Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10: 9781285195698
ISBN 13: 978-1-28519-569-8

Chapter 8 - Review Exercises - Page 386: 34b

Answer

$17.65$

Work Step by Step

We first find the radius: $A = \pi r^2 \\ 27 \pi = \pi r^2 \\ r = \sqrt{27} = 3\sqrt{3}$ We know that the perimeter contains double the radius plus 80/360 of the circumference of the circle: $ P = 2\pi(3\sqrt{3})(80/360) + 6\sqrt3 = 17.65 $
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