Radians & Arc Length - IB Mathematics Applications and Interpretation

1. A restaurant serves a slice of pie that is a sector of a circle with a central angle of $\frac{\pi}{6}$ radians. If the arc length of the outer crust of the slice is 12 cm, what is the area of the slice of pie, correct to one decimal place?
   Answer: 137.5 cm² First, find the radius $r$ of the pie using the arc length formula $s = r\theta$. We have $12 = r \times \frac{\pi}{6}$, which gives $r = \frac{72}{\pi}$ cm. Now, calculate the area of the sector using the formula $A = \frac{1}{2}r^2\theta$ or the simpler formula $A = \frac{1}{2}rs$. Using the latter, $A = \frac{1}{2} \times \frac{72}{\pi} \times 12 = \frac{432}{\pi} \approx 137.51$ cm². To one decimal place, the area is 137.5 cm².
2. A circular arc has a length of 15 cm and is subtended by a central angle of 2.5 radians. This arc is bent to form a full circle. What is the radius of this new circle?
   Answer: 2.39 cm The length of the circular arc is 15 cm. When this arc is bent to form a full circle, its length becomes the circumference of the new circle. The formula for circumference is $C = 2\pi r_{new}$. So, $15 = 2\pi r_{new}$. Solving for the new radius: $r_{new} = \frac{15}{2\pi} \approx 2.387$ cm. The closest answer is 2.39 cm. The original angle of 2.5 radians and the original radius are extra information not needed to solve the problem.
3. What is the total area, in m², that the goat can graze?
   Answer: $53\pi$ m² The grazing area consists of three sectors. First, a large sector with radius 8 m covering three-quarters of a circle: $A_1 = \frac{3}{4} \times \pi (8^2) = \frac{3}{4} \times 64\pi = 48\pi$ m². When the rope goes around the 6 m side, its effective length becomes $8 - 6 = 2$ m. This allows grazing in a quarter-circle sector of radius 2 m: $A_2 = \frac{1}{4} \times \pi (2^2) = \pi$ m². When the rope goes around the 4 m side, its effective length becomes $8 - 4 = 4$ m. This allows grazing in a quarter-circle sector of radius 4 m: $A_3 = \frac{1}{4} \times \pi (4^2) = 4\pi$ m². The total area is the sum: $A_{total} = A_1 + A_2 + A_3 = 48\pi + \pi + 4\pi = 53\pi$ m².
4. A cogwheel with a radius of 6 cm meshes with a larger cogwheel of radius 10 cm. If the smaller wheel rotates through an angle of $1.5\pi$ radians, what is the angle of rotation of the larger wheel?
   Answer: 0.90π radians When two cogwheels mesh, the arc length travelled by the teeth on their circumferences must be equal. Let the radii be $r_1=6$ and $r_2=10$, and the angles be $\theta_1=1.5\pi$ and $\theta_2$. The condition is $s_1=s_2$, which means $r_1\theta_1 = r_2\theta_2$. Plugging in the values: $6 \times 1.5\pi = 10 \times \theta_2$. $9\pi = 10\theta_2$. $\theta_2 = \frac{9\pi}{10} = 0.9\pi$ radians.
5. A wind turbine blade is 50 m long. The tip of a blade travels 15.7 km in one hour. What is the average angular velocity of the turbine in radians per minute?
   Answer: 5.24 rad/min First, convert all units to be consistent. The distance is $s = 15.7 \text{ km} = 15700 \text{ m}$. The radius is $r=50$ m. The time is $t = 1 \text{ hour} = 60 \text{ minutes}$. The total angle $\theta$ turned in one hour is found using $s=r\theta$: $\theta = \frac{s}{r} = \frac{15700}{50} = 314$ radians. The angular velocity $\omega$ is the angle turned per unit time: $\omega = \frac{\theta}{t} = \frac{314 \text{ radians}}{60 \text{ minutes}} \approx 5.236$ rad/min. This is approximately $5\pi/3$. The closest answer is 5.24 rad/min.
6. The minute hand on a clock tower is 4.5 metres long. What is the area swept by the minute hand between 10:10 am and 10:35 am?
   Answer: 26.5 m² When you see a clock hand sweeping through time, you're looking at a sector of a circle. The key insight is that the minute hand traces out a portion of a full circle, and you need to find that sector's area. First, determine how much the minute hand moves. From 10:10 to 10:35 is 25 minutes. Since the minute hand completes a full 360° rotation in 60 minutes, it moves $\frac{25}{60} = \frac{5}{12}$ of a full circle. The area of a sector is $A = \frac{\theta}{360°} \times \pi r^2$ where θ is the angle in degrees. Here, the angle is $\frac{5}{12} \times 360° = 150°$. With radius r = 4.5 m: $A = \frac{150°}{360°} \times \pi \times (4.5)^2 = \frac{5}{12} \times \pi \times 20.25 = \frac{5 \times 20.25\pi}{12} = \frac{101.25\pi}{12} ≈ 26.5 \text{ m}^2$ Looking at the wrong answers: A) 15.9 m² likely comes from using 90° instead of 150° (perhaps calculating 15 minutes of movement instead of 25). B) 53.0 m² is approximately double the correct answer, suggesting someone might have calculated the full semicircle or made an error with the fraction. C) 31.8 m² appears to use an incorrect time calculation, possibly 30 minutes instead of 25. Remember: Always convert the time interval to a fraction of the full hour first, then multiply by 360° to get your sector angle. Clock problems are really just sector area problems in disguise.
