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: 30


$14 \text{ and } 14\sqrt{2}$

Work Step by Step

Since the acute angle of the right triangle is $45^o,$ then the right triangle is isosceles. Hence, the legs both measure $14$ units. Let the right triangle have $a$ and $b$ as the legs and $c$ as the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $ a=14 $ and $ b=14 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 14^2+14^2=c^2 \\\\ 196+196=c^2 \\\\ 2(196)=c^2 \\\\ c=\sqrt{196(2)} \\\\ c=\sqrt{(14)^2\cdot2} \\\\ c=14\sqrt{2} .\end{array} Hence, the missing sides have measure $ 14 \text{ and } 14\sqrt{2} $ units.
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