demonstrating this strategy.

Others?

Readers are invited to submit further ways of thinking about the number of degrees in the interior angles of a polygon. Don’t worry about the art-describe your thinking in detail and I can draw the picture for you!

Addendum 4/20/2011

I have written a Sophia packet on angle measures in polygons-considering the general case of the argument involving exterior angles of a polygon. If you’re looking for more information on this topic, the packet is worth checking out.

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