let`ABC` be a right angled triangle at `C. `If the inscribed circle touches the side `AB `at `D` and `(AD) (BD)=11,` then find the area of triangle `ABC.`.

Answer

To find the area of triangle \( ABC \) where \( C \) is the right angle, and the inscribed circle touches side \( AB \) at point \( D \) with the condition \( AD \cdot BD = 11 \), we can follow these steps:

### Step 1: Define the segments
Let:
- \( AD = z \)
- \( BD = y \)
- \( CD = x \)

Since \( D \) is the point where the incircle touches \( AB \), we can express the sides of triangle \( ABC \) in terms of \( x \), \( y \), and \( z \):
- \( AB = AD + BD = z + y \)
- \( AC = AD + CD = z + x \)
- \( BC = BD + CD = y + x \)

### Step 2: Use the given condition
We know from the problem that:
\[ AD \cdot BD = z \cdot y = 11 \]

### Step 3: Apply the Pythagorean theorem
Since triangle \( ABC \) is a right triangle at \( C \), we can apply the Pythagorean theorem:
\[ AC^2 + BC^2 = AB^2 \]
Substituting the expressions we defined:
\[ (z + x)^2 + (y + x)^2 = (z + y)^2 \]

### Step 4: Expand the equation
Expanding both sides:
\[ (z^2 + 2zx + x^2) + (y^2 + 2xy + x^2) = (z^2 + 2zy + y^2) \]
Combining like terms:
\[ z^2 + y^2 + 2zx + 2xy + 2x^2 = z^2 + y^2 + 2zy \]

### Step 5: Simplify the equation
Cancelling \( z^2 \) and \( y^2 \) from both sides:
\[ 2zx + 2xy + 2x^2 = 2zy \]
Dividing through by 2:
\[ zx + xy + x^2 = zy \]

### Step 6: Substitute \( zy \) with 11
Since \( zy = 11 \), we can substitute this into our equation:
\[ zx + xy + x^2 = 11 \]

### Step 7: Rearranging for area
The area \( A \) of triangle \( ABC \) can be expressed as:
\[ A = \frac{1}{2} \times AC \times BC = \frac{1}{2} \times (z + x) \times (y + x) \]
Expanding this gives:
\[ A = \frac{1}{2} \times (zy + zx + xy + x^2) \]
Substituting \( zy = 11 \):
\[ A = \frac{1}{2} \times (11 + zx + xy + x^2) \]

### Step 8: Use the equation from step 6
From the equation \( zx + xy + x^2 = 11 \):
\[ A = \frac{1}{2} \times (11 + 11) = \frac{1}{2} \times 22 = 11 \]

### Conclusion
Thus, the area of triangle \( ABC \) is:
\[ \boxed{11} \]