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: 118


$a=6\sqrt 2$

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, $a^{2}+7^{2}=11^{2}$. $a^{2}+49=121$ Subtract 49 from both sides. $a^{2}=72$ Take the square root of both sides. $a=\sqrt 72=\sqrt (36\times2)=\sqrt 36\times\sqrt2=6\sqrt 2$
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