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

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