In a triangle with sides, a,b, and c a semicircle toughing the sides AC and CB is inscribed whose diameter lies on AB. Then the radius of the semicricel is

Options

A. `a//2`
B. `Delta//s`
C. `(2Delta)/(a +b)`
D. `(2abc)/((s) (a +b)) cos "" A/2 cos ""B/2 cos ""C/2`

Answer

To find the radius of the semicircle inscribed in triangle ABC, we can follow these steps:

### Step 1: Understand the Setup
We have a triangle ABC with sides:
- AC = b
- BC = a
- AB = c

A semicircle is inscribed such that its diameter lies on side AB, and it touches sides AC and BC.

### Step 2: Area of Triangle
The area of triangle ABC can be expressed as the sum of the areas of triangles ACD and BCD.

Let the area of triangle ABC be denoted as Δ.

### Step 3: Area of Triangles ACD and BCD
The area of triangle ACD can be calculated using the formula:
\[ \text{Area of } ACD = \frac{1}{2} \times AC \times h_1 = \frac{1}{2} \times b \times R \]
where \( R \) is the radius of the semicircle and \( h_1 \) is the height from point C to line AB.

Similarly, the area of triangle BCD can be calculated as:
\[ \text{Area of } BCD = \frac{1}{2} \times BC \times h_2 = \frac{1}{2} \times a \times R \]
where \( h_2 \) is the height from point C to line AB.

### Step 4: Combine the Areas
Now, we can express the area of triangle ABC as:
\[ Δ = \text{Area of } ACD + \text{Area of } BCD \]
Substituting the areas we calculated:
\[ Δ = \frac{1}{2} \times b \times R + \frac{1}{2} \times a \times R \]
Factoring out the common terms:
\[ Δ = \frac{1}{2} R (a + b) \]

### Step 5: Solve for Radius R
Now, we can solve for the radius \( R \):
\[ Δ = \frac{1}{2} R (a + b) \]
Multiplying both sides by 2:
\[ 2Δ = R (a + b) \]
Now, divide both sides by \( (a + b) \):
\[ R = \frac{2Δ}{a + b} \]

### Conclusion
Thus, the radius of the semicircle is given by:
\[ R = \frac{2Δ}{a + b} \]