A cube is built using $64$ cubic blocks of side one unit. After it is built, one cubic block is removed from every corner of the cube. The resulting surface area of the body (in square units) after the removal is?
- $56$
- $64$
- $72$
- $96$
Somewhere it explained as :
If $64$ cubic blocks are used to make the cube, it must be $4 \times 4 \times 4$ blocks.
The surface area is $6 \times 4 \times 4 = 96$ sq units. Now picture this...each corner block shows off three faces. When it is removed, three new faces are exposed, thus there appears no net change in the exposed surface area$... 96 $ sq. units
Can you explain in formal way, please?
As long as the removed corner cubes do not intersect, you can reason on each separately. The following image (cube on the left) gives you a visual insight:
For the removed small cube, its top surface projects orthogonally on the bottom (the mid gray "rhombus"), the right-most side projects on the darker gray rhombus, the front side on the light gray square.
Another way is to split the empty small cube as follows:
and see the three-sided outside corner leaves space to the same three-sided carved corner.