IB Maths AI HL Geometry Toolkit Paper 1 & 2 l = rθ, A = ½r²θ ~7 min read

Arcs & Sectors Using Radians

When the central angle is in radians, the formulas become beautifully simple: arc length is just l = rθ, and sector area is A = ½r²θ. No fractions of 360, no 2π floating around — this is the main payoff of switching from degrees to radians.

📘 What you need to know

Arc length (radians): l = rθ — just radius times angle.
Sector area (radians): A = ½r²θ — half of radius-squared times angle.
θ must be in radians for both formulas. If you’re given degrees, convert first using θ_rad = θ_deg × (π/180).
Both formulas in the formula booklet — you don’t have to memorise them, but you must recognise when to use each.
Perimeter of a sector: P = l + 2r = rθ + 2r = r(θ + 2).
Major sector: replace θ with 2π − θ — or work out the minor sector and subtract from the whole disc.

Why the radian formulas are simpler

In degrees, you have to multiply by the fraction θ/360, which buries the geometry inside arithmetic. In radians, θ already is the fraction of a full turn — just with a different scale (2π instead of 360). On a unit circle (r = 1), the arc length equals the angle in radians by definition. Scaling the circle by r scales the arc by r as well, giving l = rθ. For the sector area, the full disc has area πr², and the sector takes the fraction θ/(2π) of it, giving (θ/(2π)) × πr² = ½r²θ. Same idea as the degree version, just cleaner constants.

A sector with central angle θ (in radians) and radius r. The bold teal arc is the boundary furthest from O; the shaded region is the sector itself.

Radian formulas arc length: l = rθ · sector area: A = 12r²θ perimeter of sector: P = rθ + 2r = r(θ + 2)

Forward and reverse problems

The forward problem is direct substitution: given r and θ, compute l or A. The reverse problem rearranges one of the formulas: given l and θ, find r = l/θ; given A and r, find θ = 2A/r². If you’re ever given an angle in degrees in a “radian formulas” question, convert it to radians first — the formulas l = rθ and A = ½r²θ require θ in radians, otherwise the answer will be off by a factor of about 57 (the ratio 180/π).

Sanity check: if your sector spans a small fraction of the disc (say, π/6 rad = 1/12 of a full turn), then its arc length should be about 1/12 of the circumference, and its area about 1/12 of the disc area. Quick estimates like this catch unit errors fast.

🧭 Recipe — arcs & sectors (radians)

Check units: the angle must be in radians. Convert if necessary.
Pick the formula: l = rθ for arc, A = ½r²θ for area.
Substitute and simplify — many answers end up clean with π factors.
For perimeter, add the two radii: P = rθ + 2r.
For reverse problems, rearrange the formula for the unknown.

Worked examples

WE 1

Basic arc length (radians)

A sector has radius r = 10 cm and central angle θ = 3π5 rad. Find the arc length in exact form and to 3 s.f.

apply l = rθ l = 10 × (3π/5) = 30π/5 l = 6π cm (exact) decimal 6π ≈ 18.85 l ≈ 18.8 cm (3 s.f.)

WE 2

Basic sector area (radians)

A sector has radius r = 4 cm and central angle θ = 5π6 rad. Find the area in exact form and to 3 s.f.

apply A = ½r²θ A = ½(4)² × (5π/6) = 8 × 5π/6 simplify = 40π/6 = 20π/3 A = 20π/3 cm² (exact) decimal 20π/3 ≈ 20.94 A ≈ 20.9 cm² (3 s.f.)

WE 3

Reverse: find the radius

A sector has central angle θ = 2π9 rad and arc length 6 cm. Find the radius to 3 s.f.

rearrange l = rθ r = l/θ = 6 / (2π/9) flip the fraction r = 6 × 9/(2π) = 54/(2π) = 27/π r = 27/π ≈ 8.59 cm (3 s.f.)

WE 4

Reverse: find the angle

A sector has radius 5 cm and area 12 cm². Find the central angle in radians.

rearrange A = ½r²θ θ = 2A/r² = 2(12)/5² = 24/25 θ = 0.96 rad clean exact answer — no π involved.

WE 5

Applied: windscreen wiper

A windscreen wiper of length 50 cm sweeps through an angle of 2π3 rad. Find: (a) the arc length swept by the wiper’s tip; (b) the area cleaned; (c) the perimeter of the swept region. Give answers to 3 s.f.

(a) arc length l = rθ = 50 × (2π/3) = 100π/3 l ≈ 105 cm (b) sector area A = ½(50)²(2π/3) = 1250(2π/3) = 2500π/3 A ≈ 2620 cm² (c) perimeter = arc + 2r P = 100π/3 + 100 P ≈ 205 cm factored form: P = r(θ + 2) = 50(2π/3 + 2) ≈ 205 cm.

WE 6

Major sector of a plaza

A circular plaza of radius 20 m is divided by a path into two sectors. The smaller (seating) sector has central angle 3π4 rad. For the larger (walking) sector, find: (a) the central angle; (b) the area; (c) the perimeter. Give answers in exact form, then to 3 s.f. where appropriate.

(a) larger angle = 2π − smaller = 2π − 3π/4 = 8π/4 − 3π/4 5π/4 rad (b) area = ½r²θ = ½(20)²(5π/4) = 200 × 5π/4 = 250π A = 250π ≈ 785 m² (c) arc = rθ = 20(5π/4) = 25π P = 25π + 2(20) = 25π + 40 P ≈ 119 m sanity: smaller area 150π + larger area 250π = 400π = πr² ✓

💡 Top tips

Always check units first: the radian formulas need radians. If degrees are given, multiply by π/180 before substituting.
Keep π symbolic wherever possible — you’ll usually get the cleanest exact-form answer that way.
Factor when finding perimeter: P = r(θ + 2) is often quicker than computing arc + 2r separately.
Major sector trick: use 2π − θ in place of θ, or just subtract the minor sector’s value from the whole disc.
Pythagorean and “nice radian” recognition: when you see common angles π/6, π/4, π/3, π/2, π, expect clean simplifications.

⚠ Common mistakes

Plugging degrees into l = rθ — this gives an answer about 57 times too large; always convert first.
Confusing the two formulas — l = rθ for arc, A = ½r²θ for area. The square is on the area side.
Forgetting the ½ in the area formula — doubles your answer.
Calculator in degree mode when computing things like sin(π/6) inside a problem — check the mode setting.
Missing the two radii in the perimeter — sectors are bounded by an arc AND two radii.

Chapter complete — you now have all five Geometry Toolkit sub-topics: Coordinate Geometry, Perpendicular Bisectors, Arcs & Sectors (Degrees), Radian Measure, and Arcs & Sectors (Radians). Together they cover the foundational geometry you’ll lean on throughout AI HL.

Need help with Arcs & Sectors using Radians?

Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.

Book Free Session →