1997 AMC 8 Problem 21

Attempt Problem 21 of the 1997 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

Each corner cube is removed from this 3 cm×3 cm×3 cm3\text{ cm}\times 3\text{ cm}\times 3\text{ cm} cube. The surface area of the remaining figure is
19 sq.cm19\text{ sq.cm}
24 sq.cm24\text{ sq.cm}
30 sq.cm30\text{ sq.cm}
54 sq.cm54\text{ sq.cm}
72 sq.cm72\text{ sq.cm}
Answer: D
Concepts:surface areacube geometryinvariant
Small hint:
The total surface area stays the same
Big hint:
Removing a corner loses three faces and reveals three faces
Solution:
Note that there is one unit cube between any pair of corner cubes, so the removal of each does not affect the others.
When we remove a corner, we are losing three unit squares. We, however, gain these back from the three faces that get uncovered.
This means that removing a corner cube does not change the surface area. The surface area of the original cube is 6⋅32=6⋅9=54 6 \cdot 3^2 = 6 \cdot 9 = 54 square centimeters.
Thus, D is the correct answer.
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