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?
 A: First Solution: Let our trapezoid be $ABCD$ as in the diagram supplied by pedja.  Let the diagonals meet at $O$.
Note that $\triangle OAB$ and $\triangle OCD$ are similar.  Indeed we know the scaling factor.  Since $AB=20$ and $CD=7$, the sides of $\triangle OCD$ are $\frac{7}{20}$ times the corresponding sides of $\triangle OAB$.
That is very useful. We have $AC=13=AO+\frac{7}{20}AO$. It follows that
$$AO=\frac{(20)(13)}{27}, \quad\text{and similarly,}\quad BO=\frac{(20)(5\sqrt{10})}{27}.$$
If we want to use the usual formula for the area of a trapezoid, all we need is the height of the trapezoid.  That is $1+\frac{7}{20}$ times the height of $\triangle OAB$.  
The height of $\triangle OAB$ can  be found in various ways. For example, we can use the Heron Formula to find the area of $\triangle OAB$, since we know all three sides.  Or else we can use trigonometry. The Cosine Law can be used to compute the cosine of $\angle OAB$. Then we can find an exact (or approximate) expression for the sine of that angle. From this we can find the height of $\triangle OAB$.   
Second Solution: This is a variant of the first solution that uses somewhat more geometry.  Let $\alpha$ be the area of $\triangle OAB$. 
We first compute the area of $\triangle COB$. Triangles $OAB$ and $COB$ can be viewed as having bases $OA$ and $CO$ respectively, and the same height. But the ratio of $CO$ to $OA$ is $\frac{7}{20}$, so the area of $\triangle COB$ is $\frac{7}{20}\alpha$.  
Since triangles $ABC$ and $ABD$ have the same area, by subtraction so do $\triangle COB$ and $\triangle DOA$. And since $\triangle OCD$ is $\triangle OAB$ scaled by the linear factor $\frac{7}{20}$, the area of $\triangle OCD$ is $\left(\frac{7}{20}\right)^2\alpha$.  Putting things together, we find that the area of our trapezoid is
$$\alpha +2\frac{7}{20}\alpha +\left(\frac{7}{20}\right)^2\alpha,\quad\text{that is,}\quad \left(\frac{27}{20}\right)^2\alpha.$$
Pretty! Finally, by the similarity argument of the first solution, we know the sides of $\triangle OAB$, so we can find $\alpha$ by using Heron's Formula. 
A: 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 :
$A=\frac{a+b}{2} \cdot h$

A: Here is a very short solution: Consider a point B' on the line (A,B) to the right of point B such that |BB'|=b. Then consider the triangle ACB'. It is easy to see that the area of triangle ACB' is equal to the area of our trapezoid ABCD since the area of triangle ACD is equal to the area of trianlge BCB'. For the triangle ACB' it is easy to find the area from the Heron formula since its three sides are known: 27, 13, and 5$\sqrt{10}$. Plugging in these numbers we find that the area is 67.5.
