To solve this problem, we need to use the concept of similar solids and the ratios of their volumes and surface areas.

Given:

1. Volumes of solids:

   • Volume of solid $A$ ($V_A$) = 1125 cm³
   • Volume of solid $B$ ($V_B$) = 576 cm³

2. Surface area information:

   • Total surface area of solid $A$ and solid $B$ is 3198 cm².

We need to find the surface area of solid $B$.

Step 1: Find the Ratio of the Volumes

Since solids $A$ and $B$ are similar, the ratio of their volumes is given by: $\left(\frac{\text{Volume of } B}{\text{Volume of } A}\right) = \left(\frac{\text{Scale factor of } B}{\text{Scale factor of } A}\right)^3$

Let $k$ be the scale factor of $B$ compared to $A$. Then: $k^3 = \frac{V_B}{V_A} = \frac{576}{1125}$

Calculating this ratio: $k^3 = \frac{576}{1125} = 0.512$

Now, take the cube root to find $k$: $k = \sqrt[3]{0.512} \approx 0.8$

Step 2: Use the Scale Factor to Find the Surface Area Ratio

For similar solids, the ratio of their surface areas is the square of the scale factor: $\frac{\text{Surface Area of } B}{\text{Surface Area of } A} = k^2 = (0.8)^2 = 0.64$

Let the surface area of solid $A$ be $S_A$ and that of solid $B$ be $S_B$.

We know: $S_A + S_B = 3198$

And from the surface area ratio: $S_B = 0.64 S_A$

Step 3: Solve for $S_A$ and $S_B$

Substitute $S_B = 0.64 S_A$ into the total surface area equation: $S_A + 0.64 S_A = 3198$ $1.64 S_A = 3198$ $S_A = \frac{3198}{1.64} \approx 1950 \, \text{cm}^2$

Now, substitute $S_A$ back to find $S_B$: $S_B = 0.64 \times 1950 \approx 1248 \, \text{cm}^2$

Answer

The surface area of solid $B$ is approximately: $\boxed{1248 \, \text{cm}^2}$

Would you like more details on any step, or have further questions?

Here are 5 related questions to explore this topic further:

1. How would the answer change if the solids were not similar?
2. What if only the ratio of the volumes was given, without the actual volume values?
3. How can this problem be extended to find other properties, like height, if they are similar shapes?
4. Why do we use the square of the scale factor for surface area and the cube for volume?
5. How would this approach differ if we were working with 2D similar figures rather than 3D solids?

Tip: When working with similar figures, remember that length, area, and volume scale differently: linearly, by the square, and by the cube, respectively.