# Two cubes each of volume $27\ m^{3}$ are joined end to end. Find the surface area of the resulting cuboid.

Given: Two cubes of volume $27\ m^{3}$ are joined together.

To do: To find the surface area of the resulting cuboid.

Solution:

$\because$ Volume of a cube $=27\ m^{3}$

Let $a$ is a side of the cube.

Then, the volume of the cube$=a^{3}$

$\Rightarrow a^{3}=27$

$\Rightarrow a=\sqrt[3]{27}$

$\Rightarrow a=3\ m$

When these two cubes are joined together, width and height remains the same of the resultant cuboid but its length doubles the original.

$\therefore$ Length of the resultant cuboid $l=a+a=2a=2\times3=6\ m$

Width of the resulting cuboid $b=3\ m$

Height of the resulting cuboid $h=3\ m$

$\therefore$ Surface Area of the resulting cuboid $A=2( lb+bh+hl)$

$\Rightarrow A=2( 6\times3+3\times3+3\times6)$

$\Rightarrow A=2( 18+9+18)$

$\Rightarrow A=2( 45)$

$\Rightarrow A=90\ cm^{2}$.

Thus, The surface area of the resulting cuboid $=90\ cm^{2}$.

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