Sum of Interior angles of an n-sided polygon

 

There are many methods to find the sum of the interior angles of an n-sided convex polygon. Most books discuss only one or two ways.

 

 

Method 1

From any one of the vertices, say A1, construct diagonals to other vertices.

 

There are altogether (n-2) triangles.

 

Sum of angles of each triangle = 180°

 

Sum of interior angles of n-sided polygon

= (n-2) x 180°

 

 


 


Method 2

 

From any point P on the line segment, say A1 A2, construct lines to the vertices A3, A4, …, An.

 

There are altogether (n-1) triangles.

 

Sum of angles of each triangle = 180°

 

Please note that there is a straight angle
A1PA2 = 180°  containing angles which
are not interior angles of the given polygon.

 

Sum of interior angles of n-sided polygon

= (n-1) x 180°-  180°  = (n-2) x 180°

 

 

 

 

 

 


Method 3

From any one point P inside the polygon,

construct lines to the vertices.

 

There are altogether n triangles.

 

Sum of angles of each triangle = 180°

 

Please note that there is an angle at a point = 360° around P containing angles which are not interior angles of the given polygon.

 

Sum of interior angles of n-sided polygon

= n x 180°- 360°  = (n-2) x 180°

 

 

 

 


Method 4

 

The point P chosen may not be on the vertex, side or inside the polygon.

It can even be a point outside the polygon.

There are altogether (n-1) triangles.

 

Sum of angles of each triangle = 180°

 

Please note that the angles in triangle PA1A2 = 180° are not interior angles of the given polygon.

 

Sum of interior angles of n-sided polygon

= (n-1) x 180°- 180°  = (n-2) x 180°

 

 

 

 


Method 5

 

Before we carry on with our proof,
let us mention that the sum of the exterior angles of an n-sided convex polygon  = 360°

 

I would like to call this the Spider Theorem.

 

Imagine you are a spider and you are now in the point A1 and facing A2.

 

 


You crawl to A2 and turn an exterior angle, shown in red, and face A3.

You then crawl from A2 to A3 and turn another exterior angle and face A4.

 

You carry on with the journey and turn all exterior angles.

Lastly you come back to point A1 and face A2 again.

 

Aha! You have turn a complete circle, which is  360°.

And the intelligent spider has proved that the sum of the exterior angles of an n-sided convex polygon  = 360°

 

Now, let us come back to our interior angles theorem.

 

You can see that, by considering the red and blue angles in the diagram, the sum of any one of the interior angle and the adjacent exterior angle is 180°. (adjacent angle on straight line)

 

There are n sides in the polygon and therefore n straight angles.

 

Sum of interior angles + sum of exterior angles = n x 180°

Sum of interior angles + 360°= n x 180°

Sum of interior angles = n x 180°- 360°  = (n-2) x 180°

 

 

Method 6

 

This method needs some knowledge of difference equation. It is a bit difficult but I think you are smart enough to master it.

 

Let  xn be the sum of interior angles
of a n-sided polygon.

 

So you may say that xn-1 is the sum of interior angles of an (n-1)-sided polygon.

 

As in the diagram, if you cut away one vertex, say A1, of an n-sided polygon you can get an (n-1) sided polygon, A2A3A4…An.

 

The angle sum of the triangle A1A­2An= 180°

 

 

 

 

 

 

So, you get the difference equation:          

Similarly,                                                 

                                                                                           

                                                                                                       

                                                                      ………                                                    

                                                                     

Lastly, we get the angle sum of triangle                  

Adding up all the (n-2) equalities, and canceling all the terms, we get

           

 

 

Insight      Wow! What theorem can you see from the drawing?