Three faces of a rectangular box meet at a corner of the box. The centres of these faces form the vertices of a triangle having side lengths of $4 \mathrm{~cm}, 5 \mathrm{~cm}$ and $6 \mathrm{~cm}$. The volume of the box, in $\mathrm{cm}^{3}$, is
(A) $45 \sqrt{3}$
(B) $45 \sqrt{6}$
(C) $90 \sqrt{6}$
(D) 125
(E) $120 \sqrt{2}$
Let the edge lengths of the box be $2 a, 2 b$ and $2 c$. We label the vertices of the box $A B C D E F G H$.

Let the centres of the three faces meeting at a corner be $P, Q$ and $R$.

Then we can say that $P Q=4, Q R=5$, and $P R=6$.

We would like to try to express these distances in terms of the edge lengths of the box.

We will start by letting $M$ be the midpoint of edge $A B$, and joining $P$ to $M, Q$ to $M$ and $P$ to $Q$.

Since $P$ is the centre of face $A B C D$ and $P M$ is

perpendicular to $A B$, then $P M=\frac{1}{2} A D=b$. Similarly,

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-245.jpg?height=376&width=426&top_left_y=945&top_left_x=1359)

$M Q=a$.

Since $M P$ and $M Q$ are both perpendicular to $A B$ and they lie in perpendicular planes $A B C D$ and $A B G F$, then $M P$ is indeed perpendicular to $M Q$, and so $\triangle Q M P$ is rightangled.

By Pythagoras, $P Q^{2}=M P^{2}+M Q^{2}$ or $16=a^{2}+b^{2}$.

Similarly, looking at $P R, 36=a^{2}+c^{2}$, and, looking at $Q R, 25=b^{2}+c^{2}$.

So we now have $16=a^{2}+b^{2}, 36=a^{2}+c^{2}$, and $25=b^{2}+c^{2}$, and we want to determine the volume of the box, which is $V=(2 a)(2 b)(2 c)=8 a b c$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-245.jpg?height=374&width=422&top_left_y=1491&top_left_x=1361)

Adding the three equations, we obtain $77=2\left(a^{2}+b^{2}+c^{2}\right)$, or $a^{2}+b^{2}+c^{2}=\frac{77}{2}$.

Subtracting the above equations one by one from this last equation yields

$$
\begin{aligned}
& c^{2}=\frac{45}{2} \\
& b^{2}=\frac{5}{2} \\
& a^{2}=\frac{27}{2}
\end{aligned}
$$

Therefore, $a^{2} b^{2} c^{2}=\frac{(27)(5)(45)}{8}=\frac{6075}{8}$ and so $a b c=\sqrt{\frac{6075}{8}}=\frac{45 \sqrt{3}}{2 \sqrt{2}}$.

Thus, the volume is $V=8\left(\frac{45 \sqrt{3}}{2 \sqrt{2}}\right)=(2 \sqrt{2})(45 \sqrt{3})=90 \sqrt{6} \mathrm{~cm}^{3}$.