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 1From 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
There are
altogether (n-1) triangles. Sum of angles
of each triangle = 180° Please note
that there is a straight angle Sum of interior
angles of n-sided polygon = (n-1) x 180°- 180° = (n-2) x 180° |
|
|||||||||||||||||||||||||||||||||||||||||
Method 3From 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, 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 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?
|
||||||||||||||||||||||||||||||||||||||||||