Given that `Delta = 6, r_(1) = 3, r_(3) = 6`
Inradius is equal to

Options

A. 2
B. 1
C. 1.5
D. 2.5

Explanation

To find the inradius of the triangle given the area (Δ), \( r_1 \), and \( r_3 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Area of the triangle, \( \Delta = 6 \)
- \( r_1 = 3 \)
- \( r_3 = 6 \)

2. **Use the Formula for Inradius**:
The inradius \( r \) can be calculated using the formula:
\[ r = \frac{\Delta}{s} \]
where \( s \) is the semi-perimeter of the triangle.

3. **Relate \( r_1 \) and \( r_3 \) to \( s \)**:
We know:
\[ r_1 = \frac{\Delta}{s-a}, \quad r_2 = \frac{\Delta}{s-b}, \quad r_3 = \frac{\Delta}{s-c} \]
From the given values:
\[ 3 = \frac{6}{s-a} \quad \text{(1)} \]
\[ 6 = \frac{6}{s-c} \quad \text{(2)} \]

4. **Solve for \( s-a \) and \( s-c \)**:
From equation (1):
\[ s-a = \frac{6}{3} = 2 \implies s = a + 2 \quad \text{(3)} \]
From equation (2):
\[ s-c = \frac{6}{6} = 1 \implies s = c + 1 \quad \text{(4)} \]

5. **Set up equations for \( s-b \)**:
We can also express \( s-b \) using \( r_2 \):
\[ r_2 = \frac{\Delta}{s-b} \implies 3 = \frac{6}{s-b} \implies s-b = 2 \quad \text{(5)} \]

6. **Combine Equations**:
From equations (3), (4), and (5):
- From (3): \( s = a + 2 \)
- From (4): \( s = c + 1 \)
- From (5): \( s = b + 2 \)

7. **Express \( a, b, c \) in terms of \( s \)**:
From (3):
\[ a = s - 2 \]
From (4):
\[ c = s - 1 \]
From (5):
\[ b = s - 2 \]

8. **Calculate \( a + b + c \)**:
\[ a + b + c = (s - 2) + (s - 2) + (s - 1) = 3s - 5 \]

9. **Find \( s \)**:
The semi-perimeter \( s \) is also given by:
\[ s = \frac{a + b + c}{2} \]
Setting \( 3s - 5 = 2s \):
\[ 3s - 5 = 2s \implies s = 5 \]

10. **Calculate Inradius**:
Now substituting \( s \) back into the inradius formula:
\[ r = \frac{\Delta}{s} = \frac{6}{5} = 1.2 \]

### Final Answer:
The inradius \( r \) of the triangle is \( 1.2 \).