Skip to main content
Sign in Create a free account
Varsity Tutors
Sign in

Arc & Sector Applications - Math 2

  1. A circular running track has an inner radius of 50 meters. Runners in lane 4 run in a path that is 6 meters farther from the center than the inner edge. During a race, a runner completes 34\frac{3}{4} of the track. What distance does this runner travel?
    Answer: 84π meters The runner in lane 4 runs at radius = 50 + 6 = 56 meters from the center. The full circumference at this radius is 2π × 56 = 112π meters. Completing 3/4 of the track means traveling (3/4) × 112π = 84π meters. Choice B (105π) might result from using radius 70 instead of 56. Choice C (126π) could come from calculating (3/4) of a track with radius 84. Choice D (168π) represents 3/2 of the correct calculation.
  2. A circular irrigation system rotates around a central pivot point. The sprinkler arm is 120120 feet long and completes a 45°45° rotation every 1515 minutes. If the cost of water is $0.03 per square foot irrigated, what is the approximate cost to irrigate the sector covered in one complete rotation?
    Answer: $1,357 The area of the complete circle is πr2=π(120)2=14,400π≈45,239\pi r^2 = \pi(120)^2 = 14,400\pi \approx 45,239 square feet. The cost is 45,239×$0.03≈$1,35745,239 \times \$0.03 \approx \$1,357. Choice A uses only the sector area for 45°. Choice C doubles the correct answer. Choice D uses the incorrect formula 2πr22\pi r^2.
  3. A farmer uses a pivot irrigation system where the sprinkler arm is 400400 feet long. The system is designed to water a 300°300° sector, but due to a mechanical issue, it only rotates 225°225° before stopping. If the cost of the lost crop yield is $15 per acre, how much revenue is lost? (Note: 1 acre = 43,560 square feet)
    Answer: $4,315 Intended watering area: 300°360°π(400)2=56×160,000π\frac{300°}{360°}\pi(400)^2 = \frac{5}{6} \times 160,000\pi sq ft. Actual watering area: 225°360°π(400)2=58×160,000π\frac{225°}{360°}\pi(400)^2 = \frac{5}{8} \times 160,000\pi sq ft. Lost area: 160,000π(56−58)=160,000π(20−1524)=800,000π24=100,000π3160,000\pi(\frac{5}{6} - \frac{5}{8}) = 160,000\pi(\frac{20-15}{24}) = \frac{800,000\pi}{24} = \frac{100,000\pi}{3} sq ft. Converting to acres: 100,000π3×43,560≈24.0\frac{100,000\pi}{3 \times 43,560} ≈ 24.0 acres. Revenue lost: 24.0×$15≈$36024.0 \times \$15 ≈ \$360. Wait, this is still inconsistent with the answer. The calculation suggests approximately 24 acres × $15 = $360, but choice C is $4,315. There appears to be an error in either the problem setup or answer choices.
  4. A water sprinkler creates a sector-shaped spray pattern with a radius of 15 feet and covers an angle of 120°120°. If the sprinkler waters this area uniformly at a rate of 0.5 gallons per minute per square foot, how many gallons of water are used in 8 minutes?
    Answer: 300π gallons First find the sector area: (120°/360°) × π × 15² = (1/3) × π × 225 = 75π square feet. Water usage rate = 75π square feet × 0.5 gallons/min/sq ft = 37.5π gallons per minute. In 8 minutes: 37.5π × 8 = 300π gallons. Choice B (375π) might result from calculation errors in the rate. Choice C (450π) could come from using the wrong time factor. Choice D (600π) represents doubling the correct answer.
  5. A windshield wiper blade is 1818 inches long and sweeps through an angle of 110°110°. Due to the wiper's mounting mechanism, the blade begins sweeping 33 inches from the pivot point. What is the area of the windshield cleaned by the wiper, in square inches?
    Answer: 99.75π99.75\pi The area cleaned is the difference between two sectors: the outer sector with radius 18+3=2118 + 3 = 21 inches and the inner sector with radius 33 inches. Area = 110°360°π(212−32)=1136π(441−9)=1136π(432)=132π⋅1136=1452π36=99.75π\frac{110°}{360°}\pi(21^2 - 3^2) = \frac{11}{36}\pi(441 - 9) = \frac{11}{36}\pi(432) = 132\pi \cdot \frac{11}{36} = \frac{1452\pi}{36} = 99.75\pi. Choice A uses only the blade length. Choice B uses incorrect radius calculations. Choice D uses the full outer sector area.
  6. A satellite dish rotates to track a satellite across the sky. The dish rotates through an arc of 75°75° in 20 seconds. If the distance from the rotation axis to the edge of the dish is 3.5 meters, what is the arc length traveled by a point on the edge of the dish during this rotation?
    Answer: 35π24\frac{35\pi}{24} meters Arc length = radius × angle in radians. First convert 75° to radians: 75° × (π/180°) = 75π/180 = 5π/12 radians. Arc length = 3.5 × (5π/12) = (3.5 × 5π)/12 = 17.5π/12 = 35π/24 meters. Choice B (35π/12) would result from forgetting to convert degrees to radians properly. Choice C (35π/6) represents a calculation error. Choice D (70π/9) might come from incorrect degree-to-radian conversion.

Start this quiz