Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set - Page 677: 34


$4\sqrt{2} \text{ and } 4\sqrt{2}$

Work Step by Step

Since the given acute angle of the right triangle is $45^o,$ then the the legs ($a$ and $a$), measure $\dfrac{\sqrt{2}}{2}\cdot c,$ where $c$ is the hypotenuse. Hence, the legs measure \begin{array}{l}\require{cancel} \dfrac{\sqrt{2}}{2}\cdot c=a \\\\ \dfrac{\sqrt{2}}{2}\cdot 8=a \\\\ 4\sqrt{2}=a \\\\ a=4\sqrt{2} .\end{array} Hence, the missing sides have measure $ 4\sqrt{2} \text{ and } 4\sqrt{2} $ units.
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