Given a right-angled triangle with perimeter $18$. The sum of the squares
of the three side lengths is $128$. What is the area of the triangle?

1. Let the side lengths of the right-angled triangle be \(a\), \(b\), and \(c\), where \(c\) is the hypotenuse. We are given the following conditions:
   \[
   a + b + c = 18
   \]
   \[
   a^2 + b^2 + c^2 = 128
   \]
   \[
   a^2 + b^2 = c^2
   \]

2. From the Pythagorean theorem, we know:
   \[
   a^2 + b^2 = c^2
   \]
   Substituting this into the second equation:
   \[
   c^2 + c^2 = 128 \implies 2c^2 = 128 \implies c^2 = 64 \implies c = \sqrt{64} = 8
   \]

3. Now, substitute \(c = 8\) into the perimeter equation:
   \[
   a + b + 8 = 18 \implies a + b = 10
   \]

4. We also know from the Pythagorean theorem that:
   \[
   a^2 + b^2 = 64
   \]

5. To find \(ab\), we use the identity \((a + b)^2 = a^2 + 2ab + b^2\):
   \[
   (a + b)^2 = 10^2 = 100
   \]
   \[
   a^2 + b^2 + 2ab = 100
   \]
   Substituting \(a^2 + b^2 = 64\) into the equation:
   \[
   64 + 2ab = 100 \implies 2ab = 36 \implies ab = 18
   \]

6. The area of the right-angled triangle is given by:
   \[
   \text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} \times 18 = 9
   \]

The final answer is \(\boxed{9}\).