Geometry Section 10-1 1112

transcript

CHAPTER 10CIRCLES

Thursday, May 10, 2012

SECTION 10-1Circles and Circumference

ESSENTIAL QUESTIONS

How do you identify and use parts of circles?

How do you solve problems involving the circumference of a circle?

VOCABULARY

1. Circle:

2. Center:

3. Radius:

4. Chord:

VOCABULARY

1. Circle: The set of points that are all the same distance from a given point

2. Center:

3. Radius:

4. Chord:

VOCABULARY

2. Center: The point that all of the points of a circle are equidistant from

3. Radius:

4. Chord:

VOCABULARY

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:

VOCABULARY

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

VOCABULARY

5. Diameter:

6. Congruent Circles:

7. Concentric Circles:

8. Circumference:

9. Pi (π):

VOCABULARY

5. Diameter: A special chord that passes through the center of a circle; twice the length of the radius

6. Congruent Circles:

8. Circumference:

9. Pi (π):

VOCABULARY

6. Congruent Circles: Two or more circles with congruent radii

8. Circumference:

9. Pi (π):

VOCABULARY

7. Concentric Circles: Coplanar circles with the same center

8. Circumference:

9. Pi (π):

VOCABULARY

8. Circumference: The distance around a circle

9. Pi (π):

VOCABULARY

8. Circumference: The distance around a circle

9. Pi (π): The irrational number found from the ratio of circumference to the diameter

VOCABULARY

10. Inscribed:

11. Circumscribed:

VOCABULARY

10. Inscribed: A polygon inside a circle where all of the vertices of the polygon are on the circle

11. Circumscribed:

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

EXAMPLE 1

a. Name the circle

b. Identify a radius

c. Identify a chord d. Name the diameter

EXAMPLE 1

a. Name the circle

Circle C or ⊙C

EXAMPLE 1

a. Name the circle

Circle C or ⊙C

AC CDor

EXAMPLE 1

a. Name the circle

Circle C or ⊙C

AC CDor

c. Identify a chord

d. Name the diameter

EXAMPLE 1

a. Name the circle

Circle C or ⊙C

AC CDor

c. Identify a chord

d. Name the diameter

EXAMPLE 2

If JT = 24 in, what is KM?

EXAMPLE 2

JT = KL

EXAMPLE 2

JT = KL

KL = 24 in

EXAMPLE 2

JT = KL

KL = 24 in

KM is half of KL

EXAMPLE 2

JT = KL

KL = 24 in

KM is half of KL

KM = 12 in

EXAMPLE 3

The diameter of ⊙L is 22 cm and the diameter of ⊙P is 16 cm. MN = 5 cm. Find LP.

EXAMPLE 3

=11 MP =

EXAMPLE 3

=11 MP =

EXAMPLE 3

=11 MP =

LP = LN + MP − MN

EXAMPLE 3

=11 MP =

LP = LN + MP − MN

LP =11+ 8 − 5

EXAMPLE 3

=11 MP =

LP = LN + MP − MN

LP =11+ 8 − 5

LP =14 cm

EXAMPLE 4

Find the diameter and radius of a circle to the nearest hundredth if the circumference of the circle is 65.4 feet.

EXAMPLE 4

C = πd

EXAMPLE 4

C = πd

65.4 = πd

EXAMPLE 4

C = πd

65.4 = πdπ π

EXAMPLE 4

C = πd

65.4 = πdπ π

d ≈ 20.82 ft

EXAMPLE 4

C = πd

65.4 = πdπ π

d ≈ 20.82 ft

r ≈10.41 ft

EXAMPLE 5

Find the exact circumference of ⊙R.

BF = 3 2

EXAMPLE 5

BF = 3 2

(BR)2 + (RF )2 = (BF )2

EXAMPLE 5

BF = 3 2

(BR)2 + (RF )2 = (BF )2

r2 + r 2 = (3 2)2

EXAMPLE 5

BF = 3 2

(BR)2 + (RF )2 = (BF )2

r2 + r 2 = (3 2)2

2r 2 = (3 2)2

EXAMPLE 5

BF = 3 2

(BR)2 + (RF )2 = (BF )2

r2 + r 2 = (3 2)2

2r 2 = (3 2)2

2r 2 = 9(2)

EXAMPLE 5

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

EXAMPLE 5

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

EXAMPLE 5

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

EXAMPLE 5

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

EXAMPLE 5

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)

EXAMPLE 5

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

CHECK YOUR UNDERSTANDING

p. 687 #1-9

PROBLEM SET

p. 685 #11-41 odd, 71

“We don't know who we are until we see what we can do.” - Martha Grimes

Geometry Section 10-1 1112

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