1. **Stating the problem:** We have two right triangles with given side lengths and an unknown segment $x$. We need to find $x$ in each case. --- ### Problem 1 (Top-left triangle): - Legs: 20 and 15 - Segment $x$ is perpendicular to the hypotenuse. ### Problem 2 (Bottom-left triangle): - Legs: 12 and $x$ - Smaller inner right triangle with legs 18 and 36 --- 2. **Formula and rules:** For right triangles, the Pythagorean theorem applies: $$a^2 + b^2 = c^2$$ where $a$ and $b$ are legs and $c$ is the hypotenuse. When a segment $x$ is perpendicular to the hypotenuse, it creates two smaller right triangles similar to the original. --- ### Solution for Problem 1: 1. Calculate the hypotenuse $c$: $$c = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25$$ 2. The segment $x$ perpendicular to the hypotenuse satisfies: $$x = \frac{ab}{c}$$ where $a=20$, $b=15$, $c=25$. 3. Substitute values: $$x = \frac{20 \times 15}{25} = \frac{300}{25}$$ 4. Simplify fraction: $$x = \frac{\cancel{300}^{12} \times 25}{\cancel{25}} = 12$$ --- ### Solution for Problem 2: 1. The smaller triangle with legs 18 and 36 is similar to the larger triangle with legs 12 and $x$. 2. Set up the ratio of corresponding legs: $$\frac{12}{18} = \frac{x}{36}$$ 3. Cross multiply: $$12 \times 36 = 18 \times x$$ 4. Calculate: $$432 = 18x$$ 5. Solve for $x$: $$x = \frac{432}{18}$$ 6. Simplify fraction: $$x = \frac{\cancel{432}^{24} \times 18}{\cancel{18}} = 24$$ --- **Final answers:** - For Problem 1: $x = 12$ - For Problem 2: $x = 24$