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Circles. Parts of a Circle A B C D Radius - segment from center pt to a point on the circle. Ex. AC, BC, DC are all radiuses.

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Presentation on theme: "Circles. Parts of a Circle A B C D Radius - segment from center pt to a point on the circle. Ex. AC, BC, DC are all radiuses."— Presentation transcript:

1 Circles

2 Parts of a Circle A B C D Radius - segment from center pt to a point on the circle. Ex. AC, BC, DC are all radiuses

3 Diameter - a chord that passes through the center point of a circle. Ex. PR is a diameter Chord - segment whose endpoints are on the circle. Ex. PR, PS, are chords Parts of a Circle

4 Minor Arc - an arc that is less than 180 0 - use two letters to label a minor arc. Ex. Major Arc - an arc that is more than 180 0 - use three letters to label major arc. Ex. Arc - Part of a circle's edge Parts of a Circle

5 Central angle - an angle whose vertex is at the center of the circle. Ex. <APB Parts of a Circle Intercepted Arc – arc that is cut off by the sides of an angle. Ex. arc AB is the intercepted arc

6 Inscribed angle - an angle whose vertex is on the circle. Ex. <1 is an inscribed angle. 1 Parts of a Circle

7 Put the answers to the following on your notesheet: Radius Diameter Major Arc Minor Arc Chord Central angle Inscribed Angle T A P L C

8 Central Angle = Intercepted Arc 80 B A 1 In the picture at right arc AB = 80, so angle 1 = 80 because <1 is a central angle In the picture at right arc AC = 105, because its central angle is 105. B A 105C D 3 2 1 <1= 105 (vertical angles), <2=75 (forms a line with 105), <3 = 75 (forms a line with105). Therefore arc BD = 105, arc AB = 75, arc DC = 75 105 75

9 B 30 A C x D E 1 - Find x 2 - find x 3 - find arc AB x 110 A 127 A D B L Central Angle = Intercepted Arc Put these on your notesheet B 100 1 2 3 C A 4 - find angles 1, 2, 3 D L 132

10 B 30 A C x D E 1 - Find x 2 - find x x 110 A L Central Angle = Intercepted Arc X=110 because Central angle = intercepted arc Put these on your notesheet X=30 because Central angle = intercepted arc so <ECA = 30, x is vertical to <ECA so x=30 30 110

11 3 - find arc AB 127 A D B Central Angle = Intercepted Arc Put these on your notesheet Arc AD=127, arc AB and arc AD form a semicircle (180 degrees) 180-127=53 Or you could say the unlabeled angle next to 127 is 53 and then the arc is 53 <1=100 b/c it is central to arch AB <2=80 b/c it forms a line with <1 (180- 100 = 80) <3=48 b/c arc AD is 48 (180-132) B 100 1 2 3 C A 4 - find angles 1, 2, 3 D L 132 53 127 53 100 80 132 48

12 Place these problems on your HALF SHEET OF PAPER

13 Inscribed Angle = (Intercepted Arc)/2 Or Intercepted Arc = 2(Inscribed Angle) 70 B A 1 C Above arc AB = 70, so angle 1 = 35 because <1 is an inscribed angle 98 B A 4 C 3 2 1 120 Above arc AB=98, so <1=98 (central angle=arc), <2=49 (inscribed angle = arc/2), <4=60 (inscribed angle = arc/2) <3=71 (180-49-60), arc AC=142 (arc=2(inscribed angle) 35 98 49 60 71 142

14 A B 80 x EX1 - Find x EX2 - Find arc AB, and x A B C R angle 1 = 90 1 x A B T 145 Inscribed Angle = (Intercepted Arc)/2 EX3 - Find arc AB and arc ATB Put these on your notesheet B 100 1 2 3 A 4 - find angles 1, 2, 3 D L 132 56

15 A B 80 x EX1 - Find x EX2 - Find arc AB, and x A B C R angle 1 = 90 1 x Inscribed Angle = (Intercepted Arc)/2 Put these on your notesheet x = 40 because an inscribed angle equals half the intercepted arc 40 Arc AB=90 because an arc equals its central angle. Since angle x is inscribed and Intercepts arc AB, x = AB/2=45 90 45

16 A B T 145 Inscribed Angle = (Intercepted Arc)/2 EX3 - Find arc AB and arc ATB B 100 1 2 3 A 4 - find angles 1, 2, 3 D L 132 290 Arc ATB is a major arc (more than 180). Arc ATB=290 {because the arc is twice its inscribed angle of 145} AB=170, because AB and ATB make the whole circle so 360-290=70 70 <1=50because it is an inscribed angle and half arc AB <2=28 because it is an inscribed angle and half arc BD <3=36, Arc AL=72 because 360-100-50-132=72. <3 is inscribed so it is 72/2 56 50 28 36

17 Place these problems on your HALF SHEET OF PAPER

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