Group_D_Assignment_5 - Team Problem Set #5 Learning Team D...

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Team Problem Set #5Learning Team DMTH/535 GeometryAugust 1, 2016Dr. Karen Busch
– Ch. 7, Exercises 7.3, problems: 1, 3, 5, 11A = ()/4= (π(52¿)/4= (25/4)πsq. cmA = ()(45/360)= (π(42¿(1/8)= (2)πsq. cm= (90/180)π(5)= (5/2)πP = [(5/2)π+ 5 + 5] cm= [(5/2)π+ 10] cm= (45/180)π(4)= (180/180)π=πP = (π+ 4 + 4) cm= (π+ 8) cm
Area of Triangle = .5bh = .5(4)(4) = 8 sq. ydArea of Sector == (90/360)π(42) = 4πsq. ydArea of shaded region = Area of sector – Area of triangle= 8 - 4π= 4.6 sq. yd= (30/360)π(4.12)= 4.4 sq. cm
- Ch. 7, Exercises 7.3, problems: 13, 17, 21p. 36513.Area of asector=m360π r2,so w e'llsolvefor rdoubleit .24π=60360π r2;r2=246=144;r=12,so thediamete17.Area of the sector=12360(720)=24cm221.I will first find the area of one arc, then subtract twice what’s left over from the square:Single arcarea=9036042π=4π . Area of shaded region=162(164π)9.1yd2
- Ch. 7, Exercises 7.4, problems: 1, 3, 7, 11

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Term
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Tags
Geometry, Chapter 7, Regular polygon, Apothem

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