1. **Problem statement:** A surveyor wants to find the distance $d$ across a lake. The given measurements form a right triangle with legs 90 ft and $(75 + d)$ ft, and the hypotenuse is 225 ft. 2. **Formula used:** We apply the Pythagorean theorem for right triangles: $$a^2 + b^2 = c^2$$ where $a$ and $b$ are the legs and $c$ is the hypotenuse. 3. **Set up the equation:** $$90^2 + (75 + d)^2 = 225^2$$ 4. **Calculate squares:** $$8100 + (75 + d)^2 = 50625$$ 5. **Expand the binomial:** $$(75 + d)^2 = 75^2 + 2 \times 75 \times d + d^2 = 5625 + 150d + d^2$$ 6. **Substitute back:** $$8100 + 5625 + 150d + d^2 = 50625$$ 7. **Combine constants:** $$13725 + 150d + d^2 = 50625$$ 8. **Isolate terms:** $$d^2 + 150d + 13725 - 50625 = 0$$ $$d^2 + 150d - 36900 = 0$$ 9. **Solve quadratic equation:** Use the quadratic formula: $$d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=150$, $c=-36900$. 10. **Calculate discriminant:** $$\Delta = 150^2 - 4 \times 1 \times (-36900) = 22500 + 147600 = 170100$$ 11. **Calculate roots:** $$d = \frac{-150 \pm \sqrt{170100}}{2}$$ $$\sqrt{170100} \approx 412.4$$ 12. **Find two possible values:** $$d_1 = \frac{-150 + 412.4}{2} = \frac{262.4}{2} = 131.2$$ $$d_2 = \frac{-150 - 412.4}{2} = \frac{-562.4}{2} = -281.2$$ 13. **Interpretation:** Distance cannot be negative, so $$d = 131.2 \text{ feet}$$ **Final answer:** The distance across the lake is approximately $131.2$ feet.