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 390: 21

Answer

Perimeter = (12 + 4$\pi$) in. Area = (24$\pi$ - 36$\sqrt 3$)$in^{2}$

Work Step by Step

Find Perimeter of the segment 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 l(AB) = $\frac{60}{360}$ * 2$\pi$r = $\frac{1}{6}$ * 2$\pi$ * 12 = 4$\pi$ in. As we know $\triangle$ ABC is an equilateral triangle so AB = 12 Perimeter of segment = AB + l(AB) = (12 + 4$\pi$) in. Area of sector = Area of Segment + Area of triangle Area of segment = Area of sector - Area of triangle In a circle of radius length r , the Area A of the sector whose arc has degree measure m is A = $\frac{60}{360}$ * $\pi r^{2}$ = $\frac{60}{360}$ * $\pi 12^{2}$ = $\frac{1}{6} * \pi * 144$ = 24$\pi in^{2}$ Area of Triangle : $12^{2}\frac{\sqrt 3}{4}$ = 36$\sqrt 3 in^{2}$ Area of segment = Area of sector - Area of triangle = 24$\pi in^{2}$ - 36$\sqrt 3 in^{2}$ =(24$\pi$ - 36$\sqrt 3$)$in^{2}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.