Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 12 - Section 12.5 - Circles - Exercise - Page 410: 20



Work Step by Step

The area of the side of the tire is the area of the outer circle minus the area of the inner circle. We use the area of a circle: $A=\frac{\pi*d^{2}}{4}$ With the two different diameters and subtract the difference. Thus, we have: $A=\frac{\pi*dout^{2}}{4}-\frac{\pi*din^{2}}{4}=\frac{\pi*(23in)^{2}}{4}-\frac{\pi*(15in)^{2}}{4}=238.76in^{2}\approx239~in^2$
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