# Area of irregular polygon using side edges

I have only lengths for the sides of an irregular polygon, can anyone tell me how I can measure the area of the polygon? Remember only lengths of all the sides , no angles or coordinates.

Few forums mention about trangulation of the polygon etc But I only have side lengths.

Does anybody has any feedback?

• Unless the polygon is a 3-gon, more information is needed. – David Simmons Aug 20 '13 at 11:00

Imagine building the polygon by joining pieces of wood with hinges. Because the structure is in general not rigid, the answer to your question is indeterminate except if:

(a) One side is longer than the sum of all the other sides - in which case no polygon exists

(b) One side is equal to the sum of all the other sides, in which case the polygon is degenerate of area $0$

(c) Your polygon is a triangle - apply Heron's Formula

Here is a proof that the polygon has maximum area if all vertices lie on a circle. It also proves that for the maximum area, the order of the sides does not matter.

The area of a square with all sides $1$ is greater than the area of a parallelogram with all sides $1$, if the parallelogram is not a square.

Building on coffeemath`s answer i think i should mention this (because he uses the square which is regular polygon and you want to work with irregular ones):

It is true that for every $\varepsilon>0$ there exists parallelogram with side lengths all equal to $1$ such that its area is less than $\varepsilon$ so at least in the case $n=4$ we need more information then just the side lengths.

There is work by David Robbins, et al., on the area of cyclic polygons given the lengths of the sides.