jmc

Label points $A$, $B$, $C$ as shown below, and 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 $AB=BC=10$, we have $BH=AB/2=5$ and $AH=CH=\sqrt{10^2-5^2}=\sqrt{75}=5\sqrt{3}$. (Notice that this value is $\sqrt{3}$ times the length of $BH$, the short leg. In general, the ratio of the sides in a $30^{\circ }-60^{\circ }-90^{\circ }$ is $1:\sqrt{3}:2$, which can be shown by the Pythagorean Theorem.) Then, $AC=2\cdot 5\sqrt{3}=10\sqrt{3}$.

The shaded region is a rectangle with base length $10$ and height length $10\sqrt{3}$; its area is $10\cdot 10\sqrt{3}=\boxed{100\sqrt{3}}$ square cm.