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To solve for \( x \), let's analyze the given diagram.

### Analysis:
1. The diagram shows a semicircle with a diameter labeled as \( 4 + x + 1 = x + 5 \).
2. Inside the semicircle, there's a circle tangent to the diameter and the semicircle itself.
3. The distance \( 4 \) and \( 1 \) are the segments from the center of the semicircle to where the semicircle meets the diameter on either side of the circle.

### Steps to solve:

1. **Identify the radius of the semicircle**: 
   - The total diameter of the semicircle is \( x + 5 \), so the radius \( R \) of the semicircle is \( R = \frac{x + 5}{2} \).

2. **Determine the radius \( r \) of the inner circle**:
   - The radius of the inner circle will be equal to \( R \) minus the distance from the center of the semicircle to the point of tangency on the diameter. This distance is half of \( 4 + 1 \), so it is \( \frac{5}{2} = 2.5 \).
   - Therefore, the radius \( r \) of the inner circle is \( R - r \).

3. **Set up the equation**:
   - Since the inner circle is tangent to the semicircle and the diameter, the distance from the center of the semicircle to the center of the inner circle is \( R - r \).
   - The radius \( r \) is also equal to \( R - (R - 2.5) \).

   This results in the equation:
   \[
   r = R - 2.5
   \]

4. **Substitute \( R = \frac{x + 5}{2} \) into the equation**:
   \[
   r = \frac{x + 5}{2} - 2.5
   \]

5. **Solve for \( r \)** and match with the geometry:
   The radius \( r \) is determined by the conditions of the problem, such as the circle fitting perfectly inside the semicircle with tangency.

   Therefore, we simplify and solve the equation:
   \[
   \frac{x + 5}{2} - 2.5 = r
   \]
   The radius of the inner circle \( r \) is calculated by the tangency condition.

Given the information, the equation simplifies to determine \( x \).

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Explore how to solve for x in a geometry problem involving a semicircle and an inner circle tangent to its diameter. This problem covers circle geometry concepts and involves determining radii based on tangency conditions. Suitable for advanced high school students.

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