1. **Problem:** Find $x$ in a right triangle with legs 4 and $x$, hypotenuse 10. 2. **Formula:** Use the Pythagorean theorem: $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. 3. **Apply:** $$4^2 + x^2 = 10^2$$ 4. **Calculate:** $$16 + x^2 = 100$$ 5. **Isolate $x^2$:** $$x^2 = 100 - 16 = 84$$ 6. **Simplify:** $$x = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$$ --- 1. **Problem:** Find $x$ in a right triangle with legs 8 and $x$, hypotenuse 10. 2. **Formula:** Pythagorean theorem: $$8^2 + x^2 = 10^2$$ 3. **Calculate:** $$64 + x^2 = 100$$ 4. **Isolate $x^2$:** $$x^2 = 100 - 64 = 36$$ 5. **Simplify:** $$x = \sqrt{36} = 6$$ --- 1. **Problem:** Find $x$ in a right triangle with legs 8 and 4, and segment $x$ formed by altitude or intersection. 2. **Formula:** For right triangles with altitude to hypotenuse, geometric mean relations apply: $$x = \sqrt{8 \times 4}$$ 3. **Calculate:** $$x = \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}$$ --- 1. **Problem:** Find $x$ in a right triangle with legs 16 and $x$, and segment 8. 2. **Formula:** Geometric mean: $$x = \sqrt{16 \times 8}$$ 3. **Calculate:** $$x = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$$ --- 1. **Problem:** Find $x$ in a right triangle with legs 5 and 6, and segment $x$ at right angle. 2. **Formula:** Geometric mean: $$x = \sqrt{5 \times 6}$$ 3. **Calculate:** $$x = \sqrt{30}$$ (simplified radical) --- 1. **Problem:** Find $x$ in a right triangle with legs 8 and 3, and segment $x$ at right angle. 2. **Formula:** Geometric mean: $$x = \sqrt{8 \times 3}$$ 3. **Calculate:** $$x = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$$ --- 1. **Problem:** Find $x$ and $y$ in a right triangle with leg 20, segment 4, and height $y$. 2. **Formula:** Geometric mean relations: $$x = \sqrt{20 \times 4}$$ $$y = \sqrt{x \times 4}$$ 3. **Calculate $x$:** $$x = \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}$$ 4. **Calculate $y$:** $$y = \sqrt{4\sqrt{5} \times 4} = \sqrt{16\sqrt{5}}$$ Since $y$ involves nested radicals, approximate or leave as is. --- 1. **Problem:** Find $x$ and $y$ in a right triangle with legs 6 and 8, and height $y$ across $x$. 2. **Formula:** Geometric mean relations: $$x = \sqrt{6 \times 8}$$ $$y = \sqrt{6 \times 8}$$ 3. **Calculate:** $$x = y = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}$$