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Two 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: To find the surface area of the resulting cuboid.
Solution:
Let the edge of each cube$=x$
$\therefore x^3=64=4^3$
$\Rightarrow x=4\ cm$
Now, Length of the resulting cuboid $'l'=2x\ cm$
Breadth of the resulting cuboid $'b'=x\ cm$
Height of the resulting cuboid $'h'=x\ cm$
$\therefore$ Surface area of the cuboid $=2( lb + bh + hl)$
$=2[( 2x·x)+( x·x)+( x·2x)]$
$=2[( 2\times4\times4)+( 4\times4)+( 4\times2\times4)]\ cm^2$
$=2[32+16+32]\ cm^2$
$=2[80]\ cm^2$
$=160\ cm^2$.
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