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How to find the area of a quadrilateral in which the length of all the sides are different. For example one pair of opposite sides are 630 foot and 357 foot, and another pair of opposite sides are 587 foot and 358 foot. Furthermore, no measure of any angle is given. The figure is roughly drawn as follows. enter image description here

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    $\begingroup$ @AlanAbraham makes a geat point. Measure one of the diagonals or one of the angles, then add it to the drawing, and the area can be calculated. $\endgroup$
    – Jim Clark
    Sep 5 at 15:34
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Quadrilaterals are, in general, not determined by solely their side lengths. If you are given a quadrilateral and you only know the side lengths, there are infinite possible areas it can have.

The only case where you could determine the area is if the quadrilateral was degenerate. That is, if the sides are $a,b,c,d$ with WLOG $\text{max}(a,b,c,d)=a$, then we have iff $a=b+c+d$, the area of the quadrilateral is $0$.

The question can be solved, however, if there is more information provided. For example, an angle measure or diagonal length and it's location wrt the sides.

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There are two particular cases where the area of a quadrilateral is determined solely by its sides $a,b,c,d$:

  • when it possesses a circumscribed circle. In this case, the quadrilateral is called cyclic and we have to use Brahmagupta formula:

$$A={\frac{1}{4}}{\sqrt {(-a+b+c+d)(a-b+c+d)(a+b-c+d)(a+b+c-d)}}.$$

giving the following area...

$$A \approx 217,144 ft^2$$

This area is achievable (see quadrilateral ABCD on fig. 1)

The formula for the area of a tangential quadrilateral is:

$$\displaystyle A=\sqrt{(e+f+g+h)(efg+fgh+ghe+hef)}.$$

with $$\begin{cases}e&+&f&&&&&=&a\\ &&f&+&g&&&=&b\\ &&&&g&+&h&=&c\\ e&+&&&&&h&=&d\end{cases}$$

giving the condition $a+c=b+d$ which isn't fullfilled (the sums of opposite sidelengths aren't equal)

Remark: the area quadrilateral ABCD can have is $\approx 218,700$ (obtained experimentaly).

In order to understand how, with the same lengths, the area of quadrilateral $ABCD$ can vary, have a look at the following figure, where AB (length 630) is fixed. .

enter image description here

Fig. 1. For a fixed line segment $AB$, vertices $C$ and $D$ belong to circular arcs with $CD=357$. A particular case occurs when $B,C,D$ are aligned. Two cases are highlighted : the case where $ABCD$ is cyclic and a non convex case.

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  • $\begingroup$ Also, when a quadrilateral is cyclic you can use Brahmagupta's formula. But I don't think you can check if a quadrilateral is cyclic with only the side lengths. $\endgroup$ Sep 5 at 16:21
  • $\begingroup$ @TheBestMagician I just had the same idea ! I think you are right: there must be a restrictive condition (that should be fulfilled in this case) for a quadrilateral (given by its sidelengths) to be cyclic. $\endgroup$
    – Jean Marie
    Sep 5 at 16:27
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If you imagine your quadrilateral as built from rods of the prescribed lengths joined at the corners in a way that does not constrain the angles and fix one side as a "hoirizontal base" you can see how the other two vertices can move to change the area.

There are pictures in the answer from @JeanMarie.

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