Mathematics Advanced • Year 11 • Module 2 • Lesson 2
Arc Length & Area of Sectors
Apply l = rθ and A = ½ r² θ to realistic problems: pizza, satellite dishes, garden irrigation, running tracks, and bent-wire optimisation.
Problem 1 — Pizza slice (the lesson's hook)
A pizza has radius 20 cm. A wedge-shaped slice is cut with a central angle of 45° at the pizza's centre. The crust runs along the curved outer edge of the slice, and the topping covers the flat (sector) region.
Set up: What are we solving for?
(i) Convert 45° to radians and find the exact arc length of the crust on this slice. 2 marks
(ii) Find the exact area of the topping (the sector area). Give your answer to 1 decimal place as well, for context. 2 marks
(iii) The pizzeria sells a "double crust" slice that has the same topping area as this slice but a wider angle. If the new slice has an angle of 90°, what radius must the new pizza have? 3 marks
Stuck on (iii)? Equate the two sector areas (50π cm² from (ii) and ½ R² × π/2) and solve for R.Problem 2 — Satellite-dish sector cutting (manufacturing)
An engineer fabricates a circular satellite dish from a flat metal disc, then cuts out sectors that are bent up to form the parabolic profile. Each metal sector has radius 60 cm. The total angle removed (summed over six identical sectors) is 2π/3 rad.
Set up: What are we solving for?
(i) Find the central angle of one sector in exact radian form, and convert to degrees. 2 marks
(ii) Calculate the arc length of a single sector to 2 decimal places. Engineers manufacture to within ±0.5 mm; explain whether the lesson's claim "errors of a few mm defocus the signal" is consistent with the precision required here. 3 marks
(iii) Find the total area of metal removed across all six sectors, in exact form. 2 marks
Problem 3 — Garden sprinkler coverage (real measurement)
A rotating sprinkler at the corner of a garden bed throws water out 4.5 m and oscillates through a fixed angle. A gardener measures three different sprinkler settings and records the corresponding watered-region areas:
| Setting | Central angle θ | Watered area (m²) |
|---|---|---|
| Narrow | π/4 rad | |
| Standard | π/2 rad | |
| Wide | 3π/4 rad |
Set up: What are we solving for?
(i) Fill in the third column (exact form, in terms of π). 3 marks
(ii) Show that the area is directly proportional to θ, given fixed r. State the constant of proportionality (in m²/rad) and explain in one sentence why this proportionality follows from the formula A = ½ r² θ. 2 marks
(iii) The garden bed itself is 16 m². What is the smallest sprinkler angle (in radians, exact form) that fully waters the bed? Give a decimal approximation in degrees for the gardener's setting dial. 2 marks
Stuck on (iii)? Set A = 16 and rearrange θ = 2A / r².Problem 4 — Running-track curve (geometric)
An athletics track has two straight sides of length 100 m connected by two semicircular bends. The inner edge of each bend has radius 36.5 m (an IAAF Lane 1 standard).
Set up: What are we solving for?
(i) Find the arc length of one bend in exact form (in terms of π), and to 2 decimal places. (A semicircle subtends π rad.) 2 marks
(ii) Find the total lap distance around the inner edge of the track. 2 marks
(iii) A runner in Lane 8 runs on a track whose bend radius is 7 m greater (i.e. 43.5 m). Find by how much further the Lane 8 runner travels in one lap, assuming the straight sections are the same. Comment on whether this justifies the staggered starts you see on TV. 3 marks
Problem 5 — Bent-wire sector optimisation
A 24 cm length of stiff wire is bent to form the boundary of a sector (two radii + arc). The wire is used completely — nothing is wasted. Let r be the radius and θ the central angle.
Perimeter: P = 2r + l = 2r + rθ = 24 (constraint).
Area: A = ½ r² θ.
Set up: What are we solving for?
(i) If θ = 2 rad, find r and the resulting sector area. 2 marks
(ii) Use the constraint to express θ as a function of r, then substitute into A to express A as a function of r alone. 2 marks
(iii) What practical constraint(s) on r ensure the sector is geometrically valid (i.e. θ > 0)? Give the largest permitted r. 2 marks
Stuck on (ii)? From 2r + rθ = 24, get θ = (24 − 2r)/r, then A = ½ r² × θ.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Pizza slice
Set up. We convert 45° to radians, then apply l = rθ and A = ½ r² θ to a single slice; then equate areas to find the radius of a 90° slice of the same topping area.
