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How do you find the area of a convex quadrilateral $ABCD$, given only the length of its sides $a$, $b$, $c$ and $d$. If the length of a single diagonal is given, I could easily find it's area by dividing it into two triangles and applying Heron's formula to each.

I encountered this problem while trying to find the area of a patch of land. Sometimes, the area maybe approximated by a rectangle or a trapezium or some other simple figure, but in general opposite sides are not parallel. All the formulas I have so far seen includes knowing at least one angle or a diagonal.

So is there a formula (even a complicated one maybe) that gives the area of a quadrilateral given only its side lengths?

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Quadrilaterals are not rigid: you can fix a side and move one of the other vertices freely while respecting all side lengths. The area changes when you move the vertex.

enter image description here

(image from http://www.mathsisfun.com/definitions/rigid.html)

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  • $\begingroup$ See also math.stackexchange.com/a/832795/589. $\endgroup$ – lhf Sep 2 '14 at 11:55
  • $\begingroup$ I see. But if a diagonal is given then it'll be rigid with a unique area right? $\endgroup$ – Nabigh Sep 2 '14 at 12:04
  • $\begingroup$ @Nabigh, yes: if a diagonal is given, then you'll have two triangles, and triangles are rigid. $\endgroup$ – lhf Sep 2 '14 at 12:05
  • $\begingroup$ Thanks @lhf, that settles the problem. $\endgroup$ – Nabigh Sep 2 '14 at 12:16
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Here you see that the area is not determined. The animation uses parallelograms with side lengths $a=c=3$, $b=d=2$, so the area varies between $0$ and $6$. In general the minimum area may not be exactly zero as here, but the same thing happens always.

enter image description here

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