Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set: 15

Answer

$\sqrt{19}\approx4.359 \text{ }in$

Work Step by Step

Let the right triangle have $a$ and $b$ as the legs and $c$ as the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $ c=\sqrt{20} $ and $ a=1 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 1^2+b^2=(\sqrt{20})^2 \\\\ 1+b^2=20 \\\\ b^2=20-1 \\\\ b^2=19 \\\\ b=\sqrt{19} .\end{array} Hence, the other leg is $ \sqrt{19}\approx4.359 \text{ }in $ long.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.