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.
When you divides the trapezoid in two triangles, the area of the trapezoid is the sum of the aeras of the triangles.
The triangle $DBC$ has an area of $\frac{a*h}{2}$ ($\frac{\text{basis }*\text{ height}}{2}$) and the triangle $ABD$ an area of $\frac{b*h}{2}$, hence the area of the trapezoid : $\frac{a*h}{2} + \frac{b*h}{2} = \frac{(a+b)*h}{2}$
Best Answer
So that we will be looking at the same picture, let the vertices of the trapezoid be $A,B,C,D$, in counterclockwise order, with $AB$ and $CD$ the parallel sides, and $AB$ horizontal at the bottom. Let diagonals $AC$ and $BD$ meet at $X$.
Let the area of $\triangle ABX$ be $s^2$ and let the area of $\triangle CXD$ be $t^2$. Then corresponding sides in these two similar triangles are in the ratio $s$ to $t$. This is because if we scale linear dimensions by the factor $\rho$, areas get scaled by the factor $\rho^2$.
By similarity, we have $\frac{AX}{XC}=\frac{AB}{CD}=\frac{s}{t}$. So the area of $\triangle AXD$ is $\frac{s}{t}$ times the area of $\triangle XCD$. That's because these two triangles have the same height with respect to their two bases $AX$ and $XC$.
Thus $\triangle AXD$ has area $st$. Similarly, $\triangle BXC$ has area $st$, so the trapezoid has area $s^2+t^2+2st$.