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 8 - Section 8.5 - More Area Relationships in the Circle - Exercises - Page 390: 27



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
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.