1. 11.4 Circumference and Arc Length
Unit IIIH Day 4

2. Do Now
Find the indicated measure.
about 120º
about 81.6 cm.

3. Ex. 3 Find the indicated measure.
a. length arc PQ/ 2pi r = m arc PQ / 360º
3.82/2pi r =
3.82 = 1/6
22.92 = 2pi r = C
b. length arc XY/ 2pi r = m arc XY / 360º
18 / 2pi*7.64 = m arc XY/360º
360º *18 / 2pi*7.64 = m arc XY
135º = m arc XY

4. Ex. 4: Comparing Circumferences
Tires from two different automobiles are shown below. How many revolutions does each tire make while traveling 100 feet? Round decimal answers to one decimal place. Hint: Units!
Tire A has a diameter of (5.1), or 24.2 inches. Its circumference is π(24.2), or about inches.
Tire B has a diameter of (5.25), or 25.5 inches. Its circumference is π(25.5), or about inches.
Divide the distance traveled by the tire circumference to find the number of revolutions made. First convert 100 feet to 1200 inches.
A: 100ft. / 76.03in. = 1200in. / 76.03in. = 15.8 revolutions
B: 100ft. / 80.11in. = 1200in. / 80.11in. = 15.0 revolutions

5. Ex. 5: Finding Arc Length (Track)
The track shown has six lanes, each 1.25 meters wide. There is a 180° arc at the end of each track. The radii for the arcs in the first two lanes are given.
Find the distance around Lane 1.
Find the distance around Lane 2.
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.
a. Distance = 2s + 2r1
= 2(108.9) + 2(29.00)
 meters
b. Distance = 2s + 2r2
= 2(108.9) + 2(30.25)
 meters

6. Closure
What proportion can you use to find the arc length of a given arc?