Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 7 - Similarity - Chapter Review - Page 482: 19

Answer

$x = 6 \sqrt {2}$ $y = 6 \sqrt {6}$

Work Step by Step

The altitude of a right triangle is the geometric mean of the lengths of the two hypotenuse segments. Let's set up that 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: $\frac{6}{x} = \frac{x}{12}$ Use the cross products property to get rid of the fractions: $x^2 = 72$ Rewrite $72$ as the product of a perfect square and another factor: $x^2 = 36 • 2$ Take the positive square root of each factor to solve for $x$: $x = 6 \sqrt {2}$ 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: $\frac{a}{y} = \frac{y}{b}$, where $a$ and $b$ are the length of the hypotenuse and the length of the segment of the hypotenuse closest to the leg, and $y$ is the length of the leg. $\frac{12 + 6}{y} = \frac{y}{12}$ Use the cross products property to get rid of the fractions: $y^2 = 12(12 + 6)$ Evaluate what is in parentheses first: $y^2 = 12(18)$ Multiply to simplify: $y^2 = 216$ Rewrite $216$ as the product of a perfect square and another factor: $y^2 = 36 • 6$ Take the positive square root of each factor to solve for $y$: $y = 6 \sqrt {6}$
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