H4. The points $A, B$ and $C$ are the centres of three faces of a cuboid that meet at a vertex. The lengths of the sides of the triangle $A B C$ are 4,5 and 6 .
What is the volume of the cuboid?
Solution
Let the cuboid have centre $O$ and dimensions $2 x \times 2 y \times 2 z$, as shown. Joining $O$ to each of $A, B$ and $C$ forms three rightangled triangles with sides $x, y$ and $4 ; y, z$ and 5; and $z, x$ and 6 . Hence, using Pythagoras' Theorem, we get
$$
\begin{aligned}
x^{2}+y^{2} & =4^{2} \\
y^{2}+z^{2} & =5^{2} \\
\text { and } \quad z^{2}+x^{2} & =6^{2} .
\end{aligned}
$$

Adding all three equations, we obtain
$$
\begin{aligned}
2 x^{2}+2 y^{2}+2 z^{2} & =16+25+36 \\
& =77
\end{aligned}
$$
so that
$$
x^{2}+y^{2}+z^{2}=\frac{77}{2}
$$

Subtracting each of the first three equations in turn, we therefore get
$$
\begin{aligned}
z^{2} & =\frac{77}{2}-16=\frac{45}{2} \\
x^{2} & =\frac{77}{2}-25=\frac{27}{2} \\
\text { and } \quad y^{2} & =\frac{77}{2}-36=\frac{5}{2} .
\end{aligned}
$$

Now $2 x$ is equal to $\sqrt{4 x^{2}}$ (because $x$ is positive), with similar results for $y$ and $z$. So the volume of the cuboid $2 x \times 2 y \times 2 z$ is therefore
$$
\begin{aligned}
\sqrt{54} \times \sqrt{10} \times \sqrt{90} & =\sqrt{54 \times 10 \times 90} \\
& =\sqrt{6 \times 90 \times 90} \\
& =90 \sqrt{6}
\end{aligned}
$$