11.4 Circumference and Arc Length

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11.4 Circumference and Arc Length

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Presentation transcript:

11.4 Circumference and Arc Length Unit IIIH Day 4

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

Ex. 3 Find the indicated measure. a. length arc PQ/ 2pi r = m arc PQ / 360º 3.82/2pi r = 60.360 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

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 14 + 2(5.1), or 24.2 inches. Its circumference is π(24.2), or about 76.03 inches. Tire B has a diameter of 15 + 2(5.25), or 25.5 inches. Its circumference is π(25.5), or about 80.11 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

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)  400.0 meters b. Distance = 2s + 2r2 = 2(108.9) + 2(30.25)  407.9 meters

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