# Finding the area of a quadrilateral

I have a quadrilateral whose four sides are given

2341276, 34374833, 18278172, 17267343 units.

How can I find out its area? What would be the area?

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Unlike a triangle, a quadrilateral is not uniquely defined by its side lengths. For example, if all sides are of unit length, it could be anything from a square with unit area to a very skinny rhombus with area close to zero. You need more information to determine the area, such as the angle between the diagonals, $\theta$ in PEV's answer. –  Rahul Jan 25 '11 at 14:16

Have you seen this (MathWorld article)? In particular, the area of a planar convex quadrilateral is given by $\frac{1}{4}(b^2+d^2-a^2-c^2) \tan \theta$ where $a,b,c$ and $d$ are the side lengths.

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Isn't that only for convex quadrilaterals? –  Aryabhata Jan 25 '11 at 18:52
@Aryabhata, indeed, it seems so, by the referred article «There are a number of beautiful formulas for the area of a planar convex quadrilateral in terms of the side and diagonal lengths, including [...] $\frac{1}{4}(b^2+d^2-a^2-c^2) \tan \theta$» (my emphasis). I have suggested an edit. –  JMCF125 Apr 3 at 18:29

I can't comment (not enough reputation), but as this sort of a partial answer anyway, I'll just post it as an answer. Rahula Narain's comment is correct that the area depends on more than just the sidelengths. But assuming the sidelengths are listed in the question in order around the polygon as usual (in fact, assuming any particular order on the sidelengths), the polygon is pretty constricted in terms of its shape (note that the first side is very small compared to the others, and the last two sum to about the same as the second). So it should be possible to get bounds on the area. (Even if we don't assume an order on the sidelengths, there are only three possibilities modulo things that don't affect area. And because one side is so small and one so large, the three areas should all have pretty similar bounds.) Any takers?

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If unlike sidelengths actually restrict formation any other quad then our given quad becomes unique and even geometrically we cannot change theta values i.e angles for the given quad.

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If it is a cyclic quadrilateral (i.e. all of its vertices can be contained in one circle) , you can use the Brahmagupta Formula, giving $\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $a$, $b$, $c$ and $d$ are the lengths of the four sides of the quadrilateral, and $s =\dfrac{a + b + c + d}2$ (the semi-perimeter).