# area of a convex quadrilateral

I have a quadrilateral with sides as follows: $30, 20, 30, 15.$

I do not have any other information about the quadrilateral apart from this. Is it possible to calculate its area?

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I think you need more information. Like one angle perhaps. With one angle, you can find 1 diagonal and then use Heron's formula. – Daryl Oct 1 '12 at 6:28
but i do not have any other information apart from this.. is it not possible to find out the diagonal length from this?? – vipin8169 Oct 1 '12 at 6:30
The problem is that without an angle, I don't think the quadrilateral is even unique (it can change shape). – Daryl Oct 1 '12 at 6:33
"We cannot draw a quadrilateral uniquely with 4 elements. We need at least 5 elements to draw unique quadrilaterals." (freeganita.com/en/geo/6_7.htm) – Daryl Oct 1 '12 at 6:36
that means i am screwed.. – vipin8169 Oct 1 '12 at 6:38

Here are two quadrilaterals with the specified sides:

The areas are 261 for the brown quadrilateral, while the blue quadrilateral at 522 is twice as big. And there are many other possibilities.

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The brown one isn't convex, but illustrates my point about non-uniqueness. – Daryl Oct 1 '12 at 20:13
Fair enough - I will change it – Henry Oct 1 '12 at 20:47

Let $a,b,c,d$ be the four sides of the quadrilater, and let $p= \frac{a+b+c+d}{2}$. Then the area $S$ is given by

$$S^2=(p-a)(p-b)(p-c)(p-d)-abcd \cos^2(\frac{A+C}{2})$$

So, the four sides together with the sum of the angles $A,C$ uniquely determine the area.

As it was pointed before, the four sides cannot determine the area. To understand this, here is another simple approach:

Let $d$ be the diagonal of the quadrilateral which makes a triangle with the sides $30,20$.

Since $30,20,d$ are the sides of a triangle, we must have

$$30-20 < d < 30+20 \,.$$

Similarly, since $d$ also makes a triangle with $30,15$, you get

$$15<d<45 \,.$$

Thus, combining we have

$$15< d <45 \,.$$

Now pick any such $d$. You can build a triangle with sides $30,20, d$ and you can build a triangle with sides $30,15,d$. Glue them together along $d$ and you get a quadrilateral.

We get such a quadrilateral for each value of $d \in (15, 45)$, and it is easy to see that increasing the value of $d$ increases the opposite angle in the $30,20, d$ and $30,15,d$ triangles. Thus increasing $d$ doesn't change the $a,b,c,d$ but it changes the value of $\frac{A+C}{2}$, and hence the area.

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A quadrilateral with sides $30,20,30,15?$ two sides are equal, right? Why don't you try to draw it? Divide it into two triangles. If the two equal sides have a common edge, one of the triangles is isosceles, i.e. has equal angles. Can you find the rest of the angles and the area?

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