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 - 8-1 The Pythagorean Theorem and It's Converse - Practice and Problem-Solving Exercises - Page 498: 61

Answer

$15 \sqrt 2$

Work Step by Step

Let's rewrite the problem as a fraction so we can see what we are working with: $\frac{30}{\sqrt 2}$ We want to get rid of the radical in the denominator, so we multiply both the numerator and denominator by $\sqrt 2$: $\frac{30}{\sqrt 2} • \frac{\sqrt 2}{\sqrt 2}$ Multiply to simplify: $\frac{30 \sqrt 2}{\sqrt 4}$ Take the square root of the denominator: $\frac{30 \sqrt 2}{2}$ Divide the numerator and denominator by their greatest common factor to simplify: $15 \sqrt 2$
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