1. **State the problem:** We have a right triangle with hypotenuse length 17 cm. The height is 7 cm longer than the base. We need to find the base and height. 2. **Set variables:** Let the base be $x$ cm. Then the height is $x + 7$ cm. 3. **Use the Pythagorean theorem:** For a right triangle with legs $a$ and $b$ and hypotenuse $c$, the relation is $$a^2 + b^2 = c^2$$ 4. **Apply the formula:** Here, $$x^2 + (x+7)^2 = 17^2$$ 5. **Expand and simplify:** $$x^2 + (x+7)^2 = 289$$ $$x^2 + (x^2 + 14x + 49) = 289$$ $$2x^2 + 14x + 49 = 289$$ 6. **Bring all terms to one side:** $$2x^2 + 14x + 49 - 289 = 0$$ $$2x^2 + 14x - 240 = 0$$ 7. **Divide entire equation by 2 to simplify:** $$\cancel{2}x^2 + \cancel{2}7x - \cancel{2}120 = 0$$ $$x^2 + 7x - 120 = 0$$ 8. **Factor the quadratic:** We look for two numbers that multiply to $-120$ and add to $7$. These are $15$ and $-8$. $$x^2 + 15x - 8x - 120 = 0$$ $$(x^2 + 15x) - (8x + 120) = 0$$ $$x(x + 15) - 8(x + 15) = 0$$ $$(x - 8)(x + 15) = 0$$ 9. **Solve for $x$:** $$x - 8 = 0 \Rightarrow x = 8$$ $$x + 15 = 0 \Rightarrow x = -15$$ Since length cannot be negative, $x = 8$ cm. 10. **Find the height:** $$x + 7 = 8 + 7 = 15$$ cm. **Final answer:** The base is 8 cm and the height is 15 cm.