1. **Problem statement:** A ladder leans against a wall, touching the top of a fence 1.5 m high and 1 m away from the wall. We need to find the minimum length of the ladder. 2. **Setup and variables:** Let $x$ be the height on the wall where the ladder rests, and $L$ be the length of the ladder. The fence is 1.5 m high and 1 m from the wall. 3. **Using the Pythagorean theorem:** The ladder forms two right triangles: one from the ground to the fence top, and one from the fence top to the wall. 4. The ladder length squared is: $$L^2 = (1 + d)^2 + x^2$$ where $d$ is the horizontal distance from the fence to the wall along the ground. 5. The ladder touches the fence top at height 1.5 m, so the ladder length from ground to fence top is: $$L_1 = \sqrt{1^2 + 1.5^2} = \sqrt{1 + 2.25} = \sqrt{3.25}$$ 6. The ladder continues from fence top to wall at height $x$, so the horizontal distance $d$ satisfies: $$\frac{x - 1.5}{d} = \frac{1.5}{1} = 1.5$$ which gives $$d = \frac{x - 1.5}{1.5}$$ 7. Total horizontal distance from ground to wall is: $$1 + d = 1 + \frac{x - 1.5}{1.5} = \frac{x + 0}{1.5}$$ 8. Ladder length squared is: $$L^2 = x^2 + \left(\frac{x}{1.5}\right)^2 = x^2 + \frac{x^2}{2.25} = x^2 \left(1 + \frac{1}{2.25}\right) = x^2 \times \frac{3.25}{2.25}$$ 9. To minimize $L$, minimize $x$ subject to $x \geq 1.5$ (since ladder must reach at least fence height). 10. The minimum $L$ occurs at $x = 1.5$: $$L = x \sqrt{\frac{3.25}{2.25}} = 1.5 \times \sqrt{\frac{3.25}{2.25}} = 1.5 \times \frac{\sqrt{3.25}}{1.5} = \sqrt{3.25} \approx 1.803$$ **Final answer:** The minimum length of the ladder is approximately $1.803$ meters.