smc

The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($△MNO$) and three right triangles ($△MPN,△NQO,$ and $△ORM$). Notice that $\overline{AD}$ (not shown) and $\overline{BC}$ are parallel. $\overline{XY}$ divides transversals $\overline{AB}$ and $\overline{CD}$ into a $1:1$ ratio (This can be shown by similar triangles.). Thus, it must also divide transversal $\overline{AC}$ and transversal $\overline{CO}$ into a $1:1$ ratio. By symmetry, the same applies for $\overline{CE}$ and $\overline{EA}$ as well as $\overline{EM}$ and $\overline{AN}.$ In $△ACE,$ we see that $\frac{[MNO]}{[ACE]}=\frac{1}{4}$ and $\frac{[MPN]}{[ACE]}=\frac{1}{8}.$ Our desired area becomes $$\left(\frac{1}{4}+3\cdot \frac{1}{8}\right)\cdot \frac{(\sqrt{3}{)}^2\cdot \sqrt{3}}{4}=\boxed{\frac{15}{32}\sqrt{3}}.$$