Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 8 - Right Triangles and Trigonometry - Chapter Review - Page 535: 9


$x = 5\sqrt 2$

Work Step by Step

In a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, the hypotenuse is $\sqrt 2$ times each leg. Let's write an equation to solve for $x$, one of the legs: $10 = \sqrt 2(x)$ Divide each side of the equation by $\sqrt 2$ to solve for $x$: $x = \frac{10}{\sqrt 2}$ To simplify this fraction, we multiply both the numerator and denominator by the denominator: $x = \frac{10}{\sqrt 2} • \frac{\sqrt 2}{\sqrt 2}$ Multiply to simplify: $x = \frac{10\sqrt 2}{2}$ Divide the numerator and denominator by their greatest common factor, $2$: $x = 5\sqrt 2$
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.