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The problem involves two mathematically similar ornaments, meaning their surface areas scale according to the square of the ratio of their corresponding linear dimensions (in this case, height).

We are given the following information:
- The height of the larger ornament: 220 cm
- The surface area of the larger ornament: 31,232 cm²
- The height of the smaller ornament: 55 cm

### Steps to find the surface area of the smaller ornament:
1. **Calculate the ratio of the heights**:
   \[
   \text{Ratio of heights} = \frac{\text{height of smaller ornament}}{\text{height of larger ornament}} = \frac{55}{220} = \frac{1}{4}
   \]

2. **Square the ratio** to account for surface area scaling:
   \[
   \left( \frac{1}{4} \right)^2 = \frac{1}{16}
   \]

3. **Apply the ratio to the surface area** of the larger ornament:
   \[
   \text{Surface area of smaller ornament} = \frac{31,232}{16} = 1,952 \, \text{cm}^2
   \]

Thus, the surface area of the smaller ornament is **1,952 cm²**.

Would you like further details on this process? Here are some related questions:

1. How does the ratio of surface areas change when objects are mathematically similar?
2. How would you calculate the surface area if the shapes weren't mathematically similar?
3. Why does the surface area scale with the square of the ratio of linear dimensions?
4. Can you explain how volume scaling differs from surface area scaling?
5. How would this calculation change if the ornaments had different shapes but equal heights?

**Tip:** Remember that when two objects are similar, surface area scales as the square of the length ratio, while volume scales as the cube of the length ratio.

Calculate the surface area of the smaller ornament when the two ornaments are mathematically similar. Given that the height of the larger ornament is 220 cm, the surface area of the larger ornament is 31,232 cm², and the height of the smaller ornament is 55 cm.

Learn how to calculate the surface area of a smaller ornament based on its mathematical similarity to a larger ornament. Given heights and surface area, this problem demonstrates the use of scaling laws in geometry. Suitable for Grade 10-12 students.

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