IB Mathematics: Applications and Interpretation Quiz: Radians And Arc Length
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Radians And Arc LengthQuestion 1 of 20

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?

0.67 radians
1.33 radians
1.67 radians
2.67 radians
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Radians And Arc Length

Practice Radians And Arc Length in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Radians And Arc Length, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

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?

  1. 0.67 radians (correct answer)
  2. 1.33 radians
  3. 1.67 radians
  4. 2.67 radians
Explanation: The perimeter of a sector is composed of two radii and the arc length: P=2r+sP = 2r + s. We are given P=40P=40 m and r=15r=15 m. The arc length is s=rθs = r\theta. Substituting this into the perimeter formula gives P=2r+rθP = 2r + r\theta. So, 40=2(15)+15θ40 = 2(15) + 15\theta. 40=30+15θ40 = 30 + 15\theta. 10=15θ10 = 15\theta. θ=1015=23≈0.67\theta = \frac{10}{15} = \frac{2}{3} \approx 0.67 radians.

Question 2

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 1.5π1.5\pi radians, what is the angle of rotation of the larger wheel?

  1. 0.54π radians
  2. 0.90π radians (correct answer)
  3. 1.50π radians
  4. 2.50π radians
Explanation: When two cogwheels mesh, the arc length travelled by the teeth on their circumferences must be equal. Let the radii be r1=6r_1=6 and r2=10r_2=10, and the angles be θ1=1.5π\theta_1=1.5\pi and θ2\theta_2. The condition is s1=s2s_1=s_2, which means r1θ1=r2θ2r_1\theta_1 = r_2\theta_2. Plugging in the values: 6×1.5π=10×θ26 \times 1.5\pi = 10 \times \theta_2. 9π=10θ29\pi = 10\theta_2. θ2=9π10=0.9π\theta_2 = \frac{9\pi}{10} = 0.9\pi radians.

Question 3

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?

  1. 0.67 radians (correct answer)
  2. 1.33 radians
  3. 1.67 radians
  4. 2.67 radians
Explanation: The perimeter of a sector is composed of two radii and the arc length: P=2r+sP = 2r + s. We are given P=40P=40 m and r=15r=15 m. The arc length is s=rθs = r\theta. Substituting this into the perimeter formula gives P=2r+rθP = 2r + r\theta. So, 40=2(15)+15θ40 = 2(15) + 15\theta. 40=30+15θ40 = 30 + 15\theta. 10=15θ10 = 15\theta. θ=1015=23≈0.67\theta = \frac{10}{15} = \frac{2}{3} \approx 0.67 radians.

Question 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 1.5π1.5\pi radians, what is the angle of rotation of the larger wheel?

  1. 0.54π radians
  2. 0.90π radians (correct answer)
  3. 1.50π radians
  4. 2.50π radians
Explanation: When two cogwheels mesh, the arc length travelled by the teeth on their circumferences must be equal. Let the radii be r1=6r_1=6 and r2=10r_2=10, and the angles be θ1=1.5π\theta_1=1.5\pi and θ2\theta_2. The condition is s1=s2s_1=s_2, which means r1θ1=r2θ2r_1\theta_1 = r_2\theta_2. Plugging in the values: 6×1.5π=10×θ26 \times 1.5\pi = 10 \times \theta_2. 9π=10θ29\pi = 10\theta_2. θ2=9π10=0.9π\theta_2 = \frac{9\pi}{10} = 0.9\pi radians.

Question 5

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?

  1. 56 cm² (correct answer)
  2. 88 cm²
  3. 112 cm²
  4. 120 cm²
Explanation: The perimeter of a sector is given by P=2r+sP = 2r + s, where ss is the arc length. We are given P=30P=30 cm and r=8r=8 cm. So, 30=2(8)+s  ⟹  30=16+s  ⟹  s=1430 = 2(8) + s \implies 30 = 16 + s \implies s = 14 cm. The area of a sector can be calculated using the formula A=12rsA = \frac{1}{2}rs. Substituting the known values: A=12(8)(14)=56A = \frac{1}{2}(8)(14) = 56 cm².

Question 6

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?

  1. 2.39 cm (correct answer)
  2. 3.00 cm
  3. 4.77 cm
  4. 6.00 cm
Explanation: 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=2πrnewC = 2\pi r_{new}. So, 15=2πrnew15 = 2\pi r_{new}. Solving for the new radius: rnew=152π≈2.387r_{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.

Question 7

A restaurant serves a slice of pie that is a sector of a circle with a central angle of π6\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?

  1. 22.9 cm²
  2. 72.0 cm²
  3. 137.5 cm² (correct answer)
  4. 275.0 cm²
Explanation: First, find the radius rr of the pie using the arc length formula s=rθs = r\theta. We have 12=r×π612 = r \times \frac{\pi}{6}, which gives r=72πr = \frac{72}{\pi} cm. Now, calculate the area of the sector using the formula A=12r2θA = \frac{1}{2}r^2\theta or the simpler formula A=12rsA = \frac{1}{2}rs. Using the latter, A=12×72π×12=432π≈137.51A = \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.

