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
The formula for the distance from the base $b$ to the center of mass of a trapezoid is
$$\bar y = \frac{b+2a}{3(a+b)} h.$$
You can find this in many on-line sources such as Wolfram Mathworld.
You can prove this by integrating $\int_0^h y(b + (a-b)\frac yh) dy$ and dividing by the area of the trapezoid. But here's a derivation without calculus, using the fact that the distance from a side of a triangle to the triangle's centroid is $\frac13$ the height of the triangle.
Let $T$ be a trapezoid with bases $a$ and $b$. For the case $a < b$, extend the non-parallel sides of the trapezoid until they intersect. The base $b$ of the trapezoid and the two extended sides form a triangle $B$; the base $a$ divides this triangle into two pieces, one of which is $T$ and the other of which is a triangle which we'll call $A$.
If the height of the trapezoid $T$ is $h$, the height of $B$ is $\frac{b}{b-a}h.$ The centroid of $B$ is at a distance $\frac{b}{3(b-a)}h$ from base $b$.
But another way to find the centroid of $B$ is to balance the two figures $A$ and $T$ that compose $B$. The triangle $A$ has height $\frac{a}{b-a}h$, so its area is $\frac{a^2}{2(b-a)}h$ and its centroid is a distance $\frac{a}{3(b-a)}h + h$ from the base $b$ of the trapezoid. Trapezoid $T$ has area $\frac{a+b}{2} h$ and a centroid at the unknown distance $\bar y$ from base $b$.
To "balance" the two regions, we take a weighted average of the distance of their centroids from base $b$. The "weights" in this average are just the areas of the two regions. This weighted average is the same as the distance of the centroid of $B$ (the combined figure) from base $b$. That is,
$$\begin{eqnarray} \frac{b}{3(b-a)}h &=&\frac{\mathop{Area}(A) \cdot \left(\frac{a}{3(b-a)}h + h\right) + \mathop{Area}(T) \cdot \bar y}{\mathop{Area}(A) + \mathop{Area}(T)}\\ &=&\frac{\frac{a^2}{2(b-a)}h \cdot \left(\frac{a}{3(b-a)} + 1\right)h + \frac{a+b}{2} h \cdot \bar y}{\frac{a^2}{2(b-a)}h + \frac{a+b}{2} h}\\ \end{eqnarray}$$
Solve for $\bar y$. This looks messy, but it can be simplified if you realize that $$\frac{\mathop{Area}(T)}{\mathop{Area}(A)} = \frac{b^2 - a^2}{a^2}.$$ If you divide both the numerator and denominator on the right-hand side of the weighted average by $\mathop{Area}(A)$, you get
$$ \frac{b}{3(b-a)}h = \frac{ \frac{3b - 2a}{3(b-a)} h + \frac{b^2 - a^2}{a^2} \cdot \bar y}{1 + \frac{b^2 - a^2}{a^2}} $$
After you finish collecting all the terms in $a$, $b$, and $h$ on the left side of this equation, and factor $(b-a)^2$ out of $b^3 - 3ba + 2a^3$, it all simplifies to the formula for $\bar y$ shown above.