## Geometry: Common Core (15th Edition)

$y = 3 \sqrt 3$ $x = 6 \sqrt 3$
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: $\frac{3}{y} = \frac{y}{9}$ Use the cross products property to get rid of the fractions: $y^2 = 27$ Rewrite $48$ as the product of a perfect square and another factor: $y^2 = 9 • 3$ Take the positive square root of each factor to solve for $y$: $y = 3 \sqrt 3$ To find $x$, 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. $\frac{9 + 3}{x} = \frac{x}{9}$ Use the cross products property to get rid of the fractions: $x^2 = 12 • 9$ Multiply to simplify: $x^2 = 108$ Rewrite $108$ as the product of a perfect square and another factor: $x^2 = 36 • 3$ Take the positive square root of each factor to solve for $x$: $x = 6 \sqrt 3$