<s>[INST] 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?
 [/INST] 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. 
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