IB Maths AI HL Geometry Toolkit Paper 1 & 2 Arc length, sector area ~7 min read

Arcs & Sectors Using Degrees

An arc is part of a circle’s circumference (the crust of a pizza slice); a sector is the slice itself, bounded by two radii and an arc. Both quantities are just fractions of the full circle — the fraction is θ/360, where θ is the central angle in degrees.

📘 What you need to know

Arc length: l = (θ/360) × 2πr — fraction of the circumference.
Sector area: A = (θ/360) × πr² — fraction of the disc area.
Minor vs major: angle < 180° gives a minor arc/sector; angle > 180° gives the major arc/sector on the same circle.
Major angle = 360° − minor angle; major arc + minor arc = circumference; major area + minor area = πr².
Perimeter of a sector = arc length + 2r (the two radii bound it).
Both formulas are in the formula booklet — you don’t have to memorise them, but you must recognise when to use each one.

Fractions of a whole circle

The whole disc has circumference 2πr and area πr². A sector with central angle θ degrees occupies a fraction θ/360 of the disc — that’s the only idea behind both formulas. To find the arc length, multiply 2πr by that fraction; to find the sector area, multiply πr² by it. The two formulas share the same θ/360, so once you’ve computed the fraction you can reuse it for both. The angle θ must be in degrees for these formulas — we’ll see the radian versions later.

The shaded teal slice is a minor sector with central angle θ at O; the bold teal arc is the boundary furthest from the centre. The rest of the disc forms the major sector.

Two formulas, one fraction l = θ360 × 2πr · A = θ360 × πr² perimeter of sector = arc length + 2r; major angle = 360° − minor angle

Forward and reverse problems

The forward problem — given r and θ, find l or A — is direct substitution. The reverse problem — given an arc length and angle, find the radius; or given an area and radius, find the angle — needs you to rearrange the formula first. The most common combined problem asks for the perimeter of a sector, which is the arc length plus the two radii: P = l + 2r. For the major sector of a partial circle, use 360° − θ in place of θ, or just subtract the minor sector’s value from the whole.

Always keep θ/360 as one chunk — simplify the fraction (e.g., 60/360 = 1/6) before multiplying by 2πr or πr². Often that gives cleaner exact-form answers in terms of π.

🧭 Recipe — arcs & sectors (degrees)

Identify what’s asked: arc length, sector area, or a perimeter that combines both.
Compute the fraction θ/360 and simplify.
Pick the right formula: 2πr for arc, πr² for area.
For perimeters, add 2r to the arc length.
For reverse problems, write the formula, substitute the known values, and solve for the unknown.

Worked examples

WE 1

Basic arc length

A sector of a circle has radius r = 9 cm and central angle θ = 60°. Find the length of the arc, giving your answer to 3 s.f.

apply the formula l = (60/360) × 2π(9) simplify the fraction = (1/6) × 18π = 3π l = 3π ≈ 9.42 cm (3 s.f.) exact form 3π is cleaner — keep it unless asked for a decimal.

WE 2

Basic sector area

A sector of a circle has radius r = 10 cm and central angle θ = 72°. Find the area, giving your answer to 3 s.f.

apply the formula A = (72/360) × π(10)² simplify the fraction = (1/5) × 100π = 20π A = 20π ≈ 62.8 cm² (3 s.f.)

WE 3

Reverse: find the radius

A sector has central angle 45° and arc length 8 cm. Find the radius, giving your answer to 3 s.f.

set up the arc-length equation 8 = (45/360) × 2πr simplify the fraction 8 = (1/8) × 2πr = πr/4 solve for r r = 32/π r ≈ 10.2 cm (3 s.f.)

WE 4

Reverse: find the angle

A sector has radius 6 cm and area 30 cm². Find the central angle in degrees, to 3 s.f.

set up the area equation 30 = (θ/360) × π(6)² 30 = (θ/360) × 36π = θπ/10 solve for θ θ = 300/π θ ≈ 95.5° (3 s.f.) since θ < 180°, this is a minor sector.

WE 5

Applied: sprinkler watering a sector

A garden sprinkler waters a sector-shaped region of radius 8 m with central angle 120°. Find: (a) the arc length of the watered region; (b) its area; (c) its total perimeter. Give answers to 3 s.f.

(a) arc length l = (120/360) × 2π(8) = (1/3)(16π) l = 16π/3 ≈ 16.8 m (b) sector area A = (120/360) × π(64) = (1/3)(64π) A = 64π/3 ≈ 67.0 m² (c) perimeter = arc + 2 radii P = 16π/3 + 2(8) = 16π/3 + 16 P ≈ 32.8 m don’t forget the two radii in the perimeter.

WE 6

Major sector of a cake

A circular cake of radius 15 cm has a slice cut out with central angle 50°. Find: (a) the arc length of the slice (minor arc); (b) the area of the slice; (c) the perimeter of the remaining cake (the major sector). Give answers to 3 s.f.

(a) minor arc l = (50/360) × 2π(15) = (5/36)(30π) l = 25π/6 ≈ 13.1 cm (b) slice area A = (50/360) × π(225) = (5/36)(225π) A = 125π/4 ≈ 98.2 cm² (c) major angle = 360 − 50 = 310° major arc = (310/360) × 2π(15) = 155π/6 P = major arc + 2(15) = 155π/6 + 30 P ≈ 111 cm sanity check: minor arc + major arc = 25π/6 + 155π/6 = 30π = circumference ✓

💡 Top tips

Simplify θ/360 first: 60/360 = 1/6, 72/360 = 1/5, 120/360 = 1/3. Cleaner arithmetic and exact-form answers.
Both formulas share θ/360 — compute once, reuse for both.
Major sector trick: 360° − θ gives the major angle; or just subtract the minor value from the full disc.
Perimeter ≠ arc length: a sector’s perimeter includes the two radii. Read questions carefully.
Keep exact answers in terms of π unless told to give a decimal — many IB questions accept (and prefer) the exact form.

⚠ Common mistakes

Using radians by accident — these formulas need degrees. If the question gives radians, use the radian versions instead.
Confusing arc length with sector area — 2πr is circumference (arc), πr² is area.
Forgetting the two radii when asked for a sector’s perimeter.
Using minor angle for major sector calculations — switch to 360° − θ.
Premature rounding — keep things in exact form (with π) until the final answer.

Next up: Radian Measure — a more natural angle unit where π rad = 180°, leading to the cleaner formulas l = rθ and A = ½r²θ.

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