Elementary Geometry for College Students (5th Edition)

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

Chapter 6 - Section 6.4 - Some Constructions and Inequalities for the Circle - Exercises - Page 315: 38

Answer

5.86

Work Step by Step

We call the midpoint of AC Q. Since triangle APQ is a 30-60-90 triangle, where AP equals 8, we find: $PQ = 8/2 = 4$ And: $AC =2AQ=2 \sqrt{8^2 - 4^2} = 8 \sqrt{3} $ Since PQ equals 4, this means that QB equals 4. Thus, we use the Pythagorean theorem to find AB: $AB = \sqrt{ (4 \sqrt{3})^2 + 4^2}=8$ $AC-AB = 8 \sqrt{3} -8 \approx 5.86 $
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