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 592: 38

Answer

First mistake: solving for the unknown as if it were the hypotenuse, rather than one of the legs (should be a minus instead of a plus between the squares in the square root). Second mistake: using $m^2$ for units instead of $m$. Correct answer: $15.2$ m

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, solving for one of the legs (the unknown): $a^2=c^2-b^2$ $a=\sqrt{c^2-b^2}$ Plugging in the values, we have: $a=\sqrt{20^2-13^2}$ $a=\sqrt{400-169}$ $a=\sqrt{231}$ $a=15.2$ m However, the student made the mistake of adding instead of subtracting between the squares. The student probably assumed that the unknown was the hypotenuse, rather than one of the legs. The student's second mistake was with the units -- using $m^2$ (area unit) instead of $m$ (length unit).
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.