Question

Three cubes having a face diagonal $4\sqrt{2}\mathrm{cm}$ each are joined end to end. Find the total surface area of the resulting cuboid (in ${\mathrm{cm}}^2$ ).

• A. 160
• B. 192
• C. 224
• D. 336

Answer 1

Answer: (B)

The face diagonal of the cube $=4\sqrt{2}\mathrm{cm}$
The edge of the cube $=\frac{4\sqrt{2}}{\sqrt{2}}=4\mathrm{cm}$
$\therefore$ The dimensions of the resulting cuboid are:
$l=4\times 3=12\mathrm{cm},b=4\mathrm{cm}$, and $h=4\mathrm{cm}$.
$\therefore$ Total surface area of the cuboid
$=2(lb+bh+lh)$
$=2(12\times 4+4\times 4+12\times 4)=224{\mathrm{cm}}^2$