Difference between revisions of "1951 AHSME Problems/Problem 6"
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+ | == Problem == | ||
The bottom, side, and front areas of a rectangular box are known. The product of these areas | The bottom, side, and front areas of a rectangular box are known. The product of these areas | ||
is equal to: | is equal to: | ||
+ | |||
+ | <math> \textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}</math> | ||
+ | <math> \textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}</math> | ||
+ | |||
+ | == Solution == | ||
+ | {{solution}} | ||
+ | |||
+ | == See Also == | ||
+ | {{AHSME 50p box|year=1951|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 07:48, 29 April 2012
Problem
The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:
Solution
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See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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All AHSME Problems and Solutions |