1. **State the problem:** A 25-foot ladder is placed against a vertical wall, with the foot 7 feet from the base. The top slips down 4 feet. We need to find how far the foot slides, rounded to the nearest tenth of a foot. 2. **Identify the initial situation:** The ladder, wall, and ground form a right triangle. The ladder is the hypotenuse $c=25$ ft, the base is $b=7$ ft, and the height is $a$ (unknown). 3. **Find the initial height $a$ using the Pythagorean theorem:** $$a=\sqrt{c^2 - b^2} = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24 \text{ ft}$$ 4. **After the top slips down 4 feet, the new height is:** $$a_{new} = 24 - 4 = 20 \text{ ft}$$ 5. **Let the new base distance be $b_{new}$. Use Pythagorean theorem again:** $$b_{new} = \sqrt{c^2 - a_{new}^2} = \sqrt{25^2 - 20^2} = \sqrt{625 - 400} = \sqrt{225} = 15 \text{ ft}$$ 6. **Calculate how far the foot slides:** $$\text{slide distance} = b_{new} - b = 15 - 7 = 8 \text{ ft}$$ 7. **Answer rounded to the nearest tenth:** $$8.0 \text{ feet}$$