Answer:
The volume of smaller figure is 180 m³
Step-by-step explanation:
Consider the provided information.
The ratio of the surface areas is equal to the square of scale factor K.
let K₁ and K₂ is the scale factor
Thus [tex]\frac{k^2_1}{k^2_2} =\frac{S.A_1}{S.A_2}[/tex]
Substitute the respective values as shown.
[tex]\frac{k^2_1}{k^2_2} =\frac{192}{1728}[/tex]
[tex]\frac{k^2_1}{k^2_2} =\frac{1}{9}[/tex]
[tex]\frac{k_1}{k_2} =\frac{1}{3}[/tex]
It is given that the volume of larger figure is 4860 m³.
Let V₁ and V₂ is the volume of small and larger figure respectively.
The ratio of the volume is equal to the third power of scale factor K.
Thus [tex]\frac{k^3_1}{k^3_2} =\frac{V_1}{V_2}[/tex]
Substitute the respective values as shown.
[tex]\frac{1^3}{3^3} =\frac{V_1}{4860}[/tex]
[tex]\frac{1}{27} =\frac{V_1}{4860}[/tex]
[tex]V_1=\frac{4860}{27}[/tex]
[tex]V_1=180[/tex]
Hence, the volume of smaller figure is 180 m³
