In `Delta ABC`, if a = 10 and `b cot B + c cot C = 2(r + R)` then the maximum area of `DeltaABC` will be

Options

A. 50
B. `sqrt50`
C. 25
D. 5

Explanation

To find the maximum area of triangle ABC given that \( a = 10 \) and \( b \cot B + c \cot C = 2(r + R) \), we can follow these steps:

### Step 1: Understand the given equation
We start with the equation:
\[ b \cot B + c \cot C = 2(r + R) \]
where \( r \) is the inradius and \( R \) is the circumradius of triangle ABC.

### Step 2: Express \( b \cot B \) and \( c \cot C \)
Using the relationships:
\[ b \cot B = \frac{b \cos B}{\sin B} \quad \text{and} \quad c \cot C = \frac{c \cos C}{\sin C} \]
we can rewrite the left side of the equation.

### Step 3: Substitute \( b \) and \( c \) in terms of \( r \) and \( R \)
From the formula for the circumradius \( R \) and inradius \( r \):
\[ b = 2R \sin B \quad \text{and} \quad c = 2R \sin C \]
Now substituting these into the equation gives us:
\[ 2R \sin B \cot B + 2R \sin C \cot C = 2(r + R) \]

### Step 4: Simplify the equation
This simplifies to:
\[ 2R (\sin B \cot B + \sin C \cot C) = 2(r + R) \]
Dividing both sides by 2, we have:
\[ R (\sin B \cot B + \sin C \cot C) = r + R \]

### Step 5: Use the property of angles
Using the identity \( \cot B = \frac{\cos B}{\sin B} \) and \( \cot C = \frac{\cos C}{\sin C} \), we can express:
\[ \sin B \cot B + \sin C \cot C = \cos B + \cos C \]
Thus, our equation becomes:
\[ R (\cos B + \cos C) = r + R \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ R (\cos B + \cos C - 1) = r \]

### Step 7: Use the relationship between \( r \) and \( R \)
From the known relationship in triangles, we have:
\[ r = \frac{A}{s} \quad \text{and} \quad R = \frac{abc}{4A} \]
where \( A \) is the area and \( s \) is the semi-perimeter.

### Step 8: Find the maximum area
To maximize the area \( A \), we can use the formula for the area of a triangle:
\[ A = \frac{1}{2}ab \sin C \]
Given \( a = 10 \), we can use the property of right triangles since we deduced that \( \angle A = 90^\circ \).

### Step 9: Use the Pythagorean theorem
Since \( \angle A = 90^\circ \), we have:
\[ a^2 = b^2 + c^2 \]
Substituting \( a = 10 \):
\[ 10^2 = b^2 + c^2 \implies b^2 + c^2 = 100 \]

### Step 10: Use AM-GM inequality
Using the AM-GM inequality:
\[ \frac{b^2 + c^2}{2} \geq \sqrt{b^2c^2} \]
This implies:
\[ 50 \geq bc \]
Thus, the maximum area \( A \) can be calculated as:
\[ A = \frac{1}{2}bc \leq \frac{1}{2} \cdot 50 = 25 \]

### Conclusion
Therefore, the maximum area of triangle ABC is:
\[ \boxed{25} \]