Answer:
Step-by-step explanation:
[tex]k - \text{scale of similar}[/tex]
[tex]\text{If}\ AB\sim CD\ \text{in scale}\ k,\ \text{then}\ \dfrac{length_{AB}}{length_{CD}}=k[/tex]
[tex]\text{If}\ A\sim B\ \text{in scale}\ k,\ \text{then}\ \dfrac{area_A}{area_B}=k^2[/tex]
[tex]\text{If}\ A\sim B\ \text{in scale}\ k,\ \text{then}\ \dfrac{volume_A}{volume_B}=k^3[/tex]
[tex]\text{We have}\ B\sim A\ \text{in scale}\ k=5.\ V_A=12\ m^3,\ \text{therefore}\\\\\dfrac{V_B}{V_A}=k^3\to\dfrac{V_B}{12}=5^3[/tex]
[tex]\dfrac{V_B}{12}=125[/tex] multiply both sides by 12
[tex]V_B=1500\ m^3[/tex]