Prove that sum of interior angle of a polygon of n sides is (n-2)*Pi.?

Question

it is mathematics question .
Pi is a constant and its approxmiate value is 3.14

Answer 1

join all the vertices of the polygon to a point in its interior.
let each center angle be ai for all i from 1 to n
then, in each triangle center angle ai= pi-(angle bi+angle ci) for all i from 1 to n
adding all ai's we'll get the angle made by that point in the interior on the l.h.s. & on the r.h.s. we'll have n times pi, - (all angles of a polygon)
DRAW A FIGURE SIMULTANEOUSLY TO UNDERSTAND THIS
i.e. 2pi=n*pi-all angles
so all angles=(n-2)pi

Answer 2

Choose any point in the interior of the polygon.

Join all n vertices to that point. This creates n triangles, each with two of those segments and one edge of the polygon defining it. Each triangle has pi raidians total in interior angles. For the n triangles, that is n*pi total. The interior angles at the point are a complete circle of 2*pi radians. When they are removed, we have the measure of the interior angles of the polygon left.

n*pi - 2*pi = (n-2)*pi.

Answer 3

The way you prove it is to extend one edge and note that the angle it makes with the next edge is 2π/n so that the interior angle must be π - 2π/n which is
π*(1 - 2/n). Since there are n such angles, they must have a total of
n*π*(1 - 2/n) = π*n*(1 - 2/n) = π*(n - 2) Q.E.D.

Doug