1. Problem 26: Given a right triangle with hypotenuse 9, legs $x$ and $y$, find $x$ and $y$. 2. Use the Pythagorean theorem: $$x^2 + y^2 = 9^2 = 81$$ 3. Without additional information, $x$ and $y$ can be any pair satisfying $x^2 + y^2 = 81$. --- 1. Problem 27: Right triangle with hypotenuse $x$, legs $y$ and 6, and the leg of length 6 is divided into two segments by the right angle. 2. More information is needed to solve for $x$ and $y$. --- 1. Problem 28: Right triangle with hypotenuse 10, legs 4 and $x$. 2. Use Pythagorean theorem: $$4^2 + x^2 = 10^2$$ 3. Calculate: $$16 + x^2 = 100$$ 4. Subtract 16: $$x^2 = 100 - 16 = 84$$ 5. Take square root: $$x = \sqrt{84} = \sqrt{4 \times 21} = 2\sqrt{21}$$ --- 1. Problem 29: Right triangle with legs 6 and $y$, hypotenuse 14, and other leg $x$. 2. Since two legs are 6 and $y$, hypotenuse 14, use Pythagorean theorem: $$6^2 + y^2 = 14^2$$ 3. Calculate: $$36 + y^2 = 196$$ 4. Subtract 36: $$y^2 = 160$$ 5. Take square root: $$y = \sqrt{160} = \sqrt{16 \times 10} = 4\sqrt{10}$$ --- 1. Problem 30: Right triangle with hypotenuse $x$, legs 12 and $y$, and a segment on $y$ side of length 3, right angle split into two smaller right triangles with hypotenuse $x$ and leg $z$. 2. More information or diagram needed to solve. --- 1. Problem 31: Right triangle with hypotenuse $x$, legs 8 and $y$, other leg 12, segment on hypotenuse marked $z$. 2. More information or diagram needed to solve. --- 1. Problem 36: Altitude divides hypotenuse into segments 45 and 5 inches. Find altitude length. 2. Use geometric mean theorem: altitude $h$ satisfies $$h = \sqrt{45 \times 5}$$ 3. Calculate: $$h = \sqrt{225} = 15$$ --- 1. Problem 37: Error analysis with legs 3 and $x$, hypotenuse 5, incorrect proportions. 2. Correct proportion uses Pythagorean theorem: $$3^2 + x^2 = 5^2$$ 3. Calculate: $$9 + x^2 = 25$$ 4. Subtract 9: $$x^2 = 16$$ 5. Take square root: $$x = 4$$ --- 1. Problem 38: Right triangle with sides $6\sqrt{3}$, 6, and 12. Altitude from right angle to hypotenuse. 2. Hypotenuse is 12, legs $6\sqrt{3}$ and 6. 3. Altitude $h$ satisfies $$h = \frac{6 \times 6\sqrt{3}}{12} = \frac{36\sqrt{3}}{12} = 3\sqrt{3}$$ 4. Segment of hypotenuse adjacent to shorter leg (6) is $$\frac{6^2}{12} = \frac{36}{12} = 3$$ --- Summary of solved values: - Problem 28: $x = 2\sqrt{21}$ - Problem 29: $y = 4\sqrt{10}$ - Problem 36: altitude $= 15$ - Problem 37: $x = 4$ - Problem 38: altitude $= 3\sqrt{3}$, segment adjacent to shorter leg $= 3$ Other problems lack sufficient data or diagrams to solve.