If you have a 1 m^3 cube centered on the origin (0,0,0) the exposed area on the XY plane is 1 m^2. If you rotate that cube 45 deg then the exposed area becomes $\sqrt{2}$ (larger than the surface area one side).
It made me wonder - is it possible to have a shape where the exposed area can ever be smaller than the side of that shape with the smallest surface area?
Cheers
Take a regular tetrahedron with side length $1$ and rotate it so that one face is parallel to the $yz$-plane and one of that face's edges is parallel to the $y$-axis. All faces have area $\frac{\sqrt3}4$, but the projection in this orientation onto the $xy$-plane has area $\frac1{\sqrt6}<\frac{\sqrt3}4$.
Another possibility is to take any bipyramid and pull its pole vertices out until the areas of its faces are larger than the area of the (convex) polygon running through the middle vertices. Then, if the bipyramid's axis is parallel to the $z$-axis, the projection onto the $xy$-plane will be just the middle polygon.