For a pentagon, we divide it into a quadrilateral and a triangle. Then, we find the center of gravity of each of these figure. Then, we draw a line between these two centers. Next, we divide the pentagon even into two different quadrilateral and triangle. We then find another two centers of gravity. Then draw a line between these two centers. The intersection point is the balancing point.The rationale for this conclusion is similiar to the one that we used to find the balancing point of quadrilateral in Assignment 11. For the hexagon, we also use this method to find its balancing point.

We can prove the construction of the balancing point of a n-gon using mathematical induction. An informal proof  somehow looks like this:

First, we check that the method works for k=4, which we've already proven in the quadrilateral case.
Next, assume that for this method works for any polygon with k sides where k is an integer and greater or equal to 4.
Now we must prove that it also works for any polygon with k + 1 sides.

If we can always manage to divide the polygon into a smaller polygon and a triangle, this method will always work for any polygon
with any number of sides. For example, a pentagon is divided into a quadrilateral and a triangle. We do this because we already know
how to find the centers of gravity of a quadrilateral and a triangle. Next, we use the same reasoning for the case of a polygon with
k+1 sides. We can divide this k+1 -gon into a k-sided polygon and a triangle. Since we assume that this method works for any polygon with
k number of sides and we already know how to find the center of gravity of a triangle, the method must work for any polygon with
k +1 sides. Thus, the proof is complete.