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

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.

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.

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 = 15 + 2 (5.5) = 26 in. STEP 2 Find the circumference of the tire C = πd = 3.14(26) ≈ 81.68 in.

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. = 1225.2 in. STEP 4 Convert inches traveled to feet traveled. 1225.2 in. 1 ft 12 in. = 102.1 ft The tire travels approximately 102 feet.

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

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) ≈ 87.92 in. Find circumference 500 ft revolutions 87.92 in revolutions 87.92 in 6000 in about 68 revolutions

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

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

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

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 ≈ 23.03 cm

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 = 37.71 in

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

GUIDED PRACTICE for Examples 3 and 4 Find the indicated measure. 4. Length of PQ 9π yd 75° 360° = Length of PQ ≈ 5.89 yd

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.