Two identical cubes each of volume $64 c m^{3}$ are joined together end to end. What is the surface area of the resulting cuboid?

$160 c m^{2}$

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$120 c m^{2}$

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$145 c m^{2}$

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$180 c m^{2}$

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Solution

The correct option is
D

$160 c m^{2}$

Let the length of a side of a cube = a cm

Given , volume of the cube , $a^{3} = 64 c m^{3} \Rightarrow a = 4 c m$
On joining two cubes, we get a cuboid whose
Length , $l = 2 a$ cm
Breadth , $b = a$ cm
And height , $h = a$ cm
Now, surface area of the resulting cuboid $= 2 (l b + b h + l h)$
$= 2 (2 a \times a + a \times a + a \times 2 a)$
$= 2 (2 a^{2} + a^{2} + 2 a^{2}) = 2 (5 a^{2})$
$= 10 a^{2} = 10 (4)^{2} = 160 c m^{2}$

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