Elementary Geometry for College Students (6th Edition)

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

Chapter 8 - Section 8.4 - Circumference and Area of a Circle - Exercises - Page 374: 47



Work Step by Step

We first find the height of the circle: $h = \sqrt{8^2 - 4^2} = \sqrt{48} = 4\sqrt{3}$ Since this height is double the radius, we find: $r = 2 \sqrt{3}$ Thus, we find area: $ A = \pi(2\sqrt{3})^2 = 12\pi$
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