jmc

We draw a regular hexagon $ABCDEF$ as shown below. The desired distance can be measured by diagonal $AC$: To compute the length of $AC$, we let $H$ be the foot of the perpendicular from $B$ to $AC$: Since the hexagon is regular, $\angle ABC=120^{\circ }$ and $\angle ABH=\angle CBH=120^{\circ }/2=60^{\circ }$. Thus, $△ABH$ and $△CBH$ are congruent $30^{\circ }-60^{\circ }-90^{\circ }$ triangles. These triangles are each half an equilateral triangle, so their short leg is half as long as their hypotenuse.

Since $AB=BC=6$, we have $BH=AB/2=3$ and $AH=CH=\sqrt{6^2-3^2}=\sqrt{27}=3\sqrt{3}$. (Notice that this value is $\sqrt{3}$ times the length of $BH$, the short leg. In general, the ratio of the sides in a $30^{\circ }-60^{\circ }-90^{\circ }$ triangle is $1:\sqrt{3}:2$, which can be shown by the Pythagorean Theorem.) Then, $AC=2\cdot 3\sqrt{3}=6\sqrt{3}$. Thus, any pair of parallel sides in this regular hexagon are $\boxed{6\sqrt{3}}$ units apart.