## Geometry: Common Core (15th Edition)

$x = 20$ $y = 10 \sqrt 5$
The altitude of a right triangle is the geometric mean of the lengths of the two hypotenuse segments. Let's set up that proportion: The geometric mean of two numbers can be found using the following proportion: $\frac{a}{x} = \frac{x}{b}$, where $a$ and $b$ are the lengths of the two hypotenuse segments and $x$ is the length of the altitude. Let's plug in our numbers: $\frac{50 - 40}{x} = \frac{y}{40}$ Use the cross products property to get rid of the fractions: $x^2 = 10 • 40$ Multiply to simplify: $x^2 = 400$ Take the positive square root to solve for $y$: $x = 20$ To find $y$, we know that each leg of the triangle is the geometric mean of the hypotenuse and the hypotenuse that is adjacent to that leg: $\frac{a}{x} = \frac{x}{b}$, where $a$ and $b$ are the length of the hypotenuse and the length of the segment of the hypotenuse closest to the leg, and $x$ is the length of the leg. $\frac{50}{y} = \frac{y}{50 - 40}$ Use the cross products property to get rid of the fractions: $y^2 = 50 • 10$ Multiply to simplify: $y^2 = 500$ Rewrite $500$ as the product of a perfect square and another factor: $y^2 = 100 • 5$ Take the positive square root of each factor to solve for $y$: $y = 10 \sqrt 5$