1. **Problem statement:** A ladder 16 m long is placed with its foot 3 m from a building. We want to find how much further the foot must be moved away from the building to lower the top of the ladder by 2 m. 2. **Understanding the problem:** Initially, the ladder reaches a certain height on the building. After moving the foot further, the height decreases by 2 m. 3. **Formula used:** We use the Pythagorean theorem for right triangles: $$a^2 + b^2 = c^2$$ where $c$ is the ladder length (16 m), $a$ is the distance from the building, and $b$ is the height reached. 4. **Initial height calculation:** $$b_1 = \sqrt{16^2 - 3^2} = \sqrt{256 - 9} = \sqrt{247}$$ 5. **New height after lowering by 2 m:** $$b_2 = b_1 - 2 = \sqrt{247} - 2$$ 6. **New distance from building:** Let this be $x$. Using Pythagoras again: $$x^2 + b_2^2 = 16^2 = 256$$ 7. Substitute $b_2$: $$x^2 + (\sqrt{247} - 2)^2 = 256$$ 8. Expand the square: $$(\sqrt{247} - 2)^2 = 247 - 2 \times 2 \times \sqrt{247} + 4 = 251 - 4\sqrt{247}$$ 9. So: $$x^2 + 251 - 4\sqrt{247} = 256$$ 10. Rearranged: $$x^2 = 256 - 251 + 4\sqrt{247} = 5 + 4\sqrt{247}$$ 11. Calculate approximate value: $$\sqrt{247} \approx 15.716$$ $$x^2 \approx 5 + 4 \times 15.716 = 5 + 62.864 = 67.864$$ 12. Find $x$: $$x \approx \sqrt{67.864} \approx 8.24$$ 13. **Distance moved:** $$\text{distance moved} = x - 3 = 8.24 - 3 = 5.24$$ **Final answer:** The foot of the ladder must be moved approximately 5.24 meters further from the building to lower the top by 2 meters.