Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 8 - Section 8.5 - More Area Relationships in the Circle - Exercises - Page 378: 17

Answer

Perimeter =( 16 + $\frac{8}{3}$$\pi$) in Area = $\frac{32}{3} \pi in^{2}$

Work Step by Step

Perimeter of sector = 2*radius + l(AB) In a circle whose circumference is c, the length l of an arc whose degree measure is m is given by l =$\frac{m}{360}$ * c Therefore for given circle c = 2$\pi$ r = 2$\pi$ * 8 = 16$\pi$ in l =$\frac{60}{360}$ * 16$\pi$ in =$\frac{1}{6}$ * 16$\pi$ in = $\frac{8}{3}$$\pi$ in Perimeter of sector = 2* radius + l = 2* 8 + $\frac{8}{3}$$\pi$ in = 16 + $\frac{8}{3}$$\pi$ in In a circle of radius length r , the Area A of the sector whose arc has degree measure m is A = $\frac{m}{360}$ * $\pi r^{2}$ = $\frac{60}{360}$ * $\pi 8^{2}$ = $\frac{1}{6}$ * 64$\pi $ = $\frac{32}{3} \pi in^{2}$ Therefore the area of the sector = $\frac{32}{3} \pi in^{2}$
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