11.4 Circumference and Arc Length
Geometry
Mrs. Spitz
Spring 2006
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.
• Assignment: pp. 686-688 #1-38
Finding circumference and arc
length
• 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. This
ratio is known as  or pi.
Theorem 11.6: Circumference of
a Circle
• 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.
diameter d
Ex. 1: Using circumference
• Find (a) the circumference of a circle
with radius 6 centimeters and (b) the
radius of a circle with circumference
31 meters. Round decimal answers
to two decimal places.
Solution:
a.
C = 2r
=2••6
= 12
 37.70
So, the
circumference is
about 37.70 cm.
b.
C = 2r
31 = 2r
31 = r
2
4.93  r
So, the radius is
about 4.93 cm.
And . . .
• An arc length is a portion of the
circumference of a circle. You can
use the measure of an arc (in
degrees) to find its length (in linear
units).
Arc Length Corollary
• 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 length of AB
A
P
B

=
2r
or Arc length of


m AB
360°
=
m AB
• 2r
360°
More . . .
• The length of a
semicircle is half the
circumference, and
the length of a 90°
arc is one quarter of
the circumference.
½ • 2r
r
r
r
¼ • 2r
Ex. 2: Finding Arc Lengths
• Find the length of each arc.
E
a.
5 cm
A
50°
B
b.
7 cm
C
50°
c.
7 cm
100°
D
F
Ex. 2: Finding Arc Lengths
• Find the length of each arc.
a.

a. Arc length of AB
5 cm
# of °
=
A
50°
B

a. Arc length of AB
360°
50°
=
• 2r
• 2(5)
360°
 4.36 centimeters
Ex. 2: Finding Arc Lengths
• Find the length of each arc.
b.


b. Arc length of CD
7 cm
C
50°
D
b. Arc length of CD
# of °
=
360°
50°
=
• 2r
• 2(7)
360°
 6.11 centimeters
Ex. 2: Finding Arc Lengths
• Find the length of each arc.
E
c.


c. Arc length of EF
7 cm
100°
c. Arc length of EF
F
# of °
=
360°
100°
=
• 2r
• 2(7)
360°
 12.22 centimeters
In parts (a) and (b) in Example 2, note that the
arcs have the same measure but different
lengths because the circumferences of the
circles are not equal.
Ex. 3: Using Arc Lengths
• Find the indicated measure.

a. circumference
=
2r
3.82
P
R

Arc length of PQ
60°
3.82 m
Q
=
2r
3.82
2r
m PQ
360°
60°
360°
=
1
6
3.82(6) = 2r
22.92 = 2r
C = 2r; so using substitution, C = 22.92
meters.
Ex. 3: Using Arc Lengths
• Find the indicated measure.


Arc length of

XY
b. m XY
X
=
2r
18 in.
Z
360° •
m XY
18
=
2(7.64)
360°

m XY
360°

135°  m XY
7.64 in.
Y

So the mXY 
135°
• 360°
Ex. 4: Comparing Circumferences
• Tire Revolutions: Tires
from two different
automobiles are shown
on the next slide. How
many revolutions does
each tire make while
traveling 100 feet?
Round decimal answers
to one decimal place.
Ex. 4: Comparing Circumferences
• Reminder: C = d or
2r.
• Tire A has a diameter of
14 + 2(5.1), or 24.2
inches.
• Its circumference is
(24.2), or about 76.03
inches.
Ex. 4: Comparing Circumferences
• Reminder: C = d or
2r.
• Tire B has a diameter
of 15 + 2(5.25), or 25.5
inches.
• Its circumference is
(25.5), or about 80.11
inches.
Ex. 4: Comparing Circumferences
• Divide the distance traveled by the tire
circumference to find the number of revolutions
made. First, convert 100 feet to 1200 inches.
TIRE A:
100 ft.
76.03 in.
1200 in.
= 76.03 in.
 15.8 revolutions
TIRE B:
100 ft.
80.11 in.
1200 in.
= 80.11 in.
 15.0 revolutions
Ex. 5: Finding Arc Length
•
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.
a. Find the distance around Lane 1.
b. Find the distance around Lane 2.
Ex. 5: Finding Arc Length
a.

Find the distance around Lane 1.
The track is made up of two semicircles and two
straight sections with length s. To find the total
distance around each lane, find the sum of the
lengths of each part. Round decimal answers to
one decimal place.
Ex. 5: Lane 1
• Distance = 2s + 2r1
= 2(108.9) + 2(29.00)
 400.0 meters
Ex. 5: Lane 2
• Distance = 2s + 2r2
= 2(108.9) + 2(30.25)
 407.9 meters
Upcoming
• 11.5 Monday
• 11.6 Wednesday
• Chapter 11 Test Friday
w/review before the test.
• Binder check
• Chapter 12 Postulates/Thms.
• Chapter 12 Definitions