Two similar boxes A and B are given with box B created by dividing box A's dimensions by 2.
Recall that the scale factor, k is the ratio of lengths of two corresponding sides of similar figures.
Since the dimensions of A were divided by 2 to form B, it follows that the scale factor of box A to box B is:
[tex]k=\frac{2}{1}[/tex]
Also, recall the Similar Solids Theorem
It follows that the ratio of the volumes of the given solids is:
[tex]\frac{\text{Volume of A}}{\text{Volume of B }}=k^3=(\frac{2}{1})^3=\frac{2^3}{1^3}=\frac{8}{1}[/tex]
Let the volume of box A be V₁ and let the volume of box B be V₂, it, therefore, implies that:
[tex]\frac{V_1}{V_2}=\frac{8}{1}[/tex]
Substitute the value for the volume of box A into the equation, V₁=64:
[tex]\begin{gathered} \frac{64}{V_2}=\frac{8}{1} \\ Cross-Multiply_{} \\ \Rightarrow8V_2=64 \\ Divide\text{ both sides by 8:} \\ \Rightarrow\frac{8V_2}{8}=\frac{64}{8}\Rightarrow V_2=8\text{ cubic meters} \end{gathered}[/tex]
Hence, the volume of box B is 8 m³.
