AB is a line segment of length 48 cm and C is its mid-point. If three semicircles are drawn at AB, AC and CB using as diameters, then radius of the circle inscribed in the space enclosed by three semicircles is

Options

A. `3sqrt(2)`
B. 6
C. 8
D. 10

Explanation

To find the radius of the circle inscribed in the space enclosed by three semicircles drawn on the line segment AB (length 48 cm) and its midpoints, we can follow these steps:

### Step 1: Identify the lengths
- The length of line segment AB is given as 48 cm.
- The midpoint C divides AB into two equal segments: AC and CB, each measuring 24 cm.

### Step 2: Draw semicircles
- Draw three semicircles:
1. The first semicircle has diameter AB (48 cm), so its radius is \( R_{AB} = \frac{48}{2} = 24 \) cm.
2. The second semicircle has diameter AC (24 cm), so its radius is \( R_{AC} = \frac{24}{2} = 12 \) cm.
3. The third semicircle has diameter CB (24 cm), so its radius is \( R_{CB} = \frac{24}{2} = 12 \) cm.

### Step 3: Set up the triangle
- Consider triangle OXC, where O is the center of the semicircle on AB, X is the point where the inscribed circle touches AC, and C is the midpoint of AB.
- The lengths are:
- OC = 24 cm (radius of the semicircle on AB)
- XC = 12 cm (radius of the semicircle on AC)
- Let the radius of the inscribed circle be \( r \).

### Step 4: Apply the Pythagorean theorem
- In triangle OXC, by the Pythagorean theorem:
\[ OX^2 = OC^2 - XC^2 \]
Substituting the known values:
\[ (12 + r)^2 = 12^2 + (24 - r)^2 \]

### Step 5: Expand and simplify
- Expanding both sides:
\[ (12 + r)^2 = 144 + (576 - 48r + r^2) \]
\[ 144 + 24r + r^2 = 720 - 48r + r^2 \]

### Step 6: Cancel \( r^2 \) and rearrange
- Cancel \( r^2 \) from both sides:
\[ 144 + 24r = 720 - 48r \]
- Rearranging gives:
\[ 24r + 48r = 720 - 144 \]
\[ 72r = 576 \]

### Step 7: Solve for \( r \)
- Dividing both sides by 72:
\[ r = \frac{576}{72} = 8 \]

### Conclusion
The radius of the circle inscribed in the space enclosed by the three semicircles is \( r = 8 \) cm.