IB Maths AA HL Topic 3 — Geometry & Trigonometry Paper 1 & 2 ~6 min read

Arcs & Sectors Using Degrees

An arc is part of the circumference of a circle, and a sector is the pizza-slice region bounded by two radii and an arc. Both are simple fractions of the whole circle: divide the angle at the centre by 360° and multiply by either the circumference (for an arc) or the area (for a sector). Two formulas, one shared idea — once you’ve seen one, you’ve seen them both.

📘 What you need to know

Arc: a portion of the circumference of a circle. Sector: a slice of the disc bounded by two radii and an arc.
Minor vs major: angle at the centre < 180° → minor arc/sector; > 180° → major arc/sector.
Arc length formula: l = (θ/360) × 2πr, where θ is in degrees.
Sector area formula: A = (θ/360) × πr², where θ is in degrees.
Both formulas are fractions of the whole: θ/360 represents the fraction of the full circle the sector occupies.
Neither degree formula is in the formula booklet — only the radian versions are. Memorise these or be ready to derive.
Perimeter of a sector = arc length + 2r (don’t forget the two radii).
Reverse problems: any of θ, r, l, or A can be the unknown — rearrange and solve.

Length of an arc

The arc length is just a fraction of the circumference. The fraction is θ/360 — the angle at the centre as a proportion of a full turn.

Arc length (degrees) l = θ360 × 2πr

If θ = 360°, the fraction is 1 and you recover the full circumference 2πr. If θ = 90°, the fraction is 1/4 and you get a quarter-circle. The formula is just bookkeeping for “what fraction of the circle is this?”

Watch the wording: “arc length” is just the curved bit; “perimeter of the sector” includes the two straight radii too. Always reread the question to see which one is asked for.

Area of a sector

Same idea, applied to area: the sector is a fraction θ/360 of the whole disc.

Sector area (degrees) A = θ360 × πr²

If θ = 360°, you get the full disc area πr². If θ = 180°, you get a semicircle of area πr²/2. Same fraction-of-the-whole reasoning, applied to area instead of circumference.

Two formulas, one idea

Arc length

l = (θ/360) × 2πr

fraction of the circumference

Sector area

A = (θ/360) × πr²

fraction of the disc area

If you ever forget which one has the squared r, think about units: length is one-dimensional (so r appears once), area is two-dimensional (so r is squared). Same trick that distinguishes circumference from area in general.

🧭 Recipe — solving arc & sector problems

Sketch the situation. A quick diagram of the circle with the angle and radius labelled prevents most errors.
Identify what’s asked: arc length, sector area, perimeter of sector (= arc + 2r), or something composite.
Decide minor or major. If the question asks about the major arc/sector, use θ = 360° − (minor angle).
Pick the right formula and substitute carefully — keep θ, r, and the constant π in the right places.
Solve and round. Use 3 s.f. unless told otherwise; keep π in answers if the question asks for an exact value.
For reverse problems: write the formula, substitute the known quantities, and rearrange to isolate the unknown.

Worked examples

WE 1

Find an arc length

A sector of a circle has a central angle of 72° and radius 15 cm. Find the length of the arc.

Step 1: Substitute into l = (θ/360) × 2πr l = (72/360) × 2π(15) Step 2: Simplify the fraction 72/360 = 1/5 l = (1/5) × 30π = 6π Step 3: Convert to decimal if needed 6π ≈ 18.8496… l = 6π cm ≈ 18.8 cm (3 s.f.) leaving 6π is exact — only switch to decimal if the question demands it

WE 2

Find a sector area

A sector has a central angle of 120° and radius 9 cm. Find the area of the sector.

Step 1: Substitute into A = (θ/360) × πr² A = (120/360) × π(9)² Step 2: Simplify 120/360 = 1/3; 9² = 81 A = (1/3) × 81π = 27π Step 3: Decimal value 27π ≈ 84.823… A = 27π cm² ≈ 84.8 cm² (3 s.f.) 120° is one-third of a full turn — so the sector is one-third of the disc

WE 3

Perimeter of a sector

Find the perimeter of a sector with central angle 50° and radius 8 cm. Give your answer correct to 3 significant figures.

