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Test is next class Test is open note

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1 Test is next class Test is open note
Circle review Test is next class Test is open note

2 Find the area and circumference of the circle below that has a radius of 8.5 cm.
𝐴=πœ‹ π‘Ÿ 2 𝐴=πœ‹βˆ— 8.5 2 𝐴=72.25πœ‹ π΄β‰ˆ π‘π‘š 2 𝐢=2πœ‹π‘Ÿ 𝐢=2βˆ—πœ‹βˆ—8.5 𝐢=17πœ‹ πΆβ‰ˆ53.41π‘π‘š 8.5

3 Inscribed angles 1. 3π‘₯βˆ’3=120/2 3π‘₯βˆ’3=60 3π‘₯=63 π‘₯=21 ∠𝐡=180βˆ’50βˆ’70 ∠𝐡=60Β°
π‘₯βˆ’3=120/2 3π‘₯βˆ’3=60 3π‘₯=63 π‘₯=21 ∠𝐡=180βˆ’50βˆ’70 ∠𝐡=60Β° Arc AC=60*2 Arc AC=120Β° Arc CBA =360Β°βˆ’120Β° Arc CBA =240Β°

4 Circle vocabulary Minor arc Central angle radius Inscribed angle
Tangent line diameter chord semicircle Major arc

5 More inscribed angles π‘₯= 58+106 /2 π‘₯=82Β° 𝑦=(360βˆ’58βˆ’106)/2 𝑦=98Β°
𝑧=180βˆ’93 𝑧=87Β°

6 Length of intersecting chords
3π‘₯=6βˆ—4 3π‘₯=24 π‘₯=8

7 Area of a segment Segment = sector –triangle sector:
𝐴= βˆ—πœ‹ βˆ—24 2 𝐴= 1 3 βˆ—πœ‹ βˆ—24 2 𝐴=192πœ‹β‰ˆ603.19 Triangle: 𝐴=.5βˆ—π‘ π‘–π‘‘π‘’βˆ—π‘ π‘–π‘‘π‘’βˆ—sin⁑(𝑖𝑛𝑐𝑙𝑒𝑑𝑒𝑑 π‘Žπ‘›π‘”π‘™π‘’) 𝐴=.5βˆ—24βˆ—24βˆ—sin⁑(120Β°) π΄β‰ˆ249.42 Segment: 𝐴=603.1βˆ—βˆ’249.42= 𝑒𝑛 2

8 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 π‘Šπ‘‹ 2 = βˆ’2βˆ—10βˆ—10βˆ—cos⁑(130Β°) π‘Šπ‘‹ 2 =328.56 π‘Šπ‘‹=18.13 Choice 2= law of sines: Angle W= ( )/2=25 π‘Šπ‘‹ sin⁑(130Β°) = 10 𝑠𝑖𝑛25Β° π‘Šπ‘‹=(10𝑠𝑖𝑛130)/𝑠𝑖𝑛25 WX=18.13 Length of chord WX

9 Equations and graphing circles
Identify the center and radius on the equation below and then graph it on a separate sheet of paper. (π‘₯βˆ’4) 2 + (𝑦+2) 2 =16 Center= (4, -2) R=4


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