Question

Two cubes each of volume $64\;{\rm{c}}{{\rm{m}}^3}$ are joined end to end together. Find the surface area of the resulting cuboid.

Answer 1

Hint: Since volume is a given question for each cube, we will find the side of the cube from the expression of the volume of the cube. By using the value of the side of the cube, we will find the dimensions of the resulting cuboid. Then to find the surface area of the resulting cuboid ,we will use the expression of the surface area of the cuboid. We will substitute the value of length, breadth and height in the expression of the surface area of the cuboid to get the final answer.
Formula used:
 Volume of cube is expressed as
${\rm{Volume}} = {\left( {{\rm{side}}} \right)^3}$
The surface area of the cuboid is expressed as:
${\rm{Surface \; area \; of \; cuboid}} = 2\left( {lb + bh + hl} \right)$
Where $l$stand for length ,$b$ stands for breadth and $h$ stands for height in the above expression.

Complete step-by-step answer:
It is given that Volume of 1 cube is $V = 64\;{\rm{c}}{{\rm{m}}^{\rm{3}}}$
We know that volume of cube be written as
$V = {\left( {{\rm{side}}} \right)^3}$
 Let us assume side of 1 cube as $a$ then we can write the expression of volume as
$V = {a^3}$
We will substitute $64\;{\rm{c}}{{\rm{m}}^3}$ for $V$ in the above expression we will get
$\begin{array}{c}
{a^3} = 64\;{\rm{c}}{{\rm{m}}^3}\\
a = 4\;{\rm{cm}}
\end{array}$
Hence the side of both the cube is ${\rm{4}}\;{\rm{cm}}$. It is given in question that we need to join two cubes. It can be shown as:

We can see that from the figure that after joining two cubes, one cuboid will form whose side can be expressed as:
The length of the cuboid is:
 $\begin{array}{l}
l = a + a\\
l = 4\;{\rm{cm}} + 4\;{\rm{cm}}\\
l = 8\;{\rm{cm}}
\end{array}$
The breadth of the cuboid is $b = a = 4\;{\rm{cm}}$ and the height of the cuboid is $h = a = 4\;{\rm{cm}}$.
 We know that surface area of the cuboid can be expressed as:
${\rm{Surface \; area \; of \; cuboid}} = 2\left( {lb + bh + hl} \right)$
 We will substitute ${\rm{8}}\;{\rm{cm}}$ for $l$ , $4\;{\rm{cm}}$for $b$ and $4\;{\rm{cm}}$for $h$ in the above expression, we will get:
 $\begin{array}{l}
{\rm{Surface \; area \; of \; cuboid}} = 2\left( {\left( {{\rm{8}}\;{\rm{cm}}} \right)\left( {4\;{\rm{cm}}} \right) + \left( {4\;{\rm{cm}}} \right)\left( {4\;{\rm{cm}}} \right) + \left( {4\;{\rm{cm}}} \right)\left( {4\;{\rm{cm}}} \right)} \right)\\
{\rm{Surface \; area \; of \; cuboid}} = 2\left( {32\;{\rm{c}}{{\rm{m}}^2} + 16\;{\rm{c}}{{\rm{m}}^2} + 32\;{\rm{c}}{{\rm{m}}^2}} \right)\\
{\rm{Surface \; area \; of \; cuboid}} = 2\left( {80\;{\rm{c}}{{\rm{m}}^2}} \right)\\
{\rm{Surface \; area \; of \; cuboid}} = 160\;{\rm{c}}{{\rm{m}}^2}
\end{array}$
Hence the surface area of the resulting cuboid is ${\rm{160}}\;{\rm{c}}{{\rm{m}}^{\rm{2}}}$.

Note: Whenever we join two cubes a new cuboid is formed such that the volume never changes but the surface area always changes. The volume of the two cubes joined will always be equal to the volume of the cuboid. But we need to calculate the surface area as it always changes.