1. **State the problem:** We have a right triangle with vertices at (0,0), (4,4), and a point on the x-axis (4,0). A square is inscribed inside this triangle such that its bottom left corner lies on the x-axis and its top right corner touches the hypotenuse. We need to find the side length $x$ of this square. 2. **Identify the hypotenuse equation:** The hypotenuse connects points (0,0) and (4,4). The slope is $m=\frac{4-0}{4-0}=1$, so the line equation is $$y=x.$$ 3. **Set coordinates of the square:** Let the bottom left corner of the square be at $(a,0)$ on the x-axis. Since the square has side length $x$, its top right corner is at $(a+x,x)$. 4. **Condition for the top right corner:** This point lies on the hypotenuse, so it must satisfy $$y=x,$$ thus $$x = a + x.$$ 5. **Solve for $a$:** From the equation $$x = a + x,$$ subtract $x$ from both sides: $$\cancel{x} = a + \cancel{x} \implies 0 = a,$$ so $a=0$. 6. **Check the square fits inside the triangle:** The square's bottom left corner is at $(0,0)$, and top right corner at $(x,x)$ on the hypotenuse. The square must fit inside the triangle bounded by $x=0$, $y=0$, and $y=x$ for $0 \leq x \leq 4$. 7. **Check the top right corner does not exceed the triangle's boundary:** Since the hypotenuse ends at $(4,4)$, the maximum $x$ is 4. 8. **Check the other corner of the square on the x-axis:** The bottom right corner is at $(x,0)$, which must be less than or equal to 4. 9. **Conclusion:** The side length $x$ can be at most 4. **Final answer:** $$\boxed{4}$$