Circle Measure Applications - Math 3

A satellite orbits Earth in a circular path at altitude 400 km above Earth's surface. If Earth's radius is 6400 km and the satellite completes one orbit in 90 minutes, what is the satellite's speed in km/h?
Answer: $\frac{544\pi}{3}$ km/h The satellite's orbital radius is 6400 + 400 = 6800 km. The circumference of the orbit is 2π(6800) = 13,600π km. The satellite travels this distance in 90 minutes = 1.5 hours. Speed = distance/time = 13,600π/1.5 = 13,600π ÷ (3/2) = 13,600π × (2/3) = 27,200π/3 = 544π/3 km/h. Choice A uses only Earth's radius without the altitude. Choice B incorrectly converts 90 minutes to hours. Choice D uses Earth's radius instead of orbital radius.
Two gears are connected such that when the larger gear with radius 12 cm completes $\frac{2}{3}$ of a revolution, the smaller gear with radius 8 cm completes exactly one full revolution. What is the total distance traveled by a point on the circumference of the smaller gear during this motion?
Answer: $16\pi$ cm When the smaller gear completes one full revolution, a point on its circumference travels a distance equal to the circumference: 2π(8) = 16π cm. The information about the larger gear's motion is given to establish the gear ratio and ensure the motion is physically consistent, but the question asks specifically for the distance traveled by the point on the smaller gear, which is simply one complete circumference. Choice A gives the radius times π. Choice C uses the larger gear's radius. Choice D gives 1.5 revolutions worth of distance.
A sector of a circle has area $24\pi$ square units and arc length $8\pi$ units. What is the radius of the circle?
Answer: 6 units For a sector, Area = (1/2)r²θ and arc length s = rθ, where θ is in radians. From s = rθ, we have 8π = rθ, so θ = 8π/r. Substituting into the area formula: 24π = (1/2)r²(8π/r) = (1/2)r(8π) = 4πr. Solving: 24π = 4πr, so r = 6 units. Choice B comes from incorrectly using θ = 8π/r in Area = (1/2)rθ. Choice C assumes the arc length equals the radius times π. Choice D results from calculation errors in the substitution process.
A circular running track has two lanes. The inner lane has radius 50 meters, and the outer lane has radius 54 meters. If a runner in the outer lane maintains the same angular velocity as a runner in the inner lane, what is the ratio of their linear speeds?
Answer: $\frac{54}{50}$ For circular motion, linear speed v = rω where r is radius and ω is angular velocity. Since both runners have the same angular velocity ω, the ratio of their linear speeds is v_outer/v_inner = (r_outer × ω)/(r_inner × ω) = r_outer/r_inner = 54/50 = 27/25. However, the question asks for the ratio with the outer lane runner first, so it's 54/50. Choice A gives the reciprocal ratio. Choice B simplifies 54/50 incorrectly. Choice C is the same as choice A but not simplified.
If the sprinkler system rotates through $240°$ and each arm covers a sector with no overlap between adjacent arms, what is the total area watered by all three arms?
Answer: $128\pi$ square feet Each arm has radius $8$ feet and rotates through $240°$ . The area covered by one arm is $A = \frac{240°}{360°} \times \pi(8)^2 = \frac{2}{3} \times 64\pi = \frac{128\pi}{3}$ . Since there are three arms: Total area $= 3 \times \frac{128\pi}{3} = 128\pi$ square feet. Choice B assumes each arm covers $\frac{240°}{3} = 80°$ individually. Choice C calculates the area of a full circle with radius $8$ . Choice D incorrectly multiplies the full sector area by $3$ and then by $2$ .
A circular sector has central angle $\theta$ radians and radius $r$ . If the sector is rolled into a cone (with the arc becoming the base circumference), what is the ratio of the cone's base radius to the original sector's radius?
Answer: $\frac{\theta}{2\pi}$ When the sector is rolled into a cone, the arc length of the sector becomes the circumference of the cone's base. The arc length of the sector is rθ. If R is the radius of the cone's base, then its circumference is 2πR. Setting these equal: rθ = 2πR, so R = rθ/(2π). The ratio of the cone's base radius to the original sector's radius is R/r = θ/(2π). Choice B gives the reciprocal. Choice C omits the factor of 2. Choice D is the reciprocal of choice C.
A pizza slice has a central angle of $72°$ and the crust (arc) measures $4\pi$ inches. If the entire pizza costs $18, what is the cost per square inch of this slice?
Answer: $\frac{9}{50\pi}$ dollars per square inch First find the radius: arc length = rθ, so 4π = r(72° × π/180°) = r(2π/5), giving r = 10 inches. The slice area is (72°/360°) × π(10)² = (1/5) × 100π = 20π square inches. The slice costs (72°/360°) × $18 = (1/5) × $18 = $3.60. Cost per square inch = $3.60/(20π) = 3.6/(20π) = 18/(100π) = 9/(50π) dollars per square inch. Choice A incorrectly uses the full pizza area. Choice C uses wrong angle proportion. Choice D uses incorrect area calculation.
A bicycle wheel has a radius of $14$ inches. If the wheel makes exactly $5$ complete rotations while traveling in a straight line, and then the bike turns through a circular arc where the wheel makes an additional $2.5$ rotations, what is the total distance traveled by a point on the rim of the wheel?
