Beginning with a $3 \mathrm{~cm}$ by $3 \mathrm{~cm}$ by $3 \mathrm{~cm}$ cube, a $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ cube is cut from one corner and a $2 \mathrm{~cm}$ by $2 \mathrm{~cm}$ by $2 \mathrm{~cm}$ cube is cut from the opposite corner, as shown. In $\mathrm{cm}^{2}$, what is the surface area of the resulting solid?
(A) 42
(B) 45
(C) 48
(D) 51
(E) 54

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-041.jpg?height=293&width=304&top_left_y=1030&top_left_x=1344)
We first determine the surface area of the original cube, before the two cubes were cut from its corners.

The original cube had 6 identical square faces, each of whose area was $3 \times 3=9 \mathrm{~cm}^{2}$.

Thus, the surface area of the original cube was $6 \times 9=54 \mathrm{~cm}^{2}$.

Next we will explain why the resulting solid in question has surface area equal to that of the original cube, $54 \mathrm{~cm}^{2}$.

Consider the front, bottom right corner of the resulting solid, as shown and labeled in Figure 1.

Cutting out a $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ cube from this corner exposes 3 new square faces, $R S P Q, R S T U$ and $S P W T$, which were not a part of the surface area of the original cube.

Next we consider what surface area was part of the original cube that is not present in the resulting solid.

To visualize this, we replace the vertex $V$ of the original cube and reconstruct segments $Q V, U V$ and $W V$, as shown in Figure 2.

We note that $P Q R S T U V W$ is the $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ cube that was cut from the corner of the original cube.

In Figure 2, we see that the surface areas that were part of the original cube that are not present in the resulting solid are the 3 square faces $U T W V, Q P W V$ and $R Q V U$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-199.jpg?height=361&width=550&top_left_y=481&top_left_x=1422)

Figure 1

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-199.jpg?height=369&width=545&top_left_y=908&top_left_x=1424)

Figure 2

To summarize, removal of the $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ cube from the corner of the original cube exposes 3 new square faces, $R S P Q, R S T U$ and $S P W T$, which were not a part of the surface area of the original cube.

These 3 faces represent half of the surface area of cube PQRSTUVW (since they are 3 of the 6 identical faces of the cube).

However, in removing the $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ cube from the corner of the original cube, the 3 faces, $U T W V, Q P W V$ and $R Q V U$, are lost.

These 3 faces represent the other half of the surface area of cube PQRSTUVW.

That is, the total surface area gained by removal of the $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ by $1 \mathrm{~cm}$ cube is equal to the total surface area lost by removing this same cube.

This same argument can be made for the removal of the $2 \mathrm{~cm}$ by $2 \mathrm{~cm}$ by $2 \mathrm{~cm}$ cube from the top, back left corner of the original cube.

Thus the surface area of the resulting solid is equal to the surface area of the original cube, which was $54 \mathrm{~cm}^{2}$.

ANSWER: (E)