Area of a trapezoid from given the two bases and diagonals Find the area of trapezoid with bases $7$ cm and $20$ cm and diagonals $13$ cm and $5\sqrt{10} $ cm.
My approach:
Assuming that the bases of the trapezoid are the parallel sides, the solution I can think of is a bit ugly,


*

*Find the  other two non-parallel sides of the trapezoid by using this formula.

*Find the height using this $$ h= \frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}{2(b-a)}$$


Now, we can use $\frac12 \times$ sum of the parallel sides $\times$ height.
But, this is really messy and I am not sure if this is correct or feasible without electronic aid, so I was just wondering how else we could solve this problem?

Let's denote $a=20$ , $b=7$ ,$d_1=13$ , $d_2=5 \sqrt{10}$ , (see picture below) 
You should solve following system of equations :
$\begin{cases}
 d_1^2-(b+x)^2=d_2^2-(b+y)^2 \
 a-b=x+y
\end{cases}$
After you find values of $x$ and $y$ calculate $h$ from one of the following equations :
$h^2=d_2^2-(b+y)^2$ , or
$h^2= d_1^2-(b+x)^2$ 
Then calculate area :
$A=\frac{a+b}{2} \cdot h$