Answer: $210\pi$ inches The circumference of the wheel is $2\pi r = 2\pi(14) = 28\pi$ inches. In straight line motion: $5 \times 28\pi = 140\pi$ inches. During the turn: $2.5 \times 28\pi = 70\pi$ inches. Total distance: $140\pi + 70\pi = 210\pi$ inches. Choice A miscalculates the circumference as $196\pi/7$ . Choice C only accounts for straight-line motion. Choice D uses $6$ rotations instead of $7.5$ .
A clock's minute hand has length 6 cm. Between 2:15 PM and 2:42 PM, what distance does the tip of the minute hand travel?
Answer: $5.4\pi$ cm From 2:15 to 2:42 is 27 minutes. The minute hand moves 360° in 60 minutes, so in 27 minutes it moves (27/60) × 360° = 162°. Converting to radians: 162° × (π/180°) = 9π/10 radians. Arc length = rθ = 6 × (9π/10) = 54π/10 = 5.4π cm. Choice A uses 27° instead of 162°. Choice B calculates for 36 minutes instead of 27. Choice D uses the wrong conversion factor.
A circular irrigation sprinkler rotates $120°$ every 15 seconds and waters a sector with radius 8 meters. If the sprinkler operates for 2.5 minutes, what is the total area watered?
Answer: $64\pi$ square meters In 2.5 minutes (150 seconds), the sprinkler makes 150÷15 = 10 rotations of 120°. Total rotation = 10 × 120° = 1200° = 1200° × (π/180°) = 20π/3 radians. Since 20π/3 > 2π, the sprinkler completes more than one full rotation, watering the entire circle. Area = πr² = π(8)² = 64π square meters. Choice B gives the area of one 120° sector. Choice C incorrectly doubles the sector area. Choice D uses an incorrect radius calculation.
A circular running track has a radius of $50$ meters. A runner starts at point A and runs along the track. After running for $157$ meters, she has completed an arc that subtends a central angle of $\theta$ . What is the measure of $\theta$ in degrees?
Answer: $180°$ Using the arc length formula $s = r\theta$ where $\theta$ is in radians: $157 = 50\theta$ , so $\theta = \frac{157}{50} = 3.14$ radians. Converting to degrees: $\theta = 3.14 \times \frac{180°}{\pi} \approx 3.14 \times \frac{180°}{3.14159} \approx 180°$ . More precisely, $\frac{157}{50} = 3.14$ , and $\pi \approx 3.14159$ , so $\theta \approx \pi$ radians $= 180°$ . Choice A would correspond to $\frac{\pi}{2}$ radians or arc length $78.5$ m. Choice C would correspond to $\frac{2\pi}{3}$ radians or arc length $104.7$ m. Choice D would correspond to $\frac{5\pi}{6}$ radians or arc length $130.9$ m.
A pendulum swings through an arc of $60°$ on either side of vertical, creating a total swing angle of $120°$ . If the pendulum's length is 80 cm, what is the difference between the arc length of the complete swing and the straight-line distance between the two extreme positions?
Answer: $\frac{160\pi}{3} - 80\sqrt{3}$ cm When you encounter pendulum problems involving arc length and chord distance, you're working with circular geometry concepts. The key is distinguishing between the curved path the pendulum travels and the straight-line distance between its endpoints. First, calculate the arc length of the complete swing. The pendulum swings $120°$ total, and you need this in radians: $120° = \frac{120\pi}{180} = \frac{2\pi}{3}$ radians. With radius (pendulum length) of 80 cm, the arc length is $s = r\theta = 80 \cdot \frac{2\pi}{3} = \frac{160\pi}{3}$ cm. Next, find the straight-line distance between extreme positions. This forms an isosceles triangle with two sides of 80 cm and an included angle of $120°$ . Using the law of cosines: $c^2 = 80^2 + 80^2 - 2(80)(80)\cos(120°)$ . Since $\cos(120°) = -\frac{1}{2}$ , you get $c^2 = 6400 + 6400 + 6400 = 19200$ , so $c = 80\sqrt{3}$ cm. The difference is $\frac{160\pi}{3} - 80\sqrt{3}$ cm, which is answer D. Answer A uses $\frac{80\pi}{3}$ instead of $\frac{160\pi}{3}$ , incorrectly using $60°$ rather than the full $120°$ swing. Answer B correctly calculates the arc length but uses $40\sqrt{3}$ for the chord, likely from using only half the triangle. Answer C has both the wrong arc length and wrong chord distance. Remember: always convert degrees to radians for arc length calculations, and use the full swing angle, not just one side of the pendulum's motion.
A circular sprinkler system rotates through an angle of $120°$ and waters a sector of a lawn. If the sprinkler has a radius of $15$ feet and operates for $8$ minutes, covering the entire sector twice during this time, what is the total area watered?
Answer: $150\pi$ square feet The area of a sector is $A = \frac{\theta}{360°} \times \pi r^2$ . With $\theta = 120°$ and $r = 15$ : $A = \frac{120°}{360°} \times \pi (15)^2 = \frac{1}{3} \times 225\pi = 75\pi$ . Since the sprinkler covers this sector twice, the total area is $2 \times 75\pi = 150\pi$ square feet. Choice B gives the area for one pass only. Choice C incorrectly uses the full circle area formula. Choice D uses an incorrect angle calculation.

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