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: 115


$b=8\sqrt 2$

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, $4^{2}+b^{2}=12^{2}$. $16+b^{2}=144$ Subtract 16 from both sides. $b^{2}=128$ Take the square root of both sides. $b=\sqrt 128=\sqrt (64\times2)=\sqrt 64\times\sqrt2=8\sqrt 2$
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