Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 8 - Section 8.5 - More Area Relationships in the Circle - Exercises - Page 379: 27

Answer

A=36$\pi$$\approx$113

Work Step by Step

The diameter of the inscribed circle is equal to the height of the isosceles trapezoid. h$^2$=13$^2$-($\frac{1}{2}$(18-8))$^2$ h$^2$=169-25=144 h=12 r=.5h r=6 A=$\pi$r$^2$ A=36$\pi$$\approx$113
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