I tried to construct trapezoid with lengths of 4 sides given. Without success. Then searched in internet. I believe it is not possible, but I am not sure.
The only way is to calculate the high of the trapezoid (distance of two paralell sides).
Is my assuption correct or there is a way how to construct?
[Math] Is it possible to draw trapezoid by compass and straightedge if 4 sides given
geometry
Related Solutions
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.
$$a = b - 2 \,c \cos \alpha $$
where $ \alpha = $ angle between the big side and one of equal sides length $c$.
Best Answer
It is possible.
Let $a,b,c,d$ be the lengths of the trapezoid where $a$ and $c$ are the sides to be paralel. Without loss of generality let $a > c$.
Construct the segment $AB$ with length $a$.
Draw a circle centered in $A$ with radius $d$ and another one in $B$ with radius $b$. Call this circles $\pi_A$ and $\pi_B$.
Find the point $P$ in the segment $AB$ such that the distance $AP$ is $c$.
Find the point $Q$ in the segment $AB$ such that the distance $QB$ is $c$.
Draw a circle centered in $P$ with radius $d$. Call that circle $\pi_P$.
Draw a circle centered in $Q$ with radius $b$. Call that circle $\pi_Q$.
Now $\pi_B$ and $\pi_P$ intersect in two points. Call $C$ one of them. Now $\pi_A$ and $\pi_Q$ intersect in two points, one of wich is in the same semiplane as $C$ with respect to the segment $AB$. Call that point $D$.
Then $AB$ measures $a$ by construction, $BC$ measures $b$ by construction and $DA$ measures $d$ by construction.
It's not too hard to prove that $CD$ measures $c$ and is paralel to $AB$.