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Difference between revisions of "1984 AHSME Problems/Problem 25"

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Latest revision as of 12:52, 5 July 2013

Problem

The total area of all the faces of a rectangular solid is $22\text{cm}^2$, and the total length of all its edges is $24\text{cm}$. Then the length in cm of any one of its interior diagonals is

$\mathrm{(A) \ }\sqrt{11} \qquad \mathrm{(B) \ }\sqrt{12} \qquad \mathrm{(C) \ } \sqrt{13}\qquad \mathrm{(D) \ }\sqrt{14} \qquad \mathrm{(E) \ } \text{Not uniquely determined}$

Solution

Let the edge lengths be $a, b$, and $c$. Therefore, the total area of all its faces is $2ab+2bc+2ac$. Therefore, $2ab+2bc+2ac=22$ and $ab+bc+ac=11$. Also, the total lengths of all of its edges is $4a+4b+4c$, so $4a+4b+4c=24$, and $a+b+c=6$. Therefore, we have:

$ab+bc+ac=11$

and

$a+b+c=6$.

The length of the interior diagonal is $\sqrt{a^2+b^2+c^2}$, so if we can find $a^2+b^2+c^2$, we can find the diagonal. We square the second equation above to introduce an $a^2+b^2+c^2$:

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=36$.

However, we already know that $2ab+2bc+2ac=22$, so $a^2+b^2+c^2+22=36\implies a^2+b^2+c^2=14\implies \sqrt{a^2+b^2+c^2}=\sqrt{14}, \boxed{\text{D}}$.

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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

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