A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?

$\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}$
Let one of the mats be $ABCD$, and the center be $O$ as shown: 

Since there are $6$ mats, $\Delta BOC$ is [equilateral](https://artofproblemsolving.com/wiki/index.php/Equilateral) (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$. 
By the [Law of Cosines](https://artofproblemsolving.com/wiki/index.php/Law_of_Cosines): $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$. 
Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$.