Elementary Geometry for College Students (7th Edition)

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

Chapter 7 - Section 7.2 - Concurrence of Lines - Exercises - Page 335: 25

Answer

$\frac{10\sqrt3}{3}$

Work Step by Step

We first draw the perpendicular triangles. We then draw a vertical line from the circumcenter to the base of the isosceles triangle. This creates two 30-60-90 triangles, where the base of each triangle is 5. Using the lengths of 30-60-90 triangles, we see that the distance is: $$ 2(\frac{5}{\sqrt3}) \\ = 2(\frac{5\sqrt3}{3} )\\ = \frac{10\sqrt3}{3}$$
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