Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 4 - Test - Page 218: 3

Answer

$EB=6$

Work Step by Step

We are given that $AD=5$ and $DE=4$. Altitude $DE$ forms a right triangle, $\triangle ADE$. Using the Pythagorean Theorem... $a^{2}+b^{2}=c^{2}$ $a^{2}+4^{2}=5^{2}$ $a^{2}+16=25$ Subtract 16 from both sides... $a^{2}=9$ Square root both sides... $a=3$ or $AE=3$ Opposite sides of a parallelogram are congruent. Therefore, $DC=AB=9$. Using the Segment Addition Postulate... $AE+EB=AB$ Substitute in known lengths... $3+EB=9$ Subtract three from both sides... $EB=6$
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