Let’s solve the problem step by step:

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1. Use the surface areas to find the scale factor.

The surface area of similar figures is proportional to the square of the scale factor ($k^2$):

$\frac{\text{Surface Area of Smaller Figure}}{\text{Surface Area of Larger Figure}} = \frac{k^2}{1}.$

Substitute the given surface areas:

$\frac{192}{1728} = k^2.$

Simplify:

$k^2 = \frac{192}{1728} = \frac{1}{9}.$

Thus, the scale factor ($k$) is:

$k = \sqrt{\frac{1}{9}} = \frac{1}{3}.$

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2. What is the volume ratio?

The volume ratio is proportional to the cube of the scale factor ($k^3$):

$\text{Volume Ratio} = k^3.$

Since $k = \frac{1}{3}$:

$k^3 = \left(\frac{1}{3}\right)^3 = \frac{1}{27}.$

Thus, the volume ratio is:

$\frac{1}{27}.$

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3. Use the volume ratio to find the smaller volume.

The volumes are proportional according to the volume ratio:

$\frac{\text{Volume of Smaller Figure}}{\text{Volume of Larger Figure}} = \frac{1}{27}.$

Let $V_{\text{small}}$ be the volume of the smaller figure and substitute the larger volume ($V_{\text{large}} = 4860$):

$\frac{V_{\text{small}}}{4860} = \frac{1}{27}.$

Solve for $V_{\text{small}}$:

$V_{\text{small}} = 4860 \times \frac{1}{27}.$

Simplify:

$V_{\text{small}} = 180 \, \text{cubic meters}.$

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Final Answers:

1. Scale factor: $\frac{1}{3}$.
2. Volume ratio: $\frac{1}{27}$.
3. Volume of the smaller figure: $180 \, \text{cubic meters}$.

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Would you like additional details or clarification?
Here are 5 related questions:

1. How are scale factor calculations affected by units?
2. What if the surface areas were given in different units—how would that change the solution?
3. How does this calculation relate to real-world applications like scaling down models?
4. Can the same method be applied to irregular shapes if they are similar?
5. Why does the volume scale by the cube of the scale factor?

Tip: Always double-check units when solving ratio problems involving geometry. Misaligned units can lead to incorrect results.