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:
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<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>
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<math> \textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}</math>
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== Solution ==
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{{solution}}
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== See Also ==
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{{AHSME 50p box|year=1951|num-b=5|num-a=7}}
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[[Category:Introductory Geometry Problems]]

Revision as of 08:48, 29 April 2012

Problem

The bottom, side, and front areas of a rectangular box are known. The product of these areas is equal to:

$\textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}$ $\textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}$

Solution

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See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AHSME Problems and Solutions