ACT Math Arcs and Sectors: 10 Practice Problems with Step-by-Step Explanations
Arc and sector problems on the ACT connect two ideas: the full circle always represents 360°, and any portion of it — whether a slice of a circle graph, an arc along the edge, or a wedge-shaped sector — is just that same fraction of the whole circle's angle, arc length, or area. Circle graphs (pie charts) are the most common application: to find a sector's central angle, find what fraction of the total the category represents, then multiply that fraction by 360°.
For arc length and sector area, the same fraction idea applies to the circle's other properties: arc length = (central angle/360°) × circumference, and sector area = (central angle/360°) × πr². Getting the central angle right is almost always the hard part — the arithmetic once you have it is simple multiplication. On problems that give you several angles labeled with a variable (like 4x, 5x, and 3x) that together account for the whole circle, setting their sum equal to 360° lets you solve for that variable directly.
Common traps include forgetting that a "central angle" is measured from the circle's center, not from a point on the edge, mixing up degrees and radians when a problem specifically asks for one or the other, and on real-world sector problems (like a patio or a slice of concrete), missing whether the shape described is the small wedge itself or the rest of the circle with a wedge removed. Work through the 10 problems below, including three that reference a figure, then click to reveal each step-by-step explanation.
On this page:
- Problem 1 — Central angle from a fraction of a total
- Problem 2 — Central angle of equal slices
- Problem 3 — Central angle from category counts
- Problem 4 — Solving for a variable angle, then arc length
- Problem 5 — Central angle from a percentage
- Problem 6 — Sector area
- Problem 7 — Central angle from a survey
- Problem 8 — Central angle from an expenditure
- Problem 9 — Arc length to central angle in radians
- Problem 10 — Sector area from inscribed angles
Practice Problems
Youngstown high school is creating a circle graph to represent the amount of money different clubs raised for a school fundraiser. The clubs raised a total of $15,000. The speech club raised $1,500. What will be the central angle of the sector that displays the money the speech club raised?
- A) 25°
- B) 36°
- C) 45°
- D) 60°
Find the speech club's fraction of the total: 1,500/15,000 = 0.10 (10%).
Multiply by 360°: 0.10 × 360° = 36°.
Christine's famous chocolate cake contains 2,700 calories. If the cake is sliced evenly into 9 slices, what is the central angle of each of the slices?
- A) 40°
- B) 45°
- C) 50°
- D) 55°
Since the cake is sliced evenly, each slice takes an equal share of the full 360° circle, regardless of the calorie count.
360°/9 = 40°.
Dave is making a pie chart to represent the different fruits he currently has in the house. If he has 8 bananas, 6 apples, 4 oranges, and 6 plums, what is the central angle of the portion of the pie chart that represents the apples?
- A) 45°
- B) 60°
- C) 90°
- D) 100°
Total fruit: 8+6+4+6 = 24. Apples' fraction: 6/24 = 0.25.
Multiply by 360°: 0.25 × 360° = 90°.
In the figure below, AC is a diameter of the circle and has a length of 12. What is the length of minor arc AB?
- A) 3π
- B) 4π
- C) 5π
- D) 6π
Since the three labeled central angles (4x, 5x, and 3x) together account for the entire circle: 4x+5x+3x=360°, so 12x=360° and x=30°.
Arc AB corresponds to the 4x angle: 4(30°)=120°. Circumference = πd = π(12) = 12π. Arc length = (120/360) × 12π = 4π.
Adriana's bakery is using a pie chart to display their monthly expenses. Rent is 30% of the monthly expenses. One sector of the chart will be used to display rent. What will be the central angle for that sector?
- A) 30°
- B) 60°
- C) 90°
- D) 108°
Multiply the percentage by 360°: 0.30 × 360° = 108°.
Trey's construction company is cutting a circle of concrete into 7 slices. If the original circle of concrete has a diameter of 24 meters, what is the approximate area, in square meters, of one slice of concrete?
- A) 20
- B) 51
- C) 65
- D) 82
Radius: 24/2=12 meters. Total area: πr² = π(12)² = 144π ≈ 452.39 square meters.
Divide by 7 equal slices: 452.39/7 ≈ 64.6, which rounds to about 65.
Jin asked 80 of his classmates to pick their favorite ice cream flavor from 5 options. Jin will represent the results in a circle graph. 20 students picked rocky road, and 24 selected chocolate chip cookie dough. Strawberry, mint chip, and chocolate were chosen as favorites by an equal number of the remaining students. What is the measure of the central angle for mint chip in the circle graph?
- A) 42°
- B) 47.5°
- C) 54°
- D) 58.5°
Remaining students: 80−20−24=36, split equally among 3 flavors: 36/3=12 students chose mint chip.
Fraction: 12/80=0.15. Central angle: 0.15 × 360° = 54°.
Last month, Larissa had total expenditures of $1,200. She spent $750 on a new phone, $275 on her car, $125 on groceries, and $50 on clothing. She will create a circle graph to represent her expenditures from last month. Each sector of the circle graph will represent the percent of her income spent on that category. What will be the measure of the central angle for the car sector?
- A) 76°
- B) 82.5°
- C) 99.25°
- D) 112°
Car expense fraction: 275/1,200 ≈ 0.22917.
Multiply by 360°: 0.22917 × 360° ≈ 82.5°.
An object is traveling around a circle with center O and radius of 6 meters, as shown below. The object travels at a rate of 1/8 m per second. If the object starts at point B and travels for 18 seconds to reach point A, what is the value of θ in radians?
(Note: Figure is not drawn to scale.)
- A) 9/32
- B) 3/8
- C) 9/16
- D) 9/8
Distance traveled: rate × time = (1/8)(18) = 18/8 = 9/4 meters.
Using arc length s=rθ, solve for θ: θ = s/r = (9/4)/6 = 9/24 = 3/8 radians.
Point O is the center of the circle with the radius of 3/2. If the area of sector BOC is written as kπ, what is the value of k?
- A) 0.238
- B) 0.317
- C) 0.475
- D) 0.633
Since OA=OC (both radii), triangle OAC is isosceles, so the 19° angle at C equals the angle at A in that triangle, making angle AOC = 180°−19°−19°=142°. The same reasoning applies to triangle OAB, giving angle AOB=142°.
Angle BOC (the remaining central angle) = 360°−142°−142°=76°. Sector area = (76/360) × π(3/2)² = (76/360)(9/4)π ≈ 0.475π, so k≈0.475.
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