Elementary Geometry for College Students (5th Edition)

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

Chapter 8 - Review Exercises - Page 399: 21

Answer

$162\sqrt{3}$

Work Step by Step

Half of the length of one of the sides is part of a 30-60-90 triangle with the apothem. Thus: $ s/2 = \frac{9}{\sqrt3} \\ s/2 = 3\sqrt{3} \\ s = 6\sqrt3$ We now use the perimeter and the apothem to find area: $P = 6(6\sqrt{3})=36\sqrt{3}$ $A = 1/2aP = 1/2(9)(36\sqrt{3}) = 162\sqrt{3}$
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