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2 cubes each of volume $64\ cm^3$ are joined end to end. Find the surface area of the resulting cuboid.
"
Given:
Two cubes, each of volume $64\ cm^3$ are joined end to end.
To do:
We have to find the surface area of the resulting cuboid.
Solution:
Volume of each cube $= 64\ cm^3$
This implies,
Edge of the cube $= \sqrt[3]{64}$
$=4\ cm$
The length of the cuboid formed by joining the cubes $(l) = 4 + 4$
$= 8\ cm$
Breadth of the cuboid $(b) = 4\ cm$
Height of the cuboid $(h) = 4\ cm$
Therefore,
Surface area of the resulting cuboid $= 2(lb + bh + lh)$
$= 2(8 \times 4 + 4 \times 4 + 4 \times 8)$
$= 2(32 + 16 + 32)$
$= 2 \times 80$
$= 160\ cm^2$ .
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