Two sides of triangle ABC are 5 and 12. Area of `DeltaABC=30` Find 2R+r where R is circumradius and r is inradius.

Answer

To solve the problem, we need to find \(2R + r\) where \(R\) is the circumradius and \(r\) is the inradius of triangle \(ABC\). We are given two sides of the triangle as \(5\) and \(12\), and the area of the triangle as \(30\).

### Step-by-step Solution:

1. **Identify the triangle type using the area**:
The area of triangle \(ABC\) can be expressed using the formula:
\[ \text{Area} = \frac{1}{2} \times a \times b \times \sin(\theta) \]
where \(a = 5\), \(b = 12\), and \(\text{Area} = 30\). Therefore, we have:
\[ 30 = \frac{1}{2} \times 5 \times 12 \times \sin(\theta) \]
Simplifying this gives:
\[ 30 = 30 \sin(\theta) \implies \sin(\theta) = 1 \]
This implies that \(\theta = 90^\circ\), indicating that triangle \(ABC\) is a right triangle.

2. **Determine the third side using the Pythagorean theorem**:
Since triangle \(ABC\) is a right triangle, we can find the hypotenuse \(c\) using:
\[ c = \sqrt{a^2 + b^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \]
Thus, the sides of the triangle are \(5\), \(12\), and \(13\).

3. **Calculate the semi-perimeter \(s\)**:
The semi-perimeter \(s\) is given by:
\[ s = \frac{a + b + c}{2} = \frac{5 + 12 + 13}{2} = \frac{30}{2} = 15 \]

4. **Calculate the circumradius \(R\)**:
The circumradius \(R\) for a right triangle can be calculated using the formula:
\[ R = \frac{c}{2} = \frac{13}{2} \]

5. **Calculate the inradius \(r\)**:
The inradius \(r\) can be calculated using the formula:
\[ r = \frac{\text{Area}}{s} = \frac{30}{15} = 2 \]

6. **Find \(2R + r\)**:
Now we can find \(2R + r\):
\[ 2R = 2 \times \frac{13}{2} = 13 \]
Therefore,
\[ 2R + r = 13 + 2 = 15 \]

### Final Answer:
\[ 2R + r = 15 \]