Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.3 - Simplifying Radicals, the Distance Formula, and Circles - 7.3 Exercises - Page 460: 117


$a=2\sqrt 14$

Work Step by Step

According to the Pythagorean Theorem, if $a$ and $b$ are the lengths of the shorter sides of a right triangle and $c$ is the length of the longest side, then $a^{2}+b^{2}=c^{2}$. Therefore, $a^{2}+5^{2}=9^{2}$. $a^{2}+25=81$ Subtract 25 from both sides. $a^{2}=56$ Take the square root of both sides. $a=\sqrt 56=\sqrt (4\times14)=\sqrt 4\times\sqrt14=2\sqrt 14$
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