2. SECTION 10-1 Circles and CircumferenceFriday, May 11, 2012
3. ESSENTIAL QUESTIONS How do you identify and use parts of circles? How do you solve problems involving the circumference of a circle?Friday, May 11, 2012
5. VOCABULARY 1. Circle: The set of points that are all the same distance from a given point 2. Center: 3. Radius: 4. Chord:Friday, May 11, 2012
6. VOCABULARY 1. Circle: The set of points that are all the same distance from a given point 2. Center: The point that all of the points of a circle are equidistant from 3. Radius: 4. Chord:Friday, May 11, 2012
7. VOCABULARY 1. Circle: The set of points that are all the same distance from a given point 2. Center: The point that all of the points of a circle are equidistant from 3. Radius: A segment with one endpoint at the center and the other on the edge of the circle; also the distance from the center to the edge of the circle 4. Chord:Friday, May 11, 2012
8. VOCABULARY 1. Circle: The set of points that are all the same distance from a given point 2. Center: The point that all of the points of a circle are equidistant from 3. Radius: A segment with one endpoint at the center and the other on the edge of the circle; also the distance from the center to the edge of the circle 4. Chord: A segment with both endpoints on the edge of the circleFriday, May 11, 2012
9. VOCABULARY 5. Diameter: 6. Congruent Circles: 7. Concentric Circles: 8. Circumference: 9. Pi (π):Friday, May 11, 2012
10. VOCABULARY 5. Diameter: A special chord that passes through the center of a circle; twice the length of the radius 6. Congruent Circles: 7. Concentric Circles: 8. Circumference: 9. Pi (π):Friday, May 11, 2012
11. VOCABULARY 5. Diameter: A special chord that passes through the center of a circle; twice the length of the radius 6. Congruent Circles: Two or more circles with congruent radii 7. Concentric Circles: 8. Circumference: 9. Pi (π):Friday, May 11, 2012
12. VOCABULARY 5. Diameter: A special chord that passes through the center of a circle; twice the length of the radius 6. Congruent Circles: Two or more circles with congruent radii 7. Concentric Circles: Coplanar circles with the same center 8. Circumference: 9. Pi (π):Friday, May 11, 2012
13. VOCABULARY 5. Diameter: A special chord that passes through the center of a circle; twice the length of the radius 6. Congruent Circles: Two or more circles with congruent radii 7. Concentric Circles: Coplanar circles with the same center 8. Circumference: The distance around a circle 9. Pi (π):Friday, May 11, 2012
14. VOCABULARY 5. Diameter: A special chord that passes through the center of a circle; twice the length of the radius 6. Congruent Circles: Two or more circles with congruent radii 7. Concentric Circles: Coplanar circles with the same center 8. Circumference: The distance around a circle 9. Pi (π): The irrational number found from the ratio of circumference to the diameterFriday, May 11, 2012
15. VOCABULARY 10. Inscribed: 11. Circumscribed:Friday, May 11, 2012
16. VOCABULARY 10. Inscribed: A polygon inside a circle where all of the vertices of the polygon are on the circle 11. Circumscribed:Friday, May 11, 2012
17. VOCABULARY 10. Inscribed: A polygon inside a circle where all of the vertices of the polygon are on the circle 11. Circumscribed: A circle that is around a polygon that is inscribedFriday, May 11, 2012
18. EXAMPLE 1 a. Name the circle b. Identify a radius c. Identify a chord d. Name the diameterFriday, May 11, 2012
19. EXAMPLE 1 a. Name the circle Circle C or ⊙C b. Identify a radius c. Identify a chord d. Name the diameterFriday, May 11, 2012
20. EXAMPLE 1 a. Name the circle Circle C or ⊙C b. Identify a radius AC or CD c. Identify a chord d. Name the diameterFriday, May 11, 2012
21. EXAMPLE 1 a. Name the circle Circle C or ⊙C b. Identify a radius AC or CD c. Identify a chord d. Name the diameter EBFriday, May 11, 2012
