11.4 circumference and arc length

1. 11.4 Circumference and Arc Length

2. 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.

3. 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 p or pi.

4. Theorem 11.6: Circumference of
a Circle
•The circumference
C of a circle is C =
pd or C = 2pr,
where d is the
diameter of the
circle and r is the
radius of the circle.
diameter d

5. Ex. 1: Using circumference
•(a) Find the circumference of a circle
with radius 6 centimeters
•(b) Find the radius of a circle with
circumference 31 meters. Round
decimal answers to two decimal
places.

6. Solution:
C = 2pr
= 2 • p • 6
= 12p
» 37.70
/So, the
circumference is
about 37.70 cm.
C = 2pr
31 = 2pr
31 = r
4.93 » r
/So, the radius is
about 4.93 cm.
2p
a.
b.

7. 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).

8. 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°.
AB

P
B
A
Arc length of
2pr
=
360°
or Arc length of =
360°
• 2pr
mAB

AB

m

9. More . . .
•The length of a
semicircle is half the
circumference, and
the length of a 90°
arc is one quarter of
the circumference.
½ • 2pr
r
¼ • 2pr
r
r

10. Ex. 2: Finding Arc Lengths
•Find the length of each arc.
5 cm
B
A
50°
a.
7 cm
D
C
50°
b. 7 cm
F
E
100°
c.

11. Ex. 2: Finding Arc Lengths
•Find the length of each arc.
5 cm
B
A
50°
a.
a. Arc length of = AB
 50°
360°
• 2p(5)
a. Arc length of = AB
 # of °
360°
• 2pr
» 4.36 centimeters

12. Ex. 2: Finding Arc Lengths
•Find the length of each arc.
7 cm
D
C
50°
b. b. Arc length of = CD
 # of °
360°
• 2pr
b. Arc length of = CD
 50°
360°
• 2p(7)
» 6.11 centimeters

13. Ex. 2: Finding Arc Lengths
•Find the length of each arc.
7 cm
F
E
100°
c. c. Arc length of =
# of °
360°
• 2pr
c. Arc length of = EF
 100°
360°
• 2p(7)
EF

» 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.

14. Ex. 3: Using Arc Lengths
•Find the indicated measure.
3.82 m
R
Q
P
60°
a. circumference
PQ

Arc length of
2pr
=
PQ

m
360°
3.82
2pr 6
1
=
3.82
2pr 360°
60°
=
3.82(6) = 2pr
22.92 = 2pr
C = 2pr; so using substitution, C = 22.92
meters.

15. Ex. 3: Using Arc Lengths
•Find the indicated measure.
XY

b. m
XY

Arc length of
2pr
=
360°
18
2p(7.64)360°
=
135° » m
XY

m
XY

m
• 360°
360° •
XY

/So the m » 135°XY

7.64 in.
18 in.
Z
Y
X

16. 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.

17. •Reminder: C = pd or
2pr.
•Tire A has a diameter
of 14 + 2(5.1), or 24.2
inches.
•Its circumference is
p(24.2), or about 76.03
inches.
Ex. 4: Comparing Circumferences

18. •Reminder: C = pd or
2pr.
•Tire B has a diameter
of 15 + 2(5.25), or 25.5
inches.
•Its circumference is
p(25.5), or about 80.11
inches.
Ex. 4: Comparing Circumferences

19. •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.
=
100 ft.
80.11 in.
1200 in.
80.11 in.=
» 15.8 revolutions
TIRE B:
» 15.0 revolutions
Ex. 4: Comparing Circumferences

20. 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.

21. 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.

22. •Distance = 2s + 2pr
1
= 2(108.9) + 2p(29.00)
» 400.0 meters
•Distance = 2s + 2pr
2
= 2(108.9) + 2p(30.25)
» 407.9 meters
Ex. 5: Lane 1
Ex. 5: Lane 2