Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 6 - Section 6.2 - More Angle Measures in the Circle - Exercises - Page 296: 35

Answer

10

Work Step by Step

Since ACB is an inscribed angle and is thus equal to half of the measure of 180 degrees, it follows that ACB is a right angle. Thus, we know: $AC^2 + CB^2 = AB^2$ We now solve for AC and CB. We call the length of CM k and the length CN j. Thus, we find: $(2k)^2 + j^2 = 73$ And: $k^2 + (2j)^2 = 52$ Thus, we obtain: $4(52 - 4j^2) + j^2 = 73 \\ -15j^2 = -135 \\ j = 3$ This means: $k = \sqrt{52 - 4(9)} = 4$ Since k is half of AC and j is half of CB, we find: $AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$
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