1. Section 11.4
Circumference and Arc Length
Theorem
Theorem 11.8: Circumference of a Circle
Circumference is the distance around a circle.
C = πd , when using diameter.
C = 2πr , when using radius.
C= dπ = 2πr

2. Use the formula for circumference
EXAMPLE 1
Use the formula for circumference
Use circumference formula to find the indicated measures.
a. Exact circumference of a circle with radius 9 centimeters. Leave π in your answer.
C = 2 πr
Write circumference formula.
= 2 π 9
Substitute 9 for r.
= 18 π
Simplify.
Circumference is about 18 π cm.

3. Use the formula for circumference
EXAMPLE 1
Use the formula for circumference
b. Radius of a circle with circumference 26 meters. Approximate to nearest hundredth.
C = 2 πr
Write circumference formula. =
2 πr 26
Substitute 26 for C. = 26 2π r
Divide each side by 2π r
4.14
Use calculator to divide 26 by 6.28.
Radius is about 4.14 meters.

4. EXAMPLE 2
Use circumference to find distance traveled
Tire Revolutions
The dimensions of a car tire are shown at the right. To the nearest foot, how far does the tire travel when it makes 15 revolutions?
STEP 1
Find the diameter of the tire d
= (5.5)
= 26 in.
STEP 2
Find the circumference of the tire C
= πd
= 3.14(26)
≈ in.

5. EXAMPLE 2
Use circumference to find distance traveled
STEP 3
Find the distance traveled in 15 revolutions.
One revolution equals its circumference.
Distance traveled = revolutions  circumference. 15
81.68 in.
= in.
STEP 4
Convert inches traveled to feet traveled.
in.
1 ft
12 in.
= ft
The tire travels approximately 102 feet.

6. GUIDED PRACTICE
for Examples 1 and 2
1. Find the exact circumference of a circle with diameter 5 inches. Leave π in your answer.
C = d π
= 5 π
Exact circumference is 5π inches.
2. Find the diameter of a circle with circumference 17 feet. Approximate to the nearest hundredth.
C = d π =
d π 17 = 17
3.14 d
Diameter is about 5.41 feet. d
5.41

7. Distance traveled = revolutions  circumference.
GUIDED PRACTICE
for Examples 1 and 2
3. A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?
Distance traveled = revolutions  circumference. C
= πd
= 3.14(28)
≈ in.
Find circumference
500 ft revolutions
87.92 in
revolutions
87.92 in
6000 in
about 68 revolutions

8. Section 11.4
Circumference and Arc Length
ARC LENGTH An arc length is a _____________ of a circumference of a circle. You use the measure of the _________ in degrees to find its __________________ in linear units.
portion
arc
length

9. Section 11.4
Circumference and Arc Length
Corollary
Arc Length Corollary
The ratio of arc length to circumference is equal to the ratio of arc degree to 360. or

10. EXAMPLE 3
Find arc lengths
Find the length of the indicated arc.
a. Length of Leave π in your answer.
2π(12)
60°
360° =
Length of AB
= 4π cm

11. EXAMPLE 3
Find arc lengths
Find the length of each indicated arc.
b. Length of Approximate to nearest hundredth.
2(3.14)(11)
120°
360° =
Length of GH
≈ cm

12. EXAMPLE 4
Find arc lengths to find measures
Find the indicated measure.
a. Circumference C of Z
Arc length of XY C
360°
m XY =
4.19 C
360°
40° =
4.19 C 9 1 =
37.71 = C
C = in

13. EXAMPLE 4
Find arc lengths to find measures
Find the indicated measure.
b. m RS (to nearest whole degree measure)
Arc length of RS
2 r
360°
m RS = 44
2(3.14)(15.28)
m RS
360° = 44
95.96
360°
= m RS
165°
m RS

14. GUIDED PRACTICE
for Examples 3 and 4
Find the indicated measure.
4. Length of PQ
9π yd
75°
360° =
Length of PQ
≈ yd

15. GUIDED PRACTICE
for Examples 3 and 4
Find the indicated measure.
5. Radius of G
Arc length of EF
2πr
360°
m EF =
10.5 ft
2(3.14)r
360°
150° =
10.5 ft
6.28r 12 5 =
126 ft = 31.4r
Radius is about 4.01 ft.