area of a convex quadrilateral

Question

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

I don not have any other information about the quadrilateral apart from this. Is it possible to calculate it's area. Please help asap !

Answer 1

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.

Answer 2

A quadrilateral with sides 30,20,30,15? two sides are equal right? why dont 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.have equal angles can you find the rest of the angles and the area?

Answer 3

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.