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 1-10 - Cumulative Review - Page 696: 32



Work Step by Step

We draw a line straight through the middle of the triangle to form two right triangles. We see that the base of this right triangle will be 2 and its hypotenuse will be 4. Thus: $$ h=\sqrt{4^2-2^2}\\\ h=\sqrt{12}\\ h=2\sqrt{3}$$
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