Question 8

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?

  1. 125 km
  2. 1593 km
  3. 1718 km (correct answer)
  4. 3435 km
Explanation: First, calculate the total radius of the satellite's orbit by adding the Earth's radius and the satellite's altitude: rorbit=6370 km+500 km=6870 kmr_{orbit} = 6370\text{ km} + 500\text{ km} = 6870\text{ km}. The distance the satellite travels is the arc length, given by s=rθs = r\theta. Using θ=0.25\theta = 0.25 radians: s=6870×0.25=1717.5s = 6870 \times 0.25 = 1717.5 km. To the nearest kilometer, the distance is 1718 km.

Question 9

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?

  1. 56 cm² (correct answer)
  2. 88 cm²
  3. 112 cm²
  4. 120 cm²
Explanation: The perimeter of a sector is given by P=2r+sP = 2r + s, where ss is the arc length. We are given P=30P=30 cm and r=8r=8 cm. So, 30=2(8)+s  ⟹  30=16+s  ⟹  s=1430 = 2(8) + s \implies 30 = 16 + s \implies s = 14 cm. The area of a sector can be calculated using the formula A=12rsA = \frac{1}{2}rs. Substituting the known values: A=12(8)(14)=56A = \frac{1}{2}(8)(14) = 56 cm².

Question 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?

  1. 19.9°
  2. 39.8° (correct answer)
  3. 119.2°
  4. 238.7°
Explanation: The area of a sector is given by A=12r2θA = \frac{1}{2}r^2\theta, where θ\theta is in radians. We have A=50A=50 m² and r=12r=12 m. 50=12(122)θ  ⟹  50=72θ  ⟹  θ=507250 = \frac{1}{2}(12^2)\theta \implies 50 = 72\theta \implies \theta = \frac{50}{72} radians. To convert this to degrees, we multiply by 180π\frac{180}{\pi}: Angle in degrees = 5072×180π≈39.7887...\frac{50}{72} \times \frac{180}{\pi} \approx 39.7887...°. To one decimal place, the angle is 39.8°.

Question 11

A goat is tethered to the corner of a rectangular shed measuring 6 m by 4 m. The rope is 8 m long.

What is the total area, in m², that the goat can graze?

  1. 48π48\pi m²
  2. 50π50\pi m²
  3. 53π53\pi m² (correct answer)
  4. 64π64\pi m²
Explanation: The grazing area consists of three sectors. First, a large sector with radius 8 m covering three-quarters of a circle: A1=34×π(82)=34×64π=48πA_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=28 - 6 = 2 m. This allows grazing in a quarter-circle sector of radius 2 m: A2=14×π(22)=πA_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=48 - 4 = 4 m. This allows grazing in a quarter-circle sector of radius 4 m: A3=14×π(42)=4πA_3 = \frac{1}{4} \times \pi (4^2) = 4\pi m². The total area is the sum: Atotal=A1+A2+A3=48π+π+4π=53πA_{total} = A_1 + A_2 + A_3 = 48\pi + \pi + 4\pi = 53\pi m².

Question 12

A large pizza with a diameter of 40 cm is cut into 12 equal slices. What is the perimeter of one slice?

  1. 30.5 cm
  2. 40.0 cm
  3. 50.5 cm (correct answer)
  4. 62.8 cm
Explanation: The diameter is 40 cm, so the radius is r=20r=20 cm. A full circle is 2π2\pi radians. Since there are 12 equal slices, the angle of one slice is θ=2π12=π6\theta = \frac{2\pi}{12} = \frac{\pi}{6} radians. The arc length of one slice is s=rθ=20×π6=10π3≈10.47s = 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)+10π3=40+10.47≈50.47P = 2r + s = 2(20) + \frac{10\pi}{3} = 40 + 10.47 \approx 50.47 cm. The closest answer is 50.5 cm.

Question 13

A restaurant serves a slice of pie that is a sector of a circle with a central angle of π6\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?

  1. 22.9 cm²
  2. 72.0 cm²
  3. 137.5 cm² (correct answer)
  4. 275.0 cm²
Explanation: First, find the radius rr of the pie using the arc length formula s=rθs = r\theta. We have 12=r×π612 = r \times \frac{\pi}{6}, which gives r=72πr = \frac{72}{\pi} cm. Now, calculate the area of the sector using the formula A=12r2θA = \frac{1}{2}r^2\theta or the simpler formula A=12rsA = \frac{1}{2}rs. Using the latter, A=12×72π×12=432π≈137.51A = \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.

Question 14

A car wheel has a diameter of 60 cm. The car travels 100 metres. How many full revolutions has the wheel made?

