Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 5 - Section 5.5 - Special Right Triangles - Exercises - Page 257: 7

Answer

$XZ = 10$ $YZ= 10$

Work Step by Step

By the given Figure $\angle Z=90^{\circ}$ given $\angle X=45^{\circ}$ given So remaining$ \angle Y=45^{\circ}$ now $XY= 10 \sqrt 2$ given By Theorem 5.5.1/ (45-45-90 Theorem) $XZ = 10$ $YZ= 10$
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