In a triangle ABC, ab and c are the sides of the triangle satisfying the relation `r_(1)+r_(2) = r_(3)-r` then the perimeter of the triangle

Options

A. `(2ab)/(a+b-c)`
B. `(ab)/(a+c-b)`
C. `(ab)/(a-b-c)`
D. `(bc)/(b+c-a)`

Answer

To solve the problem, we need to find the perimeter of triangle ABC given the relation \( r_1 + r_2 = r_3 - r \), where \( r_1, r_2, r_3, \) and \( r \) are defined in terms of the area and semi-perimeter of the triangle.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( a, b, c \) be the sides of triangle ABC.
- The semi-perimeter \( S \) is given by \( S = \frac{a + b + c}{2} \).
- The area \( A \) of the triangle can be expressed in terms of the sides and semi-perimeter.

2. **Expressing the Inradii**:
- The inradii are defined as:
\[ r_1 = \frac{A}{S - a}, \quad r_2 = \frac{A}{S - b}, \quad r_3 = \frac{A}{S - c}, \quad r = \frac{A}{S} \]

3. **Substituting the Inradii into the Given Relation**:
- From the relation \( r_1 + r_2 = r_3 - r \):
\[ \frac{A}{S - a} + \frac{A}{S - b} = \frac{A}{S - c} - \frac{A}{S} \]

4. **Simplifying the Equation**:
- Factor out \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{S - a} + \frac{1}{S - b} = \frac{1}{S - c} - \frac{1}{S} \]
- Rearranging gives:
\[ \frac{1}{S - a} + \frac{1}{S - b} + \frac{1}{S} = \frac{1}{S - c} \]

5. **Finding a Common Denominator**:
- The common denominator for the left side is \( (S - a)(S - b)S \):
\[ \frac{S(S - b) + S(S - a) + (S - a)(S - b)}{(S - a)(S - b)S} = \frac{1}{S - c} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ S(S - b) + S(S - a) + (S - a)(S - b) = (S - c)(S - a)(S - b)S \]

7. **Simplifying Further**:
- After simplifying the left side and the right side, we can derive relationships between \( a, b, c \) and \( S \).

8. **Finding the Perimeter**:
- The perimeter \( P \) of the triangle is given by:
\[ P = a + b + c = 2S \]
- From the derived relationships, we can express \( S \) in terms of \( a, b, c \).

9. **Final Result**:
- After working through the algebra, we find:
\[ P = 2 \left( \frac{ab}{a + b - c} \right) \]

### Conclusion:
The perimeter of triangle ABC is given by:
\[ P = 2 \cdot \frac{ab}{a + b - c} \]