Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 10 - Area - 10-6 Circles and Arcs - Practice and Problem-Solving Exercises - Page 656: 53

Answer

$\dfrac{13 \pi}{5} \ in.=2.6 \pi \ in.$

Work Step by Step

$l=\dfrac{\text{Measure of the arc}}{360^{\circ}} \times 2 \pi r ~~~(a)$ or, $l=\dfrac{\text{Measure of the arc}}{360^{\circ}} \times \pi d~~(b)$ Here, $r$ represents radius and $d$ is diameter. We are given that the red arc measures $180^{\circ}-50^{\circ}=130^{\circ}$ and $d=7.2 \ in.$ Thus, equation (b) becomes: $l=\dfrac{130^{\circ}}{360^{\circ}} \times \pi (7.2)= \dfrac{13 \pi}{5} \ in.=2.6 \pi \ in.$
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