1984 AHSME Problems/Problem 25

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 $a^2+b^2+c^2+22=36\implies a^2+b^2+c^2=14\implies \sqrt{a^2+b^2+c^2}=\sqrt{14}, \boxed{\text{D}}$ .

1984 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1984 AHSME Problems/Problem 25

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