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The problem involves two right triangles, and we're tasked with finding the value of \( x \), which is part of the bottom side of the larger triangle.

We can approach this using similar triangles. The two triangles in the figure are similar since both have a right angle, and they share a common angle at the top vertex of the larger triangle.

### Step-by-step solution:

1. **Identify corresponding sides:**
   - The height of the smaller triangle is \( 0.8 \, \text{cm} \), and the height of the larger triangle is \( 1.0 \, \text{cm} \).
   - The bottom of the smaller triangle is \( x \), and the bottom of the larger triangle is \( 4 \, \text{cm} \).

2. **Set up the ratio of corresponding sides:**
   Since the triangles are similar, the ratio of corresponding sides is the same. Therefore, we can write the following ratio:
   \[
   \frac{0.8}{1.0} = \frac{x}{4}
   \]

3. **Solve for \( x \):**
   \[
   \frac{0.8}{1.0} = \frac{x}{4}
   \]
   Simplifying the left side:
   \[
   0.8 = \frac{x}{4}
   \]
   Multiply both sides by 4:
   \[
   x = 0.8 \times 4 = 3.2 \, \text{cm}
   \]

Thus, the value of \( x \) is \( \boxed{3.2 \, \text{cm}} \).

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Do you want more details, or do you have any questions?

Here are 5 related questions to expand on the topic:
1. How do we know when triangles are similar?
2. Can we apply the Pythagorean theorem to this problem instead of using similarity?
3. What are the properties of right triangles?
4. How would the problem change if the larger triangle's base was unknown?
5. What are other applications of similar triangles in geometry?

**Tip:** In similar triangles, all corresponding angles are equal, and corresponding sides are proportional.

In the figure above, what is the value of x?

Learn how to find the value of x using the concept of similar triangles and the ratio of corresponding sides. This problem involves geometry, specifically triangles, and is suitable for students in Grades 8-10. A detailed step-by-step solution is provided, including the application of the similarity theorem.

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