# 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?

• 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.

• 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 '14 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?

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).

• The Brahmagupta Formula is valid for cyclic quadrilaterals, but that is not an assumption, apparently. – rschwieb Oct 18 '12 at 16:19
• @rschwieb, one must make some assumption; as said in a comment above, «a quadrilateral is not uniquely defined by its side lengths». So +1. – JMCF125 Apr 3 '14 at 21:45
• Hello @JMCF125 : votes are used to help other readers gauge the value of solutions. This solution could be counted as misleading since it attempts to apply a theorem in a case where the hypotheses may not be satisfied. If this solution mentioned this condition is needed, then I would upvote it too. However it is currently not very useful. – rschwieb Apr 4 '14 at 2:20
• @rschwieb, PEV tried the same and got his answer accepted. I added the necessary 'convex' condition. – JMCF125 Apr 4 '14 at 14:18
• @JMCF125 Ok, but I don't know why you are telling me this. I was just explaining why the question probably got downvotes because it seemed you're not familiar with the way downvotes are used here. – rschwieb Apr 4 '14 at 14:52

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.