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: 28

Answer

$A=1.32~in^{2}$

Work Step by Step

The area of the cross-section of the pipe can be calculated with the formula $A=\frac{\pi}{4}d^{2}$. We apply the formula with the outside and inside diameters, and subtract: $A=\frac{\pi}{4}d_{out}^{2}-\frac{\pi}{4}d_{in}^{2}$ Where $d_{out}=3.50in$ $d_{in}=3.25in$ So, $A=\frac{\pi}{4}d_{out}^{2}-\frac{\pi}{4}d_{in}^{2}=\frac{\pi}{4}(3.50in)^{2}-\frac{\pi}{4}(3.25in)^{2}=1.32~in^{2}$
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