Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 5 - Test - Page 264: 7b


$a=2 \sqrt 7$

Work Step by Step

Using Pythagorean's Theorem... $a^{2}+b^{2}=c^{2}$ Substitute in known lengths... $a^{2}+6^{2}=8^{2}$ Simplify... $a^{2}+36=64$ Subtract 36 from both sides... $a^{2}=28$ Square root both sides... $a=\sqrt 28$ Pull out the $\sqrt4$ from the radicand and you have... $a=2 \sqrt7$
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