CAT 2024 Slot 1 · Quant · Question 16

A box has 12 edges, four of each dimension,

so 4\left( l+b+h \right)=144 and l+b+h=36.

Its surface area is 2\left( lb+lh+bh \right)=846, so lb+lh+bh=423.

The identity {\left( l+b+h \right)}^{2}={l}^{2}+{b}^{2}+{h}^{2}+2\left( lb+lh+bh \right) links the two.

Substituting, {36}^{2}={l}^{2}+{b}^{2}+{h}^{2}+846, that is 1296={l}^{2}+{b}^{2}+{h}^{2}+846.

So {l}^{2}+{b}^{2}+{h}^{2}=450.

The space diagonal of the box is \sqrt{{l}^{2}+{b}^{2}+{h}^{2}}=\sqrt{450}=15\sqrt{2}.

Since the box is inscribed in the sphere, that diagonal is the diameter, so r=\frac{15\sqrt{2}}{2}.

The volume is \frac{4}{3}\pi{r}^{3}=\frac{4}{3}\pi \times \frac{6750\sqrt{2}}{8}=1125\pi \sqrt{2} cubic cm.

Shortcut. The individual dimensions are never needed — only {l}^{2}+{b}^{2}+{h}^{2}, which the squared-sum identity delivers from the two given totals in one line.