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: 18

Answer

Perimeter = 24 + 9$\pi$ in Area = $24 \pi in^{2}$

Work Step by Step

Perimeter of sector = 2*radius + l(CD) 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$\pi$ * 12 = 24$\pi$ in l =$\frac{135}{360}$ * 24$\pi$ in = $\frac{3}{8}$ * 24$\pi$ in = 9$\pi$ in Perimeter of sector = 2* radius + l = 2* 12 +9$\pi$ in = 24 + 9$\pi$ in In a circle of radius length r , the Area A of the sector whose arc has degree measure m is A = $\frac{m}{360}$ * $\pi r^{2}$ = $\frac{60}{360}$ * $\pi 12^{2}$ = $\frac{1}{6}$ * 144$\pi $ = $24 \pi in^{2}$ Therefore the area of the sector = $24 \pi in^{2}$
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