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# CIRCLES - PowerPoint PPT Presentation

CIRCLES. Lesson 6.7 Circumference and Arc Length. Objectives/Assignment. Find the circumference of a circle and the length of a circular arc . Use circumference and arc length to solve real-life problems. Homework: Lesson 6.7/1-11, 17, 19, 22 Quiz Wednesday Chapter 6 Test Friday.

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Lesson 6.7

Circumference and Arc Length

• Find the circumference of a circle and the length of a circular arc.

• Use circumference and arc length to solve real-life problems.

• Homework:

• Lesson 6.7/1-11, 17, 19, 22

• Quiz Wednesday

• Chapter 6 Test Friday

• The circumference of a circle is the distance around the circle.

• For all circles, the ratio of the circumference to the diameter is the same:  or pi.

• The exact value of Pi = 

• The approximate value of Pi ≈ 3.14

Distance around the circle

120°

X

Y

• Minor Arc

• Use 2 letters

• Angle is less than or equal to 180

Terminology

XZ

9

• Major Arc

• Use 3 letters

• Angle is greater than 180

C

XYZ

Central Angle:Any angle whose vertex is the center of the circle

m XZ = m<XCZ = 120o

m XYZ = m<XCZ = 240o

The circumference C of a circle is

C = d or C = 2r, where

d is the diameter of the circle and

r is the radius of the circle (2r = d)

Circumference of a Circle

Tires from two different automobiles are shown.

How many revolutions does each tire make while traveling 100 feet?

Comparing Circumferences

Tire A

Tire B

C = d

diameter = 14 + 2(5.1) d = 24.2 inches

circumference = (24.2)

C ≈ 75.99 inches.

Comparing Circumferences - Tire A

C = d

diameter = 15 + 2(5.25)

d = 25.5 inches

Circumference = (25.5)

C ≈ 80.07 inches

Comparing Circumferences - Tire B

Comparing Circumferences Tire A vs. Tire B

• Divide the distance traveled by the tire circumference to find the number of revolutions made.

• First, convert 100 feet to 1200 inches.

Revolutions = distance traveled circumference

100 ft.

1200 in.

TIRE A:

TIRE B:

100 ft.

1200 in.

=

=

75.99 in.

75.99 in.

80.07 in.

80.07 in.

 15.8 revolutions

 14.99 revolutions

COMPARISON: Tire A required more revolutions to cover the same distance as Tire B.

• The length of part of the circumference.

The length of the arc depends on what two things?

1) The measure of the arc.

2) The size of the circle (radius).

An arc length measures distance while

the measure of an arc is in degrees.

An arc length is a portion of the circumference of a circle.

Portions of a Circle: Determine the Arc measure based on the portion given.

180o

120o

90o

60o

90o

180o

120o

60o

¼ ● 360

½ ● 360

1/3 ● 360

180o

1/6 ● 360

90o

120o

60o

measure of the central angle or arc

The circumference of the

entire circle!

2πr

Arc Length =

360˚

The fraction of the circle!

.

• In a circle, the ratio of the length of a given arc to the circumference is equal to the ratio of the measure of the arc to 360°.

Arc measure

m

Arc length of

=

• 2r

360°

Arc length  linear units (inches/feet/meters …)

Arc measure  degrees

• Find the length of each arc.

a.

b.

c.

50°

50°

100°

• 2(5)

a. Arc length of =

50°

360°

Finding Arc Lengths, con’t.

• Find the length of each arc.

a.

# of °

• 2r

a. Arc length of =

360°

50°

 4.36 centimeters

Arc length of

Finding Arc Lengths, con’t.

• Find the length of each arc.

b.

# of °

• 2r

b. Arc length of =

360°

50°

50°

• 2(7)

b. Arc length of =

360°

 6.11 centimeters

Arc length of

In parts (a) and (b), note that the arcs have the same measure but different lengths because the circumferences of the circles are not equal.

Finding Arc Lengths, con’t.

• Find the length of each arc.

c.

# of °

• 2r

c. Arc length of =

360°

100°

100°

• 2(7)

c. Arc length of =

360°

 12.22 centimeters

Arc length of

A

300o

120o

108o

240o

B

90o

B

120o

90o

240o

300o

12

108o

O

A

O

B

O

A

A

A

O

2.4

O

6

12

10√2

B

B

Fraction of circle:

Fraction of circle:

Fraction of circle:

Fraction of circle:

Fraction of circle:

Fraction ● circumference

Fraction ● circumference

5/6 ● 24π

¼ ● 12π

2/3 ● 24π

1/3 ● 4.8π

3/10 ● 20√2π

20π units

3π units

16π units

1.6π units

6√2π units

circumference

Find the

arc length

3.82 m

60º

5cm

50º

m

=

2r

• Find the indicated measure.

Arc length of

360°

b. m

Substitute and Solve for

m

m

18 in.

=

2(7.64)

360°

18

=

m

360° •

(15.28)

135°  m

• Race Track. The track shown has six lanes.

• Each lane is 1.25 meters wide.

• There is 180° arc at the end of each track.

• The radii for the arcs in the first two lanes are given.

• Find the distance around Lane 1. (use r1)

• Find the distance around Lane 2. (use r2)

The track is made up of

two semicircles

two straight sections with length s

Finding Arc Length, con’t

= 2(108.9) + 2(29.00)

 400.0 meters

Distance = 2s + 2r2

= 2(108.9) + 2(30.25)

 407.9 meters

Finding Arc Length, con’t

Lane 1

Lane 2

Find each arc length. Give answers in terms of  and rounded to the nearest hundredth.

FG

Use formula for area of sector.

Substitute 8 for r and 134 for m.

 5.96 cm  18.71 cm

Simplify.

Find each arc length. Give answers in terms of  and rounded to the nearest hundredth.

an arc with measure 62 in a circle with radius 2 m

Use formula for area of sector.

Substitute 2 for r and 62 for m.

 0.69 m  2.16 m

Simplify.

=  m  4.19 m

Check It Out!

Find each arc length. Give your answer in terms of and rounded to the nearest hundredth.

GH

Use formula for area of sector.

Substitute 6 for r and 40 for m.

Simplify.

Find each arc length. Give your answer in terms of and rounded to the nearest hundredth.

an arc with measure 135° in a circle with radius 4 cm

Use formula for area of sector.

Substitute 4 for r and 135 for m.

= 3 cm  9.42 cm

Simplify.

• 6.7 Monday

• 6.7 Tuesday

• Chapter Review Wednesday

• Chapter Review Thursday

• Chapter 6 Test Friday