$AB$ is a chord of length $6$ in a circle of radius $5$ and centre $O$. A square is inscribed in the sector $OAB$ with two vertices on the circumference and two sides parallel to $ AB$. Find the area of the square.
1. **Construct the initial square:**
   - Consider the square $ABCD$ with side length $6$ such that $C$ and $D$ lie outside the circle.
   - Since $OC = OD$ by symmetry, there exists a homothety about $O$ that maps $C$ and $D$ to $C'$ and $D'$ such that $C'$ and $D'$ lie on the circle.

2. **Calculate the distance $OC$ and $OD$:**
   - The distance from the center $O$ to the vertices $C$ and $D$ can be calculated using the Pythagorean theorem.
   - Since $AB = 6$, the midpoint $M$ of $AB$ is $3$ units away from $A$ and $B$.
   - The distance $OM$ (perpendicular distance from $O$ to $AB$) can be found using the Pythagorean theorem in the right triangle $OMA$:
     \[
     OM = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4
     \]
   - Therefore, $OC = OD = \sqrt{OM^2 + MC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

3. **Determine the homothety ratio:**
   - The homothety ratio $k$ is the ratio of the distances from $O$ to $C$ and $O$ to $C'$.
   - Since $OC = 5$ and $OC' = 5$ (radius of the circle), the homothety ratio is:
     \[
     k = \frac{OC'}{OC} = \frac{5}{5} = 1
     \]
   - This implies that the square $A'B'C'D'$ is congruent to the square $ABCD$.

4. **Calculate the area of the inscribed square:**
   - Since the homothety ratio is $1$, the area of the inscribed square $A'B'C'D'$ is the same as the area of the original square $ABCD$.
   - The area of the square $ABCD$ is:
     \[
     \text{Area of } ABCD = 6^2 = 36
     \]

The final answer is $\boxed{36}$