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How many non-congruent quadrilaterals are there if we specify the sides to be a, b, c, d and specify its area A?

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Are you given the order of the sides? eg. do we know that a and c are opposite each other? – Wonder Aug 6 '12 at 15:17
Let's assume that for simplicity. – Plum Aug 6 '12 at 15:25

If we know the ordering of the sides (suppose sides are a, b, c and d in that order), we can note that the quadrilateral is fixed once we fix the length of a diagonal, say x. use Heron's formula to get $4A = \sqrt{(a^2+b^2+x^2)^2-2(a^4+b^4+x^4)} + \sqrt{(c^2+d^2+x^2)^2-2(c^4+d^4+x^4)}$. A, a, b, c and d being given, solve for x.

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The answer is two in general (assuming the lengths are given cyclically). As for quadratic equations, there are special cases where the two coincide and, as in the SSS case in triangle construction, restrictions on the data are required, otherwise there are no solutions. This can be seen by assuming wlog that the vertices are $( 0, 0)$ , $ (1,0) $ , $(p,q)$ $ (r,s)$ . The conditions then reduce to a system of four quadratic equations which Mathematica can solve. The solution goes over 18 pages but only involves sums, products and square roots. For concrete values one gets simple expressions.

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