Question

the volumes of two similar solids are $210\\,\mathrm{m}^3$ and $1,680\\,\mathrm{m}^3$. the surface area of the larger solid is $856\\,\mathrm{m}^2$. what is the surface area of the smaller solid?\
\\(\circ\\) $107\\,\mathrm{m}^2$\
\\(\circ\\) $214\\,\mathrm{m}^2$\
\\(\circ\\) $1034\\,\mathrm{m}^2$\
\\(\circ\\) $6848\\,\mathrm{m}^2$

Explanation:

Step1: Find the scale factor of volumes

For similar solids, the ratio of volumes is the cube of the scale factor ($k$). Let the volume of the smaller solid be $V_1 = 210\ m^3$ and the larger be $V_2 = 1680\ m^3$. So, $\frac{V_1}{V_2}=k^3$.
$$k^3=\frac{210}{1680}=\frac{1}{8}$$
Taking the cube root, $k = \frac{1}{2}$.

Step2: Relate surface areas to scale factor

The ratio of surface areas of similar solids is the square of the scale factor. Let the surface area of the smaller solid be $S_1$ and the larger be $S_2 = 856\ m^2$. So, $\frac{S_1}{S_2}=k^2$.
Since $k=\frac{1}{2}$, $k^2=\frac{1}{4}$. Then,
$$\frac{S_1}{856}=\frac{1}{4}$$
Solving for $S_1$: $S_1=\frac{856}{4}=214\ m^2$.

Answer:

214 m² (corresponding to the option "214 m²")