In triangle ABC, let `angle C = pi//2`. If r is the inradius and R is circumradius of the triangle, then `2(r + R)` is equal to

Options

A. `a + b`
B. `b + c`
C. `c + a`
D. `a + b + c`

Explanation

To solve the problem, we need to determine the value of \(2(r + R)\) in triangle \(ABC\) where \(\angle C = \frac{\pi}{2}\) (which means triangle \(ABC\) is a right triangle with the right angle at \(C\)). Here, \(r\) is the inradius and \(R\) is the circumradius of the triangle.

### Step-by-step Solution:

1. **Identify the properties of the triangle**:
- Since \(\angle C = \frac{\pi}{2}\), triangle \(ABC\) is a right triangle. For a right triangle, the circumradius \(R\) is given by:
\[ R = \frac{c}{2} \]
where \(c\) is the length of the side opposite to angle \(C\).

2. **Calculate the inradius \(r\)**:
- The inradius \(r\) of a right triangle can be calculated using the formula:
\[ r = \frac{a + b - c}{2} \]
where \(a\) and \(b\) are the lengths of the other two sides.

3. **Calculate the semi-perimeter \(s\)**:
- The semi-perimeter \(s\) of triangle \(ABC\) is given by:
\[ s = \frac{a + b + c}{2} \]

4. **Express \(r\) in terms of \(s\)**:
- We can rewrite \(r\) using \(s\):
\[ r = s - c \]

5. **Substitute \(R\) and \(r\) into the expression \(2(r + R)\)**:
- Now we substitute \(R\) and \(r\) into the expression:
\[ 2(r + R) = 2\left((s - c) + \frac{c}{2}\right) \]

6. **Simplify the expression**:
- Distributing the 2:
\[ 2(r + R) = 2(s - c) + c = 2s - 2c + c = 2s - c \]

7. **Substituting \(s\)**:
- Recall that \(s = \frac{a + b + c}{2}\), so:
\[ 2s = a + b + c \]
Therefore:
\[ 2s - c = a + b + c - c = a + b \]

8. **Final Result**:
- Hence, we conclude that:
\[ 2(r + R) = a + b \]

### Conclusion:
The value of \(2(r + R)\) is equal to \(a + b\).