Triple Pythagoras

Let the lengths of the edges of the cuboid be $a$, $b$ and $c$, as shown in the diagram below.

 

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 The required length $\text{XA}$ is the hypotenuse of the blue right-angled triangle in the diagram on the left, whose other sides are $c$ and $k$. So, by Pythagoras' theorem, $c^2+k^2=\text{XA}^2$.

The side labelled $k$ is also the hypotenuse of the red right-angled triangle, so $a^2+b^2=k^2$.

Substituting $a^2+b^2=k^2$ into $c^2+k^2=\text{XA}^2$ gives $c^2+a^2+b^2=\text{XA}^2$, so $\text{XA}^2=a^2+b^2+c^2$. 

We can use Pythagoras' theorem on the other triangles to get more information about $a^2,$ $b^2$ and $c^2,$ and then use this to find $\text{XA}.$

 

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$\text{XY}=9$, so, by applying Pythagoras' theorem to the yellow triangle in this diagram, $a^2+b^2=81$.

 

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$\text{XZ}=8$, so, by applying Pythagoras' theorem to the yellow triangle in this diagram, $b^2+c^2=64$.

 

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$\text{YZ}=\sqrt{55}$, so, by applying Pythagoras' theorem to the yellow triangle in this diagram, $a^2+c^2=55$.

Aiming for $\text{XA}^2=a^2+b^2+c^2$, we can get all three unknowns in the same equation by adding the equations $a^2+b^2=81$, $b^2+c^2=64$ and $a^2+c^2=55$: $$\begin{array}{rl}{a}^2+b^2+b^2+c^2+a^2+c^2& =81+64+55\\ \Rightarrow 2a^2+2b^2+2c^2& =200\\ \Rightarrow a^2+b^2+c^2& =100\\ \Rightarrow {\text{XA}}^2& =100\\ \Rightarrow \text{XA}& =10\end{array}$$