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|.