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 = 2r, 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:

C = 2r = 2 • • 6 = 12 37.70So, the circumference is about 37.70 cm.

C = 2r31 = 2r31 = r24.93 rSo, the radius is about 4.93 cm.

a. b.

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°.

AB

AB

P

B

A

Arc length of

2r =

360°

or Arc length of =360°

• 2r

m

AB

m

MORE . . .

• The length of a semicircle is half the circumference, and the length of a 90° arc is one quarter of the circumference.

½ • 2r

r¼ • 2r

rr

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.

EX. 2: FINDING ARC LENGTHS

AB

• Find the length of each arc.

5 cm

B

A

50°

a.

a. Arc length of = AB 50°

360°• 2(5)

a. Arc length of = # of °

360°• 2r

4.36 centimeters

EX. 2: FINDING ARC LENGTHS• Find the length of each arc.

7 cm

D

C

50°

b. b. Arc length of = CD # of °

360°• 2r

b. Arc length of = CD 50°

360°• 2(7)

6.11 centimeters

EX. 2: FINDING ARC LENGTHS

EF

• Find the length of each arc.

7 cm

F

E

100°

c. c. Arc length of = # of °

360°• 2r

c. Arc length of = EF 100°

360°• 2(7)

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.

PQ

3.82 m

R

Q

P

60°

a. circumference Arc length of

2r =PQ

m

360°

3.82

2r 6

1=

3.82

2r 360°

60°=

3.82(6) = 2r

22.92 = 2r

C = 2r; so using substitution, C = 22.92 meters.

EX. 3: USING ARC LENGTHS• Find the indicated measure.

XY

XY

XY

b. m Arc length of

2r = 360°

18

2(7.64) 360°=

135° m

XY

m

XY

m• 360°360° •

So the m 135°XY

7.64 in.

18 in.

Z

Y

X

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 2r.

• 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 2r.

• 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.= 100 ft.

80.11 in.1200 in.

80.11 in.=

15.8 revolutions

TIRE B:

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 LENGTHa. 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 + 2r1

= 2(108.9) + 2(29.00) 400.0 meters

• Distance = 2s + 2r2

= 2(108.9) + 2(30.25) 407.9 meters

Ex. 5: Lane 2

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