Step 1: Find the arc length l = (50/360) × 2π(8) l = (5/36) × 16π = 80π/36 = 20π/9 l ≈ 6.981… Step 2: Add the two radii P = arc + 2r = 20π/9 + 16 P ≈ 6.981 + 16 = 22.98… P ≈ 23.0 cm (3 s.f.) perimeter ≠ arc length — easy mark to drop if you forget the two radii

WE 4

Find the angle from a known arc length

An arc of a circle of radius 6 cm has length 14 cm. Find the angle subtended at the centre, in degrees.

Step 1: Set up the equation 14 = (θ/360) × 2π(6) 14 = (θ/360) × 12π Step 2: Solve for θ θ/360 = 14/(12π) = 7/(6π) θ = 360 × 7/(6π) = 2520/(6π) = 420/π θ ≈ 133.69… θ ≈ 134° (3 s.f.) since θ > 90°, this is a major-ish arc — the diagram should show more than a quarter-circle

WE 5

Find the radius from a known sector area

A sector with central angle 80° has area 50 cm². Find the radius.

Step 1: Set up the equation 50 = (80/360) × π × r² 50 = (2/9)π × r² Step 2: Solve for r² r² = 50 × 9/(2π) = 450/(2π) = 225/π Step 3: Take positive square root (r > 0) r = √(225/π) = 15/√π r ≈ 8.462… r ≈ 8.46 cm (3 s.f.) always discard the negative root — radius is non-negative

WE 6

Sprinkler watering a corner of a lawn

A garden sprinkler is fixed at corner C of a rectangular lawn. The sprinkler rotates through an angle of 90° and sprays water up to 5 m from C. Find (a) the area of grass watered, and (b) the length of the curved outer boundary of the watered region.

Identify the shape: a quarter-circle (sector) centred at C θ = 90°, r = 5 m (a) Area watered A = (90/360) × π × 5² = (1/4) × 25π A = 25π/4 ≈ 19.63… A = 25π/4 m² ≈ 19.6 m² (3 s.f.) (b) Outer arc length l = (90/360) × 2π × 5 = (1/4) × 10π l = 5π/2 ≈ 7.854… l = 5π/2 m ≈ 7.85 m (3 s.f.) 90° is exactly a quarter-turn — the watered region is a quarter-disc; the curved edge is one-quarter of the full circumference

💡 Top tips

Always sketch first. The angle and radius labelled on a quick diagram catch errors before they happen.
Simplify θ/360 early: 90/360 = 1/4, 60/360 = 1/6, 45/360 = 1/8. Cleaner arithmetic, less calculator dependence.
Leave answers in terms of π when the question asks for exact values — only convert to decimal when explicitly needed.
Perimeter ≠ arc length. Read carefully and add 2r when “perimeter of the sector” is asked.
Major sector trick: if θ for the minor sector is given, the major sector uses 360° − θ. Or work out the full circle and subtract the minor.
Memorise both degree formulas — the formula booklet only gives the radian versions, and converting back is wasted time.
Check units: if r is in cm, area is in cm², length in cm. Mixing units in a single calculation is a classic trap.

⚠ Common mistakes

Using the wrong formula. Arc length has 2πr; sector area has πr². Mix them up and the answer is dimensionally wrong.
Forgetting to square the radius in the sector area formula.
Using radians instead of degrees with the degree formula. The formula l = rθ is for radians only — using it with degrees gives nonsense.
Adding 2r to the arc length when only the arc is asked for. The perimeter and the arc are different quantities.
Forgetting to take the square root when solving for the radius from a sector area. r² = 25 means r = 5, not 25.
Using the minor angle for the major sector (or vice versa). Always check: minor < 180°, major > 180°.
Premature rounding: rounding intermediate values like π or fractions of 360 introduces errors. Carry exact values until the final step.

Both formulas reduce to “fraction of the whole” once you’ve internalised the structure — which is exactly the same idea you’ll meet again in the next note, Radian Measure. Switching from degrees to radians simplifies these formulas dramatically: arc length becomes just rθ, no division by 360 required. Worth the small upfront effort to learn.

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