Answer
A=s$^2$-$\pi$(.5s)$^2$
Work Step by Step
A$_{circle}$=$\pi$(.5s)$^2$
A$_{square}$=s$^2$
A=A$_{square}$-A$_{circle}$
A=s$^2$-$\pi$(.5s)$^2$
Elementary Geometry for College Students (5th Edition)
by Alexander, Daniel C.; Koeberlein, Geralyn M.
Published by
Brooks Cole
ISBN 10:
1439047901
ISBN 13:
978-1-43904-790-3
Chapter 8 - Section 8.5 - More Area Relationships in the Circle - Exercises - Page 392: 28
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