Difference between revisions of "1950 AHSME Problems/Problem 21"
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| − | == Problem== | + | == Problem == |
The volume of a rectangular solid each of whose side, front, and bottom faces are <math> 12\text{ in}^{2} </math>, <math> 8\text{ in}^{2} </math>, and <math> 6\text{ in}^{2} </math> respectively is: | The volume of a rectangular solid each of whose side, front, and bottom faces are <math> 12\text{ in}^{2} </math>, <math> 8\text{ in}^{2} </math>, and <math> 6\text{ in}^{2} </math> respectively is: | ||
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<math> \textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these} </math> | <math> \textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these} </math> | ||
| − | ==Solution== | + | == Solution == |
If the sidelengths of the cubes are expressed as <math>a, b,</math> and <math>c,</math> then we can write three equations: | If the sidelengths of the cubes are expressed as <math>a, b,</math> and <math>c,</math> then we can write three equations: | ||
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | (ab)(bc)(ac)&=(12)(8)(6)\\ | + | (ab)(bc)(ac) &= (12)(8)(6)\\ |
| − | a^2b^2c^2&=576\\ | + | a^2b^2c^2 &= 576\\ |
| − | abc&=\boxed{\mathrm{(B)}\ 24 | + | abc &= \boxed{\mathrm{(B)}\ 24} |
\end{align*}</cmath> | \end{align*}</cmath> | ||
| − | ==See Also== | + | == See Also == |
{{AHSME 50p box|year=1950|num-b=20|num-a=22}} | {{AHSME 50p box|year=1950|num-b=20|num-a=22}} | ||
Latest revision as of 00:54, 12 October 2020
Problem
The volume of a rectangular solid each of whose side, front, and bottom faces are 
, 
, and 
respectively is:
Solution
If the sidelengths of the cubes are expressed as 
and 
then we can write three equations:
![\[ab=12, bc=8, ac=6.\]](http://latex.artofproblemsolving.com/4/2/3/423226df7be3000ec5f1ded053b7c2595014f3bf.png)
The volume is 
Notice symmetry in the equations. We can find 
my multiplying all the equations and taking the positive square root.
See Also
| 1950 AHSC (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 • 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
