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: 62

Answer

$\frac{16 \sqrt 3}{3}$

Work Step by Step

$\frac{16}{\sqrt 3}$ We want to get rid of the radical in the denominator, so we multiply both the numerator and denominator by $\sqrt 3$: $\frac{16}{\sqrt 3} • \frac{\sqrt 3}{\sqrt 3}$ Multiply to simplify: $\frac{16 \sqrt 3}{\sqrt 9}$ Take the square root of the denominator: $\frac{16 \sqrt 3}{3}$
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