Elementary Geometry for College Students (5th Edition)

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

Chapter 8 - Section 8.1 - Area and Initial Postulates - Exercises - Page 360: 27

Answer

$156 + 24\sqrt10$

Work Step by Step

In order to do this, we add up the surface areas of the sides. Thus, we first add the three sides of known lengths: $ SA = 2\times(6)\times(8) \times(6) \times (1/2) +6(12)=156$ We now find the length of the missing side: $ l = \sqrt{6^2+2^2}= \sqrt{40}$ Using this, we add the lengths of the unknown surface areas: $SA= 156 + \sqrt{40}(12) = $$156 + 24\sqrt10$
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