A cuboid has sides of lengths \(22\), \(2\) and \(10\). It is contained in a sphere of the smallest possible radius. What is the side-length of the largest cube that will fit inside the same sphere?

(A) \(10 \quad\) (B) \(11 \quad\) (C) \(12 \quad\) (D) \(13 \quad\) (E) \(14\)

Any cuboid fitting snugly inside a sphere will have all \(8\) vertices touching the sphere, and by symmetry, the diagonal of the cuboid will be a diameter of the sphere.

Let the diagonal of the given cuboid be \(d\). Then by Pythagoras’ theorem we have \[d^2=22^2+2^2+10^2 = 588.\]

If the cube has side length \(x\) then we also have \[d^2=x^2+x^2+x^2\] and so \[3x^2=588 \quad\implies\quad x=14.\]

The answer is (E).