Elementary Geometry for College Students (6th Edition)

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

Chapter 8 - Section 8.5 - More Area Relationships in the Circle - Exercises - Page 380: 45



Work Step by Step

We know that the center of the circumscribed circle is the midpoint of the line BA. Assuming that point C of the triangle is the origin, this means that the midpoint is (4,3). We now consider the midpoint of the inscribed circle, which is given by: $r =\frac{ab}{a+b+c} \\ r = \frac{8\times 6}{8+6+10} = 2$ Thus, we use the distance formula to find the distance: $ d = \sqrt{(4-2)^2 + (3-2)^2} = \sqrt{5}$
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