smc

Draw a perpendicular through the midpoint of the line of length $12$ such that it passes through a vertex. We now have created $2$ $30-60-90$ triangles. Using the ratios, we get that the hypotenuse is $6\times \frac{2}{\sqrt{3}}$ $=\frac{12}{\sqrt{3}}$ $=4\sqrt{3}\to \boxed{4\sqrt{3}}$