7. A car wheel has a diameter of 60 cm. The car travels 100 metres. How many full revolutions has the wheel made?
   Answer: 53 First, ensure units are consistent. The wheel diameter is 60 cm = 0.6 m, so its radius is $r=0.3$ m. The distance travelled is $d=100$ m. The circumference of the wheel, which is the distance covered in one revolution, is $C = 2\pi r = 2\pi(0.3) = 0.6\pi$ m. The number of revolutions is the total distance divided by the circumference: $N = \frac{100}{0.6\pi} \approx 53.05$. The question asks for the number of full revolutions, which is 53.
8. A radar scans a sector-shaped region. The radar can detect objects between 5 km and 25 km away. If the radar sweeps through an angle of 2.4 radians, what is the area of the scanned region?
   Answer: 720 km² This question tests your understanding of sector area calculations, which combine concepts of circles and angles measured in radians. When you see a "sector-shaped region" with given radii and angle, you're looking at finding the area between two concentric circles over a specific angular span. The area of a sector is calculated using the formula: Area = $\frac{1}{2}r^2\theta$, where $r$ is the radius and $\theta$ is the angle in radians. However, since this radar scans between two distances, you need to find the area of the larger sector and subtract the area of the smaller sector. For the outer sector (radius = 25 km): Area = $\frac{1}{2}(25)^2(2.4) = \frac{1}{2}(625)(2.4) = 750$ km² For the inner sector (radius = 5 km): Area = $\frac{1}{2}(5)^2(2.4) = \frac{1}{2}(25)(2.4) = 30$ km² The scanned region area = 750 - 30 = 720 km², confirming answer D. Looking at the wrong answers: A) 30 km² is just the inner sector area that should be subtracted. B) 780 km² likely comes from calculation errors, possibly adding instead of subtracting or using incorrect radius values. C) 750 km² is the outer sector area before subtracting the inner "dead zone." Remember: sector problems involving two radii always require subtraction to find the ring-shaped area. Don't forget to subtract the inner region that the radar cannot scan.
9. A large pizza with a diameter of 40 cm is cut into 12 equal slices. What is the perimeter of one slice?
   Answer: 50.5 cm The diameter is 40 cm, so the radius is $r=20$ cm. A full circle is $2\pi$ radians. Since there are 12 equal slices, the angle of one slice is $\theta = \frac{2\pi}{12} = \frac{\pi}{6}$ radians. The arc length of one slice is $s = r\theta = 20 \times \frac{\pi}{6} = \frac{10\pi}{3} \approx 10.47$ cm. The perimeter of one slice is $P = 2r + s = 2(20) + \frac{10\pi}{3} = 40 + 10.47 \approx 50.47$ cm. The closest answer is 50.5 cm.
10. A security camera on a wall rotates horizontally. It can scan a sector-shaped area of 50 m². If the maximum range of the camera is 12 m, what is its angle of scan, in degrees, correct to one decimal place?
    Answer: 39.8° The area of a sector is given by $A = \frac{1}{2}r^2\theta$, where $\theta$ is in radians. We have $A=50$ m² and $r=12$ m. $50 = \frac{1}{2}(12^2)\theta \implies 50 = 72\theta \implies \theta = \frac{50}{72}$ radians. To convert this to degrees, we multiply by $\frac{180}{\pi}$: Angle in degrees = $\frac{50}{72} \times \frac{180}{\pi} \approx 39.7887...$°. To one decimal place, the angle is 39.8°.
11. A reel of thread has a radius of 2 cm. A length of 5 metres of thread is unwound from it. How many full rotations did the reel make, assuming the radius does not change?
    Answer: 39 The length of thread unwound corresponds to the arc length. First, convert units: $s = 5$ m = 500 cm. The radius is $r = 2$ cm. The total angle of rotation $\theta$ is given by $s = r\theta$, so $\theta = \frac{s}{r} = \frac{500}{2} = 250$ radians. One full rotation is $2\pi$ radians. The number of rotations is $N = \frac{\text{total angle}}{\text{angle per rotation}} = \frac{250}{2\pi} \approx 39.7887...$. The number of full rotations is 39.
12. A metal fabricator has a 30 cm piece of wire that needs to be shaped into the perimeter of a sector for a decorative art piece. If the radius of the sector must be 8 cm, what is the area of the resulting sector?
    Answer: 56 cm² The perimeter of a sector is given by $P = 2r + s$, where $s$ is the arc length. We are given $P=30$ cm and $r=8$ cm. So, $30 = 2(8) + s \implies 30 = 16 + s \implies s = 14$ cm. The area of a sector can be calculated using the formula $A = \frac{1}{2}rs$. Substituting the known values: $A = \frac{1}{2}(8)(14) = 56$ cm².
13. A decorative circular garden has a radius of 15 meters. A paved path forms a sector of this garden with a total perimeter of 40 meters. What is the central angle of the sector in radians, to two decimal places?
    Answer: 0.67 radians The perimeter of a sector is composed of two radii and the arc length: $P = 2r + s$. We are given $P=40$ m and $r=15$ m. The arc length is $s = r\theta$. Substituting this into the perimeter formula gives $P = 2r + r\theta$. So, $40 = 2(15) + 15\theta$. $40 = 30 + 15\theta$. $10 = 15\theta$. $\theta = \frac{10}{15} = \frac{2}{3} \approx 0.67$ radians.
14. A satellite in a circular orbit 500 km above the Earth's surface travels through an angle of 0.25 radians with respect to the center of the Earth. The radius of the Earth is approximately 6370 km. How far did the satellite travel, to the nearest kilometer?
    Answer: 1718 km First, calculate the total radius of the satellite's orbit by adding the Earth's radius and the satellite's altitude: $r_{orbit} = 6370\text{ km} + 500\text{ km} = 6870\text{ km}$. The distance the satellite travels is the arc length, given by $s = r\theta$. Using $\theta = 0.25$ radians: $s = 6870 \times 0.25 = 1717.5$ km. To the nearest kilometer, the distance is 1718 km.

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