jmc

Draw a line segment from $A$ to $B,$ cutting the shaded diamond region in half. Next, draw the altitude from point $E$ to segment $AB.$ The new figure is shown below: $ABCD$ is a rectangle by symmetry of the square over line $AB.$ Thus, $\angle BAD=90$ degrees. Since $\angle BAD=\angle BFE,$ we have $△BFE\sim △BAD.$ Since the diagonals of $ABCD$ bisect each other, $BE=BD/2,$ so the triangles are similar in a $1:2$ ratio. Thus, $FE$ is half the length of $AD,$ or $4/2=2$ cm.

The area of triangle $ABE$ is $$\frac{AB\cdot FE}{2}=\frac{8\cdot 2}{2}=8.$$ The other half of the shaded region is identical and has the same area, so the entire shaded region has area $2\cdot 8=\boxed{16}$ square cm.

We also might take a clever rearrangement approach. The two red pieces below can be rearranged to form a quadrilateral that is congruent to the gray quadrilateral, as can the two blue pieces, and as can the two green pieces. So, the area of the gray quadrilateral is $\frac{1}{4}$ of the area of the square.