1. **Problem statement:** Find $x$, $y$, and $z$ in the given figures. 2. **Step 1: Triangle with hypotenuse 5, legs 3 and $x$** - Use the Pythagorean theorem: $$x^2 + 3^2 = 5^2$$ - Calculate: $$x^2 + 9 = 25$$ - Simplify: $$x^2 = 25 - 9 = 16$$ - Since $x > 0$, $$x = \sqrt{16} = 4$$ 3. **Step 2: Smaller right triangle inside with legs 2 and 3, base $y$** - Use the Pythagorean theorem: $$2^2 + y^2 = x^2$$ - Substitute $x=4$: $$4 + y^2 = 16$$ - Simplify: $$y^2 = 16 - 4 = 12$$ - Since $y > 0$, $$y = \sqrt{12} = 2\sqrt{3}$$ 4. **Step 3: Quadrilateral with diagonal $x$, sides 13, 8, 5, and height $y$** - Use Pythagorean theorem on right triangle with sides 5, $y$, and diagonal $x$: $$y^2 + 5^2 = x^2$$ - Use Pythagorean theorem on right triangle with sides 8, $y$, and side 13: $$8^2 + y^2 = 13^2$$ - Calculate: $$64 + y^2 = 169$$ - Simplify: $$y^2 = 169 - 64 = 105$$ - Calculate $x$ from first equation: $$x^2 = y^2 + 25 = 105 + 25 = 130$$ - So, $$x = \sqrt{130}$$ and $$y = \sqrt{105}$$ 5. **Step 4: Polygon with three adjacent right angles and sides 1, $x$, $y$, $z$** - Since three adjacent right angles form a corner, the polygon is an L-shape. - Using the right angles and side lengths, the unknown sides satisfy: $$x = 1$$ $$y = 1$$ $$z = \sqrt{2}$$ (by Pythagorean theorem on the diagonal of a unit square) **Final answers:** - For the first triangle: $$x = 4, y = 2\sqrt{3}$$ - For the quadrilateral: $$x = \sqrt{130}, y = \sqrt{105}$$ - For the polygon: $$x = 1, y = 1, z = \sqrt{2}$$