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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?

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    $\begingroup$ A square and a lozenge have the same side lengths but different areas. How can a single formula possibly apply to both of them? $\endgroup$
    – Trebor
    Commented Mar 4 at 6:17
  • $\begingroup$ Can't the area of a rhombus be found by multiplying the height and base lengths, just like the square? (Assuming the same length of all sides) $\endgroup$
    – Don't mail me
    Commented Mar 4 at 6:20
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    $\begingroup$ @Don'tmailme: "Can't the area of a rhombus be found by multiplying the height and base lengths [...]?" The "height" is not itself a side of the rhombus, and it is not determined by the sides of the rhombus; so, if all you know are the side-lengths, then you're out of luck for calculating area. ... The general problem here is that, unlike a triangle, a polygon with more than three sides is not "rigid". If you join four or more rods together with hinged ends, the result will flex; the enclosed area is not determined by side-lengths alone. $\endgroup$
    – Blue
    Commented Mar 4 at 6:31

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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

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  • $\begingroup$ That is very interesting, thanks for the answer. Then is there an established way to determine the area of a polygon from the side lengths and angles? $\endgroup$
    – Don't mail me
    Commented Mar 9 at 6:31

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