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 - Review Exercises - Page 396: 29


A=$\frac{8}{3}$$\pi$-4$\sqrt 3$

Work Step by Step

height of the triangle=.5s$\sqrt 3$ h=2$\sqrt 3$ base of the triangle=r=4 A=$\frac{1}{6}$$\pi$4$^2$-$\frac{1}{2}$(4)(2$\sqrt 3$) A=$\frac{8}{3}$$\pi$-4$\sqrt 3$
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