Find the length $x$ in $cm$ in the quadrilateral $ABCD$. This problem appeared in a remedial math class from my college last week. Someone I know, who attended the class, sent me this problem claiming that it was the toughest problem on the test. I'm sharing this problem here to see if there are any solutions that can be simpler than mine, which I'll also post.
Edit: While I'm aware of a similar question already having been asked. The issue is that it is a general case, unlike this version of the problem, and thus lacks a Euclidean geometrical approach as I have demonstrated in my answer.

 A: This is my approach. I'll add an explanation below.

Here's how I go about it:
1.) Label the quadrilateral as $ABCD$ and mark all the points. Drop a perpendicular $BE$ from point $B$ onto segment $AD$. It's easy to see that $\triangle AEB$ is a "$30-60-90$ triangle". Therefore, segment $AE$ is half of segment $AB$, therefore $AE=4cm$.
2.) This implies that the remaining segment $ED=x-4$. Drop another perpendicular from point $C$ onto segment $BE$ at point $F$. Notice that since $\angle CBF=60$, $\triangle CBF$ is also a "$30-60-90$ triangle". This implies that segment $CF$ is half of segment $BC$ times $\sqrt{3}$. Therefore $CF=3cm$. Since $DEFC$ is a rectangle, we can conclude that $x-4=3$, therefore $x=7cm$.
A: 
$$BD=\sqrt{BC^2+CD^2}=\sqrt{8^2+(2\sqrt3)^2}=\sqrt{76}=2\sqrt{19}$$
$$\angle DBC=\arctan\frac{\sqrt3}{4}$$$$\implies \angle ABD=60^\circ-\arctan\frac{\sqrt3}{4}$$ so $$\frac{x}{2\sqrt{19}}=\cos\angle ABD=\cos \left(60^\circ-\arctan\frac{\sqrt3}{4}\right)$$$$=\dfrac12\cdot\frac{4}{\sqrt{19}}+\frac{\sqrt3}{2}\cdot \frac{\sqrt3}{\sqrt{19}}=\frac{7}{2\sqrt{19}}$$ so $x=7$ cm.
