I am aware of the formula for the area of a regular polygon: $A=([Side Count] \times [Side Length] \times [Apothem Length])/2$
However, I could not find an equation for the area of a non-regular polygon form the list of its (different) side lengths like the equation for the area of any triangle: $A=\sqrt{Semiperimeter \times [Semiperimeter-Side1] \times [Semiperimeter-Side2] \times [Semiperimeter-Side3]}$
Can either of these equations be appropriated to find the area of a polygon using the lengths of its sides? Is there another equation?
Putting together both the original question with its mentioned solution for general triangles and the argument about flexibility of (general) polygons with more than 3 sides, it might become interesting however to restate a similar quest, which now includes not only the various side lengths only, but rather all the distances between all vertex pairs.
And this reformulation then clearly can be solved as desired, at least if the polygon would be assumed to be convex (and even within several re-entrant cases too):
The set of all those vertex pair sides or sectors clearly contains as subset a triangulation of that polygon. By means of the already mentioned formula the area of each of those triangles can be calculated. The remainder then would be just to add all these triangular area's numbers.
--- rk