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Test is next class Test is open note
Circle review Test is next class Test is open note
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Find the area and circumference of the circle below that has a radius of 8.5 cm.
π΄=π π 2 π΄=πβ 8.5 2 π΄=72.25π π΄β ππ 2 πΆ=2ππ πΆ=2βπβ8.5 πΆ=17π πΆβ53.41ππ 8.5
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Inscribed angles 1. 3π₯β3=120/2 3π₯β3=60 3π₯=63 π₯=21 β π΅=180β50β70 β π΅=60Β°
π₯β3=120/2 3π₯β3=60 3π₯=63 π₯=21 β π΅=180β50β70 β π΅=60Β° Arc AC=60*2 Arc AC=120Β° Arc CBA =360Β°β120Β° Arc CBA =240Β°
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Circle vocabulary Minor arc Central angle radius Inscribed angle
Tangent line diameter chord semicircle Major arc
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More inscribed angles π₯= 58+106 /2 π₯=82Β° π¦=(360β58β106)/2 π¦=98Β°
π§=180β93 π§=87Β°
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Length of intersecting chords
3π₯=6β4 3π₯=24 π₯=8
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Area of a segment Segment = sector βtriangle sector:
π΄= βπ β24 2 π΄= 1 3 βπ β24 2 π΄=192πβ603.19 Triangle: π΄=.5βπ πππβπ πππβsinβ‘(πππππ’πππ πππππ) π΄=.5β24β24βsinβ‘(120Β°) π΄β249.42 Segment: π΄=603.1ββ249.42= π’π 2
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Length of a chord Step 1: connect end points of chord to center of circle to make a triangle Step 2: WZ=XZ=20/2=10 Step 3: angle Z= 130 Length of chord: choice 1= law of cosines ππ 2 = β2β10β10βcosβ‘(130Β°) ππ 2 =328.56 ππ=18.13 Choice 2= law of sines: Angle W= ( )/2=25 ππ sinβ‘(130Β°) = 10 π ππ25Β° ππ=(10π ππ130)/π ππ25 WX=18.13 Length of chord WX
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Equations and graphing circles
Identify the center and radius on the equation below and then graph it on a separate sheet of paper. (π₯β4) 2 + (π¦+2) 2 =16 Center= (4, -2) R=4
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