We have the following boxes:
We know that the volume of both boxes is the same, then we have the following expression:
[tex]V_h=V_H[/tex]
then, in this case we have the following:
[tex]\begin{gathered} V_h=(5)^2\cdot h \\ V_H=(10)^2\cdot H \\ V_h=V_H \\ \Rightarrow(5)^2\cdot h=(10)^2\cdot H \\ \Rightarrow h=\frac{100H}{25}=4H \\ h=4H \end{gathered}[/tex]
therefore, the box with the smaller base is 4 times taller than the box with the bigger base