  1. 26
  2. 53 (correct answer)
  3. 55
  4. 106
Explanation: First, ensure units are consistent. The wheel diameter is 60 cm = 0.6 m, so its radius is r=0.3r=0.3 m. The distance travelled is d=100d=100 m. The circumference of the wheel, which is the distance covered in one revolution, is C=2πr=2π(0.3)=0.6πC = 2\pi r = 2\pi(0.3) = 0.6\pi m. The number of revolutions is the total distance divided by the circumference: N=1000.6π≈53.05N = \frac{100}{0.6\pi} \approx 53.05. The question asks for the number of full revolutions, which is 53.

Question 15

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?

  1. 19.9°
  2. 39.8° (correct answer)
  3. 119.2°
  4. 238.7°
Explanation: The area of a sector is given by A=12r2θA = \frac{1}{2}r^2\theta, where θ\theta is in radians. We have A=50A=50 m² and r=12r=12 m. 50=12(122)θ  ⟹  50=72θ  ⟹  θ=507250 = \frac{1}{2}(12^2)\theta \implies 50 = 72\theta \implies \theta = \frac{50}{72} radians. To convert this to degrees, we multiply by 180π\frac{180}{\pi}: Angle in degrees = 5072×180π≈39.7887...\frac{50}{72} \times \frac{180}{\pi} \approx 39.7887...°. To one decimal place, the angle is 39.8°.

Question 16

A goat is tethered to the corner of a rectangular shed measuring 6 m by 4 m. The rope is 8 m long.

What is the total area, in m², that the goat can graze?

  1. 48π48\pi m²
  2. 50π50\pi m²
  3. 53π53\pi m² (correct answer)
  4. 64π64\pi m²
Explanation: The grazing area consists of three sectors. First, a large sector with radius 8 m covering three-quarters of a circle: A1=34×π(82)=34×64π=48πA_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=28 - 6 = 2 m. This allows grazing in a quarter-circle sector of radius 2 m: A2=14×π(22)=πA_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=48 - 4 = 4 m. This allows grazing in a quarter-circle sector of radius 4 m: A3=14×π(42)=4πA_3 = \frac{1}{4} \times \pi (4^2) = 4\pi m². The total area is the sum: Atotal=A1+A2+A3=48π+π+4π=53πA_{total} = A_1 + A_2 + A_3 = 48\pi + \pi + 4\pi = 53\pi m².

Question 17

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?

  1. 125 km
  2. 1593 km
  3. 1718 km (correct answer)
  4. 3435 km
Explanation: First, calculate the total radius of the satellite's orbit by adding the Earth's radius and the satellite's altitude: rorbit=6370 km+500 km=6870 kmr_{orbit} = 6370\text{ km} + 500\text{ km} = 6870\text{ km}. The distance the satellite travels is the arc length, given by s=rθs = r\theta. Using θ=0.25\theta = 0.25 radians: s=6870×0.25=1717.5s = 6870 \times 0.25 = 1717.5 km. To the nearest kilometer, the distance is 1718 km.

Question 18

A car wheel has a diameter of 60 cm. The car travels 100 metres. How many full revolutions has the wheel made?

  1. 26
  2. 53 (correct answer)
  3. 55
  4. 106
Explanation: First, ensure units are consistent. The wheel diameter is 60 cm = 0.6 m, so its radius is r=0.3r=0.3 m. The distance travelled is d=100d=100 m. The circumference of the wheel, which is the distance covered in one revolution, is C=2πr=2π(0.3)=0.6πC = 2\pi r = 2\pi(0.3) = 0.6\pi m. The number of revolutions is the total distance divided by the circumference: N=1000.6π≈53.05N = \frac{100}{0.6\pi} \approx 53.05. The question asks for the number of full revolutions, which is 53.

Question 19

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?

  1. 15.9 m²
  2. 53.0 m²
  3. 31.8 m²
  4. 26.5 m² (correct answer)
Explanation: 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 2560=512\frac{25}{60} = \frac{5}{12} of a full circle. The area of a sector is A=θ360°×πr2A = \frac{\theta}{360°} \times \pi r^2 where θ is the angle in degrees. Here, the angle is 512×360°=150°\frac{5}{12} \times 360° = 150°. With radius r = 4.5 m: A=150°360°×π×(4.5)2=512×π×20.25=5×20.25π12=101.25π12≈26.5 m2A = \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.

Question 20

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?

  1. 30 km²
  2. 780 km²
  3. 750 km²
  4. 720 km² (correct answer)
Explanation: 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 = 12r2θ\frac{1}{2}r^2\theta, where rr is the radius and θ\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 = 12(25)2(2.4)=12(625)(2.4)=750\frac{1}{2}(25)^2(2.4) = \frac{1}{2}(625)(2.4) = 750 km² For the inner sector (radius = 5 km): Area = 12(5)2(2.4)=12(25)(2.4)=30\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.