Rahul,
I am going to assume that you are considering regular polygons. That is a polygon which has all it sides of the same length and all the interior angles congruent. I am also going to assume that you know the side length s, and the number of sides n. Here is a piece of one such polygon.
C is the centre of the polygon, that is the point that is equidistant from each of the vertices. Join each vertex to C to partition the polygon into n congruent triangles. CPQ is one such triangle and hence
the polygon area is n times the area of triangle CPQ.
Since these triangles are congruent
the measure of the angle QCP is ^{360o}/_{n}
Let R be the midpoint of PQ then CRQ is a right angle and the measure of angle QCR is ^{360o}/_{2n} and the length of RQ is ^{s}/_{2} . Hence
^{RQ}/_{CR} = tan( ^{360o}/_{2n} )
Since RQ = ^{s}/_{2} you can find CR and the area of the triangle CPQ is ^{1}/_{2} s CR.
Penny
