Question

Consider a cubic box with a side length of $2$ feet.

How many of these boxes could fit inside a larger cubic box with the base having a perimeter of $24$ feet?

Answer 1

Step-1: Find the volume of the small box having a side length of $2$ feet:

Knowing, the formula for the volume of the cube,

The volume of cube$=S^3$ ( Where $S$ is the length of the side of the cube.) …(I)

The volume of a small cube $=2^3$

$=8$

The volume of a small cube having a side length of $2$ feet is $8{\mathrm{ft}}^3$ (Where $\mathrm{ft}$ is short-form for feet)

Step-2: Finding the side of a large cubic box having a base with a perimeter of $24$ feet:

Knowing cubes have all sides in square shape and given that perimeter of the base is $24$ feet.

One can find the side of the cube using the perimeter of the square formula.

The perimeter of the square$=4s$ (Where s is the side of the square)

$$\begin{gathered} 24=4s \\ \Rightarrow s=\frac{24}{4} \\ \Rightarrow s=6\mathrm{ft} \end{gathered}$$

…(II)

The side of the cube with the base having perimeter $24$ feet is $6\mathrm{ft}$.

Step-3: Find the volume of a large cubic box having a base with a perimeter of $24$ feet:

The volume of the large cube $=6^3$ …(By using (I) )

$=216{\mathrm{ft}}^3$ …(III)

The volume of the large cube is $216{\mathrm{ft}}^3$.

Step-4: Find the number of boxes with a side length of $2$ feet could that fit inside a larger cubic box with the base having a perimeter of $24$ feet.

Consider the volume of a large box is $V$ and the volume of a small box is $v$

The required number of boxes $=\frac{V}{v}$

$$\begin{gathered} =\frac{216}{8} \\ =27 \end{gathered}$$ …(From (I) and (III) )

Hence, the number of boxes with a side length of $2$ feet could fit inside a larger cubic box with the base having a perimeter of $24$ feet is $27$.