Angle Sum Formula for Polygons
We are all familiar with the fact that the sum of all the angles of a triangle is 180 degrees. And for quadrilaterals the sum of its angles comes to be 360 degrees. This can be generalized to prove that the sum of angles of polygon of n sides is (n - 2) * 180 degrees.
A simple proof of this angle sum formula can be provided in two ways. First we can divide the polygon into (n - 2) triangles using (n - 3) diagonals and then the sum of the angles is clearly (n - 2) * 180 degrees.
Second approach is to take a point in the interior of the polygon and join this point with every vertex of the polygon. This gives us n triangles and so the sum of angles of all these triangles is n * 180 degrees. This also includes the angles made at the chosen point in the interior which add up to 360 degrees. Again this means that the sum of angles of a polygon is n * 180 - 360 = (n - 2) * 180 degrees.
Both the above approaches make use of the fact that the sum of angles of a triangle is 180 degrees. We will present here a third approach which does not use this fact but rather relies on the simpler fact that the angles in a linear pair add up to 180 degrees (this is more of an axiom, or the definition of degree measure).
The proof can be illustrated by the following diagram:
The idea here is to note that the sum of exterior angles add upto 360 degrees. This is very clear if we observe that if we rotate the ray AA' by the exterior angle at B we get a ray parallel to ray BB'. Similarly if BB' is rotated by the exterior angle at C then we get a ray parallel to ray CC'. Thus if the ray AA' is rotated by an angle which is sum of exterior angles at B and C then we reach parallel to ray CC'. Continuing this way we can see that if the ray AA' is rotated by an angle which is the sum of all exterior angles then we get back to the direction of ray AA' (full circle!). Thus the sum of exterior angles must be 360 degrees (this is the definition of degree measure, one complete revolution is 360 degrees).
Now if we add up the interior and exterior angles at a vertex we get a linear pair of angles. And so if we add all interior and exterior angles of a polygon we get n linear pairs of angles. So the total sum of the interior and exterior angles is n * 180 degrees. Subtracting the sum of exterior angles from it we see that the sum of interior angles of a polygon is n * 180 - 360 = (n - 2) * 180 degrees.
Note how smartly the proof avoids the use of the special case of sum of angles of a triangle. The idea of this proof came to me some months ago when I was looking at a problem in NCERT 8th standard text where one was supposed to measure all exterior angles of a quadrilateral and show that it is 360 degrees.
A simple proof of this angle sum formula can be provided in two ways. First we can divide the polygon into (n - 2) triangles using (n - 3) diagonals and then the sum of the angles is clearly (n - 2) * 180 degrees.
Second approach is to take a point in the interior of the polygon and join this point with every vertex of the polygon. This gives us n triangles and so the sum of angles of all these triangles is n * 180 degrees. This also includes the angles made at the chosen point in the interior which add up to 360 degrees. Again this means that the sum of angles of a polygon is n * 180 - 360 = (n - 2) * 180 degrees.
Both the above approaches make use of the fact that the sum of angles of a triangle is 180 degrees. We will present here a third approach which does not use this fact but rather relies on the simpler fact that the angles in a linear pair add up to 180 degrees (this is more of an axiom, or the definition of degree measure).
The proof can be illustrated by the following diagram:
The idea here is to note that the sum of exterior angles add upto 360 degrees. This is very clear if we observe that if we rotate the ray AA' by the exterior angle at B we get a ray parallel to ray BB'. Similarly if BB' is rotated by the exterior angle at C then we get a ray parallel to ray CC'. Thus if the ray AA' is rotated by an angle which is sum of exterior angles at B and C then we reach parallel to ray CC'. Continuing this way we can see that if the ray AA' is rotated by an angle which is the sum of all exterior angles then we get back to the direction of ray AA' (full circle!). Thus the sum of exterior angles must be 360 degrees (this is the definition of degree measure, one complete revolution is 360 degrees).Now if we add up the interior and exterior angles at a vertex we get a linear pair of angles. And so if we add all interior and exterior angles of a polygon we get n linear pairs of angles. So the total sum of the interior and exterior angles is n * 180 degrees. Subtracting the sum of exterior angles from it we see that the sum of interior angles of a polygon is n * 180 - 360 = (n - 2) * 180 degrees.
Note how smartly the proof avoids the use of the special case of sum of angles of a triangle. The idea of this proof came to me some months ago when I was looking at a problem in NCERT 8th standard text where one was supposed to measure all exterior angles of a quadrilateral and show that it is 360 degrees.
posted by Paramanand at 1:39 PM
2 Comments:
I think this approach is used to prove the sum of angles of a triangle in the first place - they just pass a line parallel to one of the sides through the opposite vertex for easier visualization. Once it is proven for triangle, and since all polygons can be reduced to triangles.
Or, you can just do induction. If sum of angles of n sided polygon is X then the sum of angles of a n + 1 sided polygon will be X + 180, because you can go from n to n+1 by adding a triangle.
The constant nature of the sum is the fun part.
That induction thing is great, but I believe not suitable for 7th graders.
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