1997 AJHSME Problems/Problem 21

Problem

Each corner cube is removed from this $3\text{ cm}\times 3\text{ cm}\times 3\text{ cm}$ cube. The surface area of the remaining figure is

$[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0)); draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3,1)--(5.7,1.99)); draw((0,2)--(3,2)--(5.7,2.99)); draw((0,3)--(3,3)--(5.7,3.99)); draw((0,3)--(3,3)--(3,0)); draw((0.9,3.33)--(3.9,3.33)--(3.9,0.33)); draw((1.8,3.66)--(4.8,3.66)--(4.8,0.66)); draw((2.7,3.99)--(5.7,3.99)--(5.7,0.99)); [/asy]$

$\text{(A)}\ 19\text{ sq.cm} \qquad \text{(B)}\ 24\text{ sq.cm} \qquad \text{(C)}\ 30\text{ sq.cm} \qquad \text{(D)}\ 54\text{ sq.cm} \qquad \text{(E)}\ 72\text{ sq.cm}$

Solution

The original cube has $6$ square surfaces that each have an area of $3^2 = 9$ , for a toal surface area of $6\cdot 9 = 54$ .

Since no two corner cubes touch, we can examine the effect of removing each corner cube individually.

Each corner cube contribues $3$ faces each of surface area $1$ to the big cube, so the surface area is decreased by $3$ when the cube is removed.

However, when the cube is removed, $3$ faces on the 3x3x3 cube will be revealed, increasing the surface area by $3$ .

Thus, the surface area does not change with the removal of a corner cube, and it remains $54$ , which is answer $\boxed{D}$ .

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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1997 AJHSME Problems/Problem 21

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