Geometry: Common Core (15th Edition)

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

Chapter 8 - Right Triangles and Trigonometry - 8-2 Special Right Triangles - Practice and Problem-Solving Exercises - Page 503: 8


$x = \sqrt 2$ $y = 2$

Work Step by Step

In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, we know that the legs are equal to one another. Since one leg measures $\sqrt 2$, $x$ also equals $\sqrt 2$. The hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $y$, the length of the hypotenuse: $y = \sqrt 2(\sqrt 2)$ Multiply the radicals: $y = \sqrt 4$ Take the positive square root to solve for $y$: $y = 2$
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