Elementary Geometry for College Students (5th Edition)

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

Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 250: 8b

Answer

$\overline{DE}=6$

Work Step by Step

Let $b=\overline{DE}$ $12^{2}=(6\sqrt 3)^{2}+b^{2}$ $144=108+b^{2}$ Subtract 108 on both sides $36=b^{2}$ $6=b$
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