# Surface area of composite cuboids

Consider the three figures pictured below, each of which is made of 16 identical cubes. Does any one have less surface area than the others?

Perhaps the easiest way to solve the problem, is to count the number of shared unit cube faces. The cubes we must focus on, are arranged in a 3D checkerboard pattern. We count the number of shared faces, for each of these cubes, writing the number of shared faces on each, as seen here:

We find 18 shared faces in total, for each figure. Thus, every figure must have the same surface area. To find the surface area, start with the surface area of the unit cube, which is 6, multiply that by the number of unit cubes, which is 16, then subtract twice the number of shared faces, as each shared face is comprised of 2 faces glued together. That gives us

$$6 \cdot 16 - 18 \cdot 2 = 96 - 36 = 60\text{unit squares}$$

Going further:

If the nth iteration of a Menger sponge is made from unit cubes, what is its surface area?