22. EXAMPLE 1 a. Name the circle Circle C or ⊙C b. Identify a radius AC or CD c. Identify a chord d. Name the diameter EB ADFriday, May 11, 2012
23. EXAMPLE 2 If JT = 24 in, what is KM?Friday, May 11, 2012
24. EXAMPLE 2 If JT = 24 in, what is KM? JT = KLFriday, May 11, 2012
25. EXAMPLE 2 If JT = 24 in, what is KM? JT = KL KL = 24 inFriday, May 11, 2012
26. EXAMPLE 2 If JT = 24 in, what is KM? JT = KL KL = 24 in KM is half of KLFriday, May 11, 2012
27. EXAMPLE 2 If JT = 24 in, what is KM? JT = KL KL = 24 in KM is half of KL KM = 12 inFriday, May 11, 2012
28. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.Friday, May 11, 2012
29. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 LN = 2Friday, May 11, 2012
30. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 LN = =11 2Friday, May 11, 2012
31. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 16 LN = =11 MP = 2 2Friday, May 11, 2012
32. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 16 LN = =11 MP = = 8 2 2Friday, May 11, 2012
33. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 16 LN = =11 MP = = 8 2 2 LP = LN + MP − MNFriday, May 11, 2012
34. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 16 LN = =11 MP = = 8 2 2 LP = LN + MP − MN LP =11+ 8 − 5Friday, May 11, 2012
35. EXAMPLE 3 The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP. 22 16 LN = =11 MP = = 8 2 2 LP = LN + MP − MN LP =11+ 8 − 5 LP =14 cmFriday, May 11, 2012
36. EXAMPLE 4 Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet.Friday, May 11, 2012
37. EXAMPLE 4 Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet. C = πdFriday, May 11, 2012
38. EXAMPLE 4 Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet. C = πd 65.4 = π dFriday, May 11, 2012
39. EXAMPLE 4 Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet. C = πd 65.4 = π d π πFriday, May 11, 2012
40. EXAMPLE 4 Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet. C = πd 65.4 = π d π π d ≈ 20.82 ftFriday, May 11, 2012
41. EXAMPLE 4 Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet. C = πd 65.4 = π d π π d ≈ 20.82 ft r ≈10.41 ftFriday, May 11, 2012
42. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2Friday, May 11, 2012
43. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2Friday, May 11, 2012
44. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r + r = (3 2 ) 2 2 2Friday, May 11, 2012
45. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r + r = (3 2 ) 2 2 2 2r = (3 2 ) 2 2Friday, May 11, 2012
46. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r + r = (3 2 ) 2 2 2 2r = (3 2 ) 2 2 2r = 9(2) 2Friday, May 11, 2012
47. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r + r = (3 2 ) 2 2 2 2r = (3 2 ) 2 2 2r = 9(2) 2 r =9 2Friday, May 11, 2012
48. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r = 9 2 r + r = (3 2 ) 2 2 2 2r = (3 2 ) 2 2 2r = 9(2) 2 r =9 2Friday, May 11, 2012
49. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r = 9 2 r + r = (3 2 ) 2 2 2 r =3 2r = (3 2 ) 2 2 2r = 9(2) 2 r =9 2Friday, May 11, 2012
50. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r = 9 2 r + r = (3 2 ) 2 2 2 r =3 2r = (3 2 ) 2 2 C = 2π r 2r = 9(2) 2 r =9 2Friday, May 11, 2012
51. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r = 9 2 r + r = (3 2 ) 2 2 2 r =3 2r = (3 2 ) 2 2 C = 2π r 2r = 9(2) 2 C = 2π (3) r =9 2Friday, May 11, 2012
52. EXAMPLE 5 Find the exact circumference of ⊙R. BF = 3 2 (BR) + (RF ) = (BF ) 2 2 2 r = 9 2 r + r = (3 2 ) 2 2 2 r =3 2r = (3 2 ) 2 2 C = 2π r 2r = 9(2) 2 C = 2π (3) r =9 2 C = 6πFriday, May 11, 2012
53. CHECK YOUR UNDERSTANDING p. 687 #1-9Friday, May 11, 2012
54. PROBLEM SETFriday, May 11, 2012
55. PROBLEM SET p. 687 #11-41 odd, 71 “We dont know who we are until we see what we can do.” - Martha GrimesFriday, May 11, 2012
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