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Q.
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 $\mathrm{cm}^2$ ).
Mensuration
Solution:
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
$=2(l b+b h+l h)$
$=2(12 \times 4+4 \times 4+12 \times 4)=224 \mathrm{~cm}^2$