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.
Hint: for any trapezoid, show there is a rectangle of equal area and smaller perimeter. Therefore, by stretching, there is a rectangle of equal perimeter but greater area.
Best Answer
This problem is more subtle than some of the other answers here let on. A great deal hinges on whether "trapezoid" is defined inclusively (i.e. as a quadrilateral with at least one pair of parallel sides) or exclusively (i.e. as a quadrilateral with exactly one pair of parallel sides). The former definition is widely considered more mathematically sophisticated, but the latter definition is more traditional, is still extensively used in K-12 education in the United States, and has some advantages.
As the other responses have pointed out, if one defines "trapezoid" inclusively, then any parallelogram is automatically a trapezoid, and as the side-lengths of a parallelogram do not determine its area, it is not possible (even conceptually) that there could be a formula for the area of a trapezoid in terms of its side lengths.
However, if "trapezoid" is defined exclusively, then things are quite different. Consider a trapezoid with parallel bases of length $a$ and $b$ with $b>a$. Let $\theta$ and $\phi$ respectively denote the angles formed by the legs $c$ and $d$ with the base $b$. Then we have the following relationships: $$c\cos\theta + d\cos\phi = b-a$$ $$c\sin\theta = d\sin\phi$$ These conditions uniquely determine $\theta$ and $\phi$, and therefore among non-parallelogram trapezoids, choosing the lengths of the parallel sides and the lengths of the bases uniquely determines the figure. In particular we would have $$\cos\theta = \frac{(b-a)^2+c^2-d^2}{2c(b-a)}$$.
The height of the trapezoid would then be $h=c\sin\theta$ (or if you prefer $h=d\sin\phi$, which is equal to it), so the area of the trapezoid can (in principal) be computed. If you really want to carry it out, you would have
$$\sin\theta = \sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$ so the area would be $$A=\frac{a+b}{2}c\sqrt{1-\left( \frac{(b-a)^2+c^2-d^2}{2c(b-a)} \right)^2}$$ I am not sure if there is a simpler expression, however.