Answer:
On each face, there are $2$ diagonals like $x$. There are $6$ faces on a cube. Thus, there are $2\times 6 = 12$ diagonals that are "x-like".
Every "y-like" diagonal must connect the bottom of the cube to the top of the cube. Thus, for each of the $4$ bottom vertices of the cube, there is a different "y-like" diagonal. So there are $4$ "y-like" diagonals.
This gives a total of $12 + 4 = 16$ diagonals on the cube, which is answer {E}
Step-by-step explanation:
There are $8$ vertices on a cube(A, B, C, D, E, F, G, H). If you pick one vertice, it will have 7 line segments associated with it, so you will have $\frac{8\cdot7}{2} = 28$ segments within the cube. The division by $2$ is necessary because you counted both the segment from $A$ to $B$ and the segment from $B$ to $A$.
But not all $28$ of these segments are diagonals. Some are edges. There are $4$ edges on the top, $4$ edges on the bottom, and $4$ edges that connect the top to the bottom. So there are $12$ edges total, meaning that there are $28 - 12 = 16$ segments that are not edges. All of these segments are diagonals, and thus the answer is $\boxed{E}$.
