## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Chapter 8 - Review Exercises - Page 386: 36

#### Answer

$A_{ring} = \pi BC^2$

#### Work Step by Step

We call r the radius of the smaller circle. The radius of the larger circle is: $R = \sqrt{BC^2 + r^2}$ Thus, the area of the ring is: $A = \pi (R^2 - r^2) \\ A = \pi (BC^2 + r^2 - r^2) = \pi BC^2$

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