Elementary Geometry for College Students (6th Edition)

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

Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 241: 32



Work Step by Step

In $\triangle ABC$ $AC^{2}=AB^{2}+BC^{2}$ $AC^{2}=4^{2}+3^{2}$ so AC = 5 In $\triangle ACD$ $AD^{2}=AC^{2}+DC^{2}$ $AD^{2}=5^{2}+12^{2}$ $AD^{2}=25+144$ $AD^{2}=169$ $AD=13$
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