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 8 - Review Exercises - Page 396: 28

Answer

A=$\frac{49\pi-49\sqrt 3}{2}$

Work Step by Step

Diameter=2(7)=14 Long leg of the triangle=shortleg($\sqrt 3$)=7$\sqrt 3$ A=.5$\pi$7$^2$-(.5)(7)(7$\sqrt 3$) A=$\frac{49\pi}{2}$-$\frac{49\sqrt 3}{2}$ A=$\frac{49\pi-49\sqrt 3}{2}$
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