We will use octagons to demonstrate how to measure the interior angles of a polygon with different formulas. These formulas are:
180*n - 360
(n-2)*180
((n-1)*180)-180
All of these formulas include dividing the octagon (or polygon) into triangles. When using regular concave polygons, all of these formulas/applications work. When looking at irregular, more specifically, convex polygons our class ran into some problems. We will explore the options to make these formulas apply to all polygons.
By placing a point in the center of the octagon, then connecting that point to the vertices, we create 8 different triangles. The sum of all of these triangles does not reflect the sum of interior angles. To get the sum of ONLY interior angles we must subtract the 360 degrees from the center. Therefor giving us the equation:
(n*180)-360
N = the number of sides in the polygon you are working with.
This same equation is reflected in this octagon as well:
Here we get the 360 degrees from the exterior angles sum, instead of the sum in the center of the octagon.
The next equation we can use is:
(n-2)*180
This can be demonstrated in this octagon:
Here, we pick any vertex on the octagon (or whatever polygon you are working with), then connect this vertex to every other vertex on the polygon. This gives us 2 less triangles then the number of sides in the polygon, thus (n-2). We then multiply this by 180 to get the sum of interior angles.
The last equation we can use on ALL regular polygons is:
((n-1)*180)-180
In this octagon, we add one more vertex. Then we connect this new vertex to the existing ones in the octagon. This creates one less triangle compared to number of sides in the octagon, thus (n-1). Because we added a vertex, we also added an additional 180 degrees. That is why we must subtract 180 from (n-1)*180.
We know that the above formulas work for all regular polygons. But what if we wanted to use them on an irregular, and more specifically, a convex polygon? Which formulas would work? And what methods would we need to modify?
First lets look at two irregular octagons and try to apply the first method of the first formula to them:
The first method works on the first octagon, only because the center point is able to reach all of the vertexes within the polygon. This does not work with the second octagon however, because point R cannot have a line segment within the octagon to each vertex.
Lets see if the other method to this formula will work:
We can see that this method does not work either. The angles that are pointed inward are larger than 180 degrees, so we cannot use these methods for an irregular or convex polygon.
Let's see if formula two will work: (n-2)*180
So, this formula works for our first octagon, but just barely. We can see that point G breaks up the octagon into 6 different triangles. If we would have used point D, this would not have worked. Our second octagon did not work at all. Even if we would have switched the vertex we started from, there is no way all the line segments would fit into the polygon.
I still think this formula can work in a different way. What if we connected all of the vertexes, but not from the same point? Could this work for the second octagon?
Here we do not use the same vertex, but each vertex has at least one line connecting to another vertex. We still get 6 triangles when we do this, so this MODIFIED version of (n-2)*180 WORKS FOR ALL POLYGONS.
Let's look at the last formula, ((n-1)*180)-180.
Just like in the first part of formula two, adding another vertex works for only the first octagon.
