To solve this problem, we will use the concept of similarity between solids. The ratio of the volumes of two similar solids is related to the cube of the ratio of their corresponding linear dimensions. Similarly, the ratio of the surface areas of two similar solids is related to the square of the ratio of their corresponding linear dimensions.

Step 1: Find the ratio of the volumes

The volumes of the two similar solids are given as 512 cm³ and 2,197 cm³. The ratio of their volumes is: $\frac{V_{\text{larger}}}{V_{\text{smaller}}} = \frac{2197}{512}$ Let's simplify this ratio: $\frac{2197}{512} = \left(\frac{h_{\text{larger}}}{h_{\text{smaller}}}\right)^3$ where $h_{\text{larger}}$ and $h_{\text{smaller}}$ represent the corresponding linear dimensions of the solids.

Step 2: Solve for the ratio of linear dimensions

To find the ratio of the linear dimensions, we take the cube root of the volume ratio: [ \frac{h_{\text{larger}}}{h_{\text{smaller}}} = \sqrt[3]{\frac{2197}{512}} = \frac{\sqrt[3]{2197}}{\sqrt[3]{512}} = \frac{13}{8} ] Thus, the ratio of the linear dimensions is $\frac{13}{8}$.

Step 3: Use the ratio of linear dimensions to find the surface area ratio

The ratio of the surface areas of two similar solids is the square of the ratio of their corresponding linear dimensions: $\frac{A_{\text{larger}}}{A_{\text{smaller}}} = \left(\frac{13}{8}\right)^2 = \frac{169}{64}$

Step 4: Find the surface area of the larger solid

We know that the surface area of the smaller solid is 960 cm². Using the surface area ratio: $\frac{A_{\text{larger}}}{960} = \frac{169}{64}$ Solve for $A_{\text{larger}}$: $A_{\text{larger}} = 960 \times \frac{169}{64} = \frac{960 \times 169}{64} = \frac{162,240}{64} = 2,535 \, \text{cm}^2$

Thus, the surface area of the larger solid is 2,535 cm².

Would you like more details on any step?

Related Questions:

1. How do you calculate the surface area of similar solids using their volume ratio?
2. How is the ratio of linear dimensions related to the ratio of surface areas in similar solids?
3. Can the method of ratios be used to calculate areas or volumes for irregular shapes?
4. What are the properties of similar figures in three-dimensional geometry?
5. How do volume and surface area scale for irregular objects?

Tip:

When comparing geometric figures, remember that volume scales by the cube of the linear dimension, and surface area scales by the square of the linear dimension.