Difference between revisions of "1994 AHSME Problems/Problem 11"
(→Problem) |
(→Problem) |
||
| Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
| − | Three cubes of volume <math>1</math>, <math>8</math> and <math>27</math> are glued together at their faces. The smallest possible surface area of the resulting | + | Three cubes of volume <math>1</math>, <math>8</math> and <math>27</math> are glued together at their faces. The smallest possible surface area of the resulting configuration is |
<math> \textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74 </math> | <math> \textbf{(A)}\ 36 \qquad\textbf{(B)}\ 56 \qquad\textbf{(C)}\ 70 \qquad\textbf{(D)}\ 72 \qquad\textbf{(E)}\ 74 </math> | ||
Latest revision as of 20:50, 22 February 2024
Problem
Three cubes of volume 
, 
and 
are glued together at their faces. The smallest possible surface area of the resulting configuration is
See Also
| 1994 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
