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?
 A: 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.
A: 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.
A: 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?
