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: 21

Answer

$y = 4 \sqrt {3}$ $x = 2 \sqrt {21}$

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: $\frac{a}{y} = \frac{y}{b}$, where $a$ and $b$ are the lengths of the two hypotenuse segments and $y$ is the length of the altitude. Let's plug in our numbers: $\frac{14 - 8}{y} = \frac{y}{8}$ Use the cross products property to get rid of the fractions: $y^2 = 6 • 8$ Multiply to simplify: $y^2 = 48$ Rewrite $48$ as the product of a perfect square and another factor: $y^2 = 16 • 3$ Take the positive square root of each factor to solve for $y$: $y = 4 \sqrt {3}$ To find $x$, we know that each leg of the triangle is the geometric mean of the hypotenuse and the hypotenuse segment 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. $\frac{14}{x} = \frac{x}{14 - 8}$ Use the cross products property to get rid of the fractions: $x^2 = 14 • 6$ Multiply to simplify: $x^2 = 84$ Rewrite $84$ as the product of a perfect square and another number: $x^2 = 4 • 21$ Take the positive square root to solve for $x$: $x = 2 \sqrt {21}$
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