Elementary Geometry for College Students (5th Edition)

Published by Brooks Cole
ISBN 10: 1439047901
ISBN 13: 978-1-43904-790-3

Chapter 9 - Section 9.2 - Pyramids, Area, and Volume - Exercises - Page 423: 36a

Answer

$h = \frac{\sqrt2}{\sqrt3}e$

Work Step by Step

To find h, we consider the lengths of each of the sides. We know that h is part of a 30-60-90 triangle, where one of the sides is e. Thus, we find that the other side is $\frac{2e}{\sqrt3}$. The other aspect of a triangle that h is a part of is a 45-45-90 triangle with sides of e, meaning that the length is $\sqrt2 e$ We use the pythagorean theorem: $h = \sqrt{-(\frac{2e}{\sqrt3})^2 + (\sqrt2 e)^2 } = \sqrt{\frac{2e^2}{3}}=$$ \frac{\sqrt2}{\sqrt3}e$
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