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c059741
NotesMath AI HLTopic 3.4
Unit 3 · Geometry & Trigonometry · Topic 3.4

IB Math AI HL — Circle arc length and area of sector (degrees)

IB Mathematics AI SL topic covering core concepts and exam-style applications.

Higher Level students should use this topic hub as a map: start with the shared sub-topics, then follow the HL-only extensions and exam-skill links where this topic asks for deeper analysis.

Exam technique guidePractice questions

Key concepts in Circle arc length and area of sector (degrees)

Key Idea: A sector is a 'pie slice' of a circle — defined by two radii and an arc. Topic 3.4 is about calculating the length of the arc and the area of the sector using the angle θ (in degrees). These formulas also combine to give the perimeter of a sector and the area of a circular segment.

✅ Core formulas

QuantityFormulaNote
Arc lengthl = (θ/360) × 2πrThe fraction of the full circumference
Sector areaA = (θ/360) × πr²The fraction of the full circle area
Sector perimeterP = l + 2rArc length plus both radii (the straight edges)
Segment areaA = sector area − triangle areaTriangle area = ½r² sinθ (the chord cuts off the segment)

Example: Arc length: radius = 8 cm, angle = 135°. l = (135/360) × 2π(8) = (3/8) × 16π = 6π ≈ 18.8 cm Sector area: radius = 5 cm, angle = 72°. A = (72/360) × π(5²) = (1/5) × 25π = 5π ≈ 15.7 cm² Segment area: radius = 10 cm, angle = 60°. Sector area = (60/360) × π(100) = 100π/6 ≈ 52.36 cm² Triangle area = ½(10²)sin60° = 50 × (√3/2) ≈ 43.30 cm² Segment area ≈ 52.36 − 43.30 = 9.06 cm²
The angle θ must be in degrees when using these formulas. If you are given radians, convert first: θ (degrees) = θ (radians) × 180/π. A segment is NOT the same as a sector. Sector = pie slice (includes both radii). Segment = region between an arc and its chord (no straight radii visible).
Paper 1 (GDC allowed): Leave answers in terms of π when possible (e.g., 6π cm). This is exact and avoids rounding errors. Paper 2 (GDC allowed): Evaluate to 3 s.f. For segment area questions, calculate sector and triangle areas in sequence and show both — the examiner wants to see both values.

IB-style question [6 marks]

A landscaped flower bed is in the shape of a sector of a circle. It has two straight edges (radii) of length 7 m and a central angle of 120°. (a) Find the length of the curved edge of the flower bed. (b) Find the perimeter of the flower bed. (c) Find the area of the flower bed.

Step by step:

  1. (a) Write the arc-length formula and substitute r = 7 m, θ = 120°.

    l=θ360×2πr=120360×2π(7)=13×14π=14π3≈14.7 ml = \frac{\theta}{360} \times 2\pi r = \frac{120}{360} \times 2\pi(7) = \frac{1}{3} \times 14\pi = \frac{14\pi}{3} \approx 14.7 \text{ m}l=360θ​×2πr=360120​×2π(7)=31​×14π=314π​≈14.7 m
  2. (b) The perimeter is the arc plus the two straight radii.

    P=l+2r=14.66+2(7)=14.66+14=28.66≈28.7 mP = l + 2r = 14.66 + 2(7) = 14.66 + 14 = 28.66 \approx 28.7 \text{ m}P=l+2r=14.66+2(7)=14.66+14=28.66≈28.7 m
  3. (c) Write the sector-area formula. Area uses r², not r.

    A=θ360×πr2A = \frac{\theta}{360} \times \pi r^2A=360θ​×πr2
  4. Substitute r = 7 m and θ = 120°.

    A=120360×π(7)2=13×49π=49π3≈51.3 m2A = \frac{120}{360} \times \pi(7)^2 = \frac{1}{3} \times 49\pi = \frac{49\pi}{3} \approx 51.3 \text{ m}^2A=360120​×π(7)2=31​×49π=349π​≈51.3 m2
Final answer:

(a) Arc = 14π/3 ≈ 14.7 m. (b) Perimeter ≈ 28.7 m. (c) Area = 49π/3 ≈ 51.3 m².

What you'll learn in Topic 3.4

  • 3.4.1 Arc Length
  • 3.4.2 Area of Sector
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 3.4 Circle arc length and area of sector (degrees)

3.4.1

Arc Length

Notes
3.4.2

Area of Sector

Notes

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Topic 3.4 Circle arc length and area of sector (degrees) forms a core part of Unit 3: Geometry & Trigonometry in IB Math AI HL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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