Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole
ISBN 10:
ISBN 13:

Chapter 8 - Test - Page 388: 15

Answer

36$\pi$-72 in$^2$

Work Step by Step

A$_{sector}$=$\frac{90}{360}$($\pi$12$^2$) A$_{sector}$=$\frac{1}{4}$(144$\pi$) A$_{sector}$=36$\pi$ A$_{triangle}$=.5(12)(12) A$_{triangle}$=72 A=36$\pi$-72
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