geometry section 10-1 1112
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Circles and CircumferenceTRANSCRIPT
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CHAPTER 10CIRCLES
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SECTION 10-1Circles and Circumference
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ESSENTIAL QUESTIONS
How do you identify and use parts of circles?
How do you solve problems involving the circumference of a circle?
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VOCABULARY
1. Circle:
2. Center:
3. Radius:
4. Chord:
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VOCABULARY
1. Circle: The set of points that are all the same distance from a given point
2. Center:
3. Radius:
4. Chord:
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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:
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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:
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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
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VOCABULARY
5. Diameter:
6. Congruent Circles:
7. Concentric Circles:
8. Circumference:
9. Pi (π):
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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):
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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):
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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):
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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):
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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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VOCABULARY
10. Inscribed:
11. Circumscribed:
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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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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
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EXAMPLE 1
a. Name the circle
b. Identify a radius
c. Identify a chord d. Name the diameter
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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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EXAMPLE 1
a. Name the circle
Circle C or ⊙C
b. Identify a radius
AC CDor
c. Identify a chord d. Name the diameter
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EXAMPLE 1
a. Name the circle
Circle C or ⊙C
b. Identify a radius
AC CDor
c. Identify a chord
EB
d. Name the diameter
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EXAMPLE 1
a. Name the circle
Circle C or ⊙C
b. Identify a radius
AC CDor
c. Identify a chord
EB
d. Name the diameter
AD
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EXAMPLE 2
If JT = 24 in, what is KM?
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EXAMPLE 2
If JT = 24 in, what is KM?
JT = KL
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EXAMPLE 2
If JT = 24 in, what is KM?
JT = KL
KL = 24 in
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EXAMPLE 2
If JT = 24 in, what is KM?
JT = KL
KL = 24 in
KM is half of KL
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EXAMPLE 2
If JT = 24 in, what is KM?
JT = KL
KL = 24 in
KM is half of KL
KM = 12 in
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
=11
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
=11 MP =
162
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
=11 MP =
162
= 8
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
=11 MP =
162
= 8
LP = LN + MP − MN
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
=11 MP =
162
= 8
LP = LN + MP − MN
LP =11+ 8 − 5
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EXAMPLE 3
The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.
LN =
222
=11 MP =
162
= 8
LP = LN + MP − MN
LP =11+ 8 − 5
LP =14 cm
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EXAMPLE 4
Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet.
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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
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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 = πd
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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 = πdπ π
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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 = πdπ π
d ≈ 20.82 ft
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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 = πdπ π
d ≈ 20.82 ft
r ≈10.41 ft
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
r2 = 9
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
r2 = 9
r 2 = 9
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
r2 = 9
r 2 = 9 r = 3
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
r2 = 9
r 2 = 9 r = 3
C = 2π r
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
r2 = 9
r 2 = 9 r = 3
C = 2π r
C = 2π(3)
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EXAMPLE 5
Find the exact circumference of ⊙R.
BF = 3 2
(BR)2 + (RF )2 = (BF )2
r2 + r 2 = (3 2)2
2r 2 = (3 2)2
2r 2 = 9(2)
r2 = 9
r 2 = 9 r = 3
C = 2π r
C = 2π(3)
C = 6π
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CHECK YOUR UNDERSTANDING
p. 687 #1-9
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PROBLEM SET
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PROBLEM SET
p. 685 #11-41 odd, 71
“We don't know who we are until we see what we can do.” - Martha Grimes
Thursday, May 10, 2012