Video solution:

Click to load, then click again to play

Written solution:

Draw a line through $A$ parallel to line $CD,$ and observe that some lengths are automatically equal to each other:

That is because $\angle AEB = \angle DCB = 60^\circ,$ and so triangle $ABE$ is equilateral. All lengths labeled $x$ are always equal.

Also, $ADCE$ is a parallelogram (not necessarily a rhombus), so the remaining lengths labeled $y$ are always equal to each other, but not necessarily equal to $x.$

The perimeter is $3x + 2y,$ but it is also supposed to be $30.$

The problem then amounts to counting how many positive integer solutions there are to the equation $3x + 2y = 30.$

As we consider positive integers for $x,$ observe that it is precisely the even integers which give an integer solution for $y,$ because $30 - 3x$ is the even number $2y.$

And, once we get to $x = 10,$ the $y$ that satisfies the equation is $y = \frac{30 - 3 \times 10}{2} = 0,$ which is not a positive integer.

So, the values that work for $x$ are the positive even integers less than $10,$ or $\{2, 4, 6, 8\}.$ There are $4$ options, which gives the answer of E.