Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10: 0134467795
ISBN 13: 978-0-13446-779-5

Chapter 8 - Geometry - 8.8 Pythagorean Theorem - 8.8 Exercises - Page 591: 23


$12$ mm

Work Step by Step

We know that for a right triangle the Pythagorean Theorem states: $leg_1^2+leg_2^2=hypotenuse^2$ $a^2+b^2=c^2$ Thus we have: $a^2+16^2=20^2$ $a^2+256=400$ $a^2=400-256$ $a^2=144$ $a=\sqrt{144}$ $a=12$ mm
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