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.
Ok, here is one approach that works. Since diagonals are equal, and bases are parallel, it is easy to show that top and bottom triangle are similar, using alternate interior and vertical angles theorem. Let's call the diagonal intersection point S and vertices of the trapezoid A,B,C,D with A at bottom left vertex. Now BS is a multiple of DS (say factor k) and AS is a multiple of CS (same factor k) Now use area formula for triangle ½*b*c*sin(enclosed angle) on left and right triangle to show that these areas are equal. You will in both cases arrive at area ½*AS*DS*sin(angle) for left and right triangle of the trapezoid. KNowing that these triangles are congruent, and top and bottom triangles are similar, I think you can finish the proof
Best Answer
The area will be $\frac12\cdot 10\cdot (y+x+y+z)=5(x+2y+z)$
Now, $(y+z)^2+10^2=50^2$ and $(x+y)^2+10^2=30^2$
$(y+z)=\sqrt{50^2-10^2} CM=20\sqrt6 CM$
$(x+y)=\sqrt{30^2-10^2} CM=20\sqrt2 CM$
SO, the area will be $5(20\sqrt2(\sqrt3+1)) CM^2$