(i) 45° = π/4 rad. l = 20 × π/4 = 5π cm (≈ 15.71 cm).
(ii) A = ½ × 400 × π/4 = 200π/4 = 50π cm² (≈ 157.1 cm²).
(iii) Set sector areas equal: 50π = ½ R² × π/2 = π R² / 4. Then R² = 200, so R = √200 = 10√2 cm (≈ 14.14 cm). The new pizza must have radius about 14.14 cm to produce a 90° slice with the same 50π cm² topping area.
Problem 2 — Satellite-dish sector cutting
Set up. The total cut-out angle is 2π/3 rad, distributed evenly across six sectors; we apply l = rθ for the cut length and A = ½ r² θ for the metal removed.
(i) Single-sector angle: (2π/3) / 6 = 2π/18 = π/9 rad. In degrees: (π/9) × (180/π) = 180/9 = 20°.
(ii) l = 60 × π/9 = 60π/9 = 20π/3 cm ≈ 20.94 cm. The lesson's claim is consistent — a 0.5 mm tolerance on a 20.94 cm arc is a relative precision of ≈ 0.024%, demonstrating that even tiny absolute errors matter for the bend geometry. (Award the mark for any reasonable justification linking the absolute error to manufacturing precision.)
(iii) Area per sector = ½ × 3600 × π/9 = 1800π/9 = 200π cm². Total across 6 sectors: 6 × 200π = 1200π cm² (≈ 3770 cm²). Faster alternative: apply A = ½ r² θ with θ = 2π/3 directly: ½ × 3600 × 2π/3 = 1200π cm². ✓
Problem 3 — Garden sprinkler coverage
Set up. Sprinkler reach r = 4.5 m, so r² = 20.25. Apply A = ½ r² θ for each θ.
(i) ½ × 20.25 = 10.125. So A = 10.125 θ.
Narrow (θ = π/4): A = 10.125 × π/4 = 10.125π/4 m² = 81π/32 m² (≈ 7.95 m²).
Standard (θ = π/2): A = 10.125 × π/2 = 81π/16 m² (≈ 15.90 m²).
Wide (θ = 3π/4): A = 10.125 × 3π/4 = 243π/32 m² (≈ 23.86 m²).
(ii) A = (10.125) θ = (½ r²) θ; with r fixed the bracketed factor is constant, so A is directly proportional to θ with constant of proportionality k = ½ r² = 81/8 = 10.125 m²/rad. This proportionality is built into A = ½ r² θ — doubling θ doubles A when r is fixed.
(iii) θ = 2A / r² = 2 × 16 / 20.25 = 32/20.25 = 128/81 rad ≈ 1.58 rad. In degrees: (128/81) × (180/π) ≈ 90.5°. The gardener should set the dial to slightly more than the Standard setting.
Problem 4 — Running-track bend
Set up. Each bend is a semicircle of radius r = 36.5 m, so θ = π rad; the full lap is two straights + two bend arcs.
(i) lbend = rθ = 36.5 × π = 36.5π m ≈ 114.67 m.
(ii) Lap = 2 × 100 + 2 × 36.5π = 200 + 73π m ≈ 429.35 m. (Close to the 400 m standard; the IAAF lap is measured 30 cm out from Lane 1's inside edge, giving exactly 400 m.)
(iii) Lane 8 bend radius = 43.5 m, so each bend = 43.5π m. Extra per bend = (43.5 − 36.5)π = 7π m; two bends per lap give 14π m extra ≈ 43.98 m. This is exactly why outer-lane runners are staggered forward at the start of races run in lanes — the ≈ 44 m difference would otherwise be unfair.
Problem 5 — Bent-wire sector optimisation
Set up. A 24 cm wire forms the entire boundary 2r + l = 2r + rθ. We use this constraint with A = ½ r² θ.
(i) 2r + 2r = 24 ⇒ 4r = 24 ⇒ r = 6 cm. A = ½ × 36 × 2 = 36 cm².
(ii) From 2r + rθ = 24: θ = (24 − 2r) / r = 24/r − 2. Substitute into A: A(r) = ½ r² (24/r − 2) = ½ (24r − 2r²) = 12r − r².
(iii) Need θ > 0: 24/r − 2 > 0 ⇒ 24/r > 2 ⇒ r < 12. Also r > 0. So the largest permitted r is anything less than 12 cm (at r = 12 the sector degenerates to θ = 0). The valid domain is 0 < r < 12.