Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

Chapter 5 - Test - Page 264: 9

Answer

$AD=\sqrt 89$

Work Step by Step

Using Pythagorean's Theorem (for the small triangle in the diagram... $a^{2}+b^{2}=c^{2}$ Substitute in known lengths... $4^{2}+3^{2}=c^{2}$ Simplify... $16+9=c^{2}$ $25=c^{2}$ Square root both sides... $c=5$ or $AC=5$ Using Pythagorean's Theorem for the larger triangle... $a^{2}+b^{2}=c^{2}$ Substitute in known lengths... $5^{2}+8^{2}=c^{2}$ Simplify... $25+64=c^{2}$ $89=c^{2}$ Square root both sides... $c=\sqrt 89$ or $AD=\sqrt 89$

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