The surface areas of two similar figures are given. The volume of the larger figure is given. Find the volume of the smaller figure. S.A. = 192 m S.A. = 1728 m² V = 4860 m3

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