APK
5-MINUTE CHECK
11.4 CIRCUMFERENCE AND ARC
LENGTH
OBJECTIVES
• Find the circumference of a circle and the length of
a circular arc.
• Why? So you can use circumference and arc
length to solve real-life problems.
• Mastery is 80% or better on 5-minute checks and
problems.
CONCEPT DEV - FINDING
CIRCUMFERENCE & 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.
CONCEPT DEV-THEOREM 11.8:
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
SKILL DEV - 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.
b.
C = 2r
=2••6
= 12
 37.70
So, the
circumference is
about 37.70 cm.
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
60°
3.82 m
Q

Arc length of PQ
=
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
2r
b. m XY
X
18 in.
360° •
18

m XY
=
=
2(7.64)
360°

m XY
360°

Z
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
EXIT SLIPS
• What was the Objective for today?
• Students will analyze and determine arc lengths
and other measures.
• Why? So you can find a running distance, as in
example 5.
• Mastery is 80% or better on 5-min checks and
practice problems.
HOMEWORK
• Page 749-750
• #3 – 28 ALL