2. SECTION 10-1
Circles and Circumference
Friday, 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 circle
Friday, 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 diameter
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15. VOCABULARY
10. Inscribed:
11. Circumscribed:
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16. VOCABULARY
10. Inscribed: A polygon inside a circle where all of
the vertices of the polygon are on the circle
11. Circumscribed:
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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 inscribed
Friday, May 11, 2012
18. EXAMPLE 1
a. Name the circle
b. Identify a radius
c. Identify a chord d. Name the diameter
Friday, 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 diameter
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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 diameter
Friday, 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
EB
Friday, 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 AD
Friday, 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 = KL
Friday, May 11, 2012
25. EXAMPLE 2
If JT = 24 in, what is KM?
JT = KL
KL = 24 in
Friday, May 11, 2012
26. EXAMPLE 2
If JT = 24 in, what is KM?
JT = KL
KL = 24 in
KM is half of KL
Friday, 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 in
Friday, 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 =
2
Friday, 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
2
Friday, 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 2
Friday, 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 2
Friday, 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 − MN
Friday, 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 − 5
Friday, 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 cm
Friday, 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 = πd
Friday, 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 d
Friday, 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 π
Friday, 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 ft
Friday, 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 ft
Friday, May 11, 2012
42. EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
Friday, May 11, 2012
43. EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR) + (RF ) = (BF )
2 2 2
Friday, 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 2
Friday, 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 2
Friday, 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)
2
Friday, 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
2
Friday, 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
2
Friday, 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
2
Friday, 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
2
Friday, 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
2
Friday, 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 = 6π
Friday, May 11, 2012
53. CHECK YOUR
UNDERSTANDING
p. 687 #1-9
Friday, May 11, 2012