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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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Let the length of the edges of this box have lengths <math>a</math>, <math>b</math>, and <math>c</math>. We're given <math>ab</math>, <math>bc</math>, and <math>ca</math>. The product of these values is <math>a^2b^2c^2</math>, which is the square of the volume of the box. <math>\boxed{\textbf{(D)}}</math>
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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]]
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{{MAA Notice}}

Latest revision as of 11:20, 5 July 2013

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

Let the length of the edges of this box have lengths $a$, $b$, and $c$. We're given $ab$, $bc$, and $ca$. The product of these values is $a^2b^2c^2$, which is the square of the volume of the box. $\boxed{\textbf{(D)}}$

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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