Mathematics Advanced • Year 11 • Module 2 • Lesson 2
Arc Length & Area of Sectors
Practise HSC-style writing on arc length and sector area, including a derivation and a structured optimisation problem.
1. Short-answer questions
1.1 A sector has radius 9 cm and central angle 2π/3 radians. Find, in exact form: (a) the arc length; (b) the area of the sector. 3 marks Band 3
1.2 A piece of wire 40 cm long is bent to form the perimeter of a sector (two radii + arc). If the radius of the sector is 12 cm, find the angle of the sector in radians (exact form). 3 marks Band 3-4
1.3 Two sectors have the same area. Sector A has radius 6 cm and central angle π/2 radians. Sector B has radius 4 cm. Find the central angle of Sector B in radians (exact form). Show all working. 4 marks Band 4
Stuck on 1.3? Compute the area of A, set it equal to the formula for B, then solve for θB.2. Extended response
2.1 A circular flower bed has radius r metres. A landscaper builds a sector-shaped fence around a wedge of the bed; the fence runs along two straight radii and a curved arc, with total length P metres. Let the central angle of the wedge be θ radians.
(a) By starting from the definition of a radian (θ = l / r), derive the formula A = ½ r² θ for the area of the sector. Use the relationship between the sector's area and the full circle's area.
(b) Given the fixed fence length P = 24 m, express A as a function of r alone (i.e. eliminate θ), stating any restriction on r.
(c) Hence, by completing the square (or otherwise), find the value of r that maximises the area A, and state the maximum area. Verify that the corresponding θ is sensible (i.e. positive and not too large). 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — identifies the sector's area as the fraction θ / (2π) of the full circle area π r².
• 1 mark — simplifies π r² × (θ / 2π) = ½ r² θ, showing π cancels.
Part (b) — 2 marks
• 1 mark — uses 2r + rθ = 24 to get θ = (24 − 2r)/r and substitutes into A.
• 1 mark — simplifies to A(r) = 12r − r², with restriction 0 < r < 12 (so that θ > 0).
Part (c) — 3 marks
• 1 mark — completes the square: A(r) = −(r − 6)² + 36 (or equivalent method).
• 1 mark — identifies r = 6 m maximises A, with Amax = 36 m².
• 1 mark — verifies θ = (24 − 12)/6 = 2 rad, which lies in (0, 2π) so the sector is geometrically valid.
Your response:
Stuck on (c)? Write A = 12r − r² = −(r² − 12r) = −[(r − 6)² − 36].How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sector with r = 9, θ = 2π/3 (3 marks)
Sample response.
(a) l = rθ = 9 × 2π/3 = 18π/3 = 6π cm.
(b) A = ½ r² θ = ½ × 81 × 2π/3 = 81π/3 = 27π cm².
Marking notes. (a) 1.5 marks: 0.5 for formula, 0.5 for substitution, 0.5 for simplified exact answer with unit. (b) 1.5 marks: same breakdown. Award 0.5 even if the ½ is forgotten (Trap 03 in the lesson costs the simplification mark).
1.2 — Wire perimeter = 40 cm, r = 12 cm (3 marks)
Sample response. Perimeter = 2r + l = 40, so l = 40 − 2(12) = 40 − 24 = 16 cm. Then θ = l/r = 16/12 = 4/3 rad.
Marking notes. 1 mark — correct perimeter equation 2r + l = 40. 1 mark — correct arc length l = 16. 1 mark — correct angle 4/3 rad (exact form, simplified). Common error: students forget the perimeter includes the two radii and use l = 40 (loses 2 marks).
1.3 — Equal areas (4 marks)
Sample response. Area of A: AA = ½ × 6² × π/2 = ½ × 36 × π/2 = 9π cm².
Setting equal to B's area: 9π = ½ × 4² × θB = 8 θB. Solving: θB = 9π/8 rad.
Answer: θB = 9π/8 rad.
Marking notes. 1 mark — computes AA = 9π. 1 mark — writes equation AA = ½ rB² θB with correct substitution. 1 mark — rearranges correctly to θB = 2A / rB² or equivalent. 1 mark — correct simplified exact answer 9π/8. Common error: cancelling π on only one side or losing the factor of ½.
2.1 — Extended response (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
Part (a) — Derivation of A = ½ r² θ. Consider a sector with central angle θ rad inside a circle of radius r. The sector occupies the fraction θ / (2π) of the full circle (since a full revolution is 2π rad). The full circle has area π r². Therefore the sector's area is:
A = (full area) × (fraction occupied) = π r² × θ / (2π). [1 mark]
The π in the numerator cancels the π in the denominator, giving:
A = r² θ / 2 = ½ r² θ. [1 mark]
This formula is valid only when θ is in radians, because the "fraction of a full revolution" rule above uses 2π rad — if θ were in degrees we would need to divide by 360 instead, producing a different (and less elegant) formula.
Part (b) — Express A as a function of r. The fence constraint gives 2r + l = 24, and since l = rθ:
2r + rθ = 24 ⇒ rθ = 24 − 2r ⇒ θ = (24 − 2r) / r. [1 mark]
Substitute into A = ½ r² θ:
A(r) = ½ r² × (24 − 2r) / r = ½ r (24 − 2r) = ½ (24r − 2r²) = 12r − r².
Restriction: θ > 0 requires 24 − 2r > 0, i.e. r < 12. Also r > 0. So 0 < r < 12. [1 mark]
Part (c) — Maximise the area. Complete the square:
A(r) = −r² + 12r = −(r² − 12r) = −[(r − 6)² − 36] = −(r − 6)² + 36. [1 mark]
This is a downward-opening parabola in r with vertex at r = 6. Since r = 6 lies inside the domain (0, 12), the maximum is attained there:
Amax = 36 m² at r = 6 m. [1 mark]
Check the corresponding angle: θ = (24 − 12) / 6 = 12/6 = 2 rad (≈ 114.6°), which is in (0, 2π) so the sector is geometrically valid. [1 mark] ▮
Total: 7/7.
Band descriptors for marker.
Band 3: Quotes A = ½ r² θ without derivation in (a); attempts (b) but does not substitute correctly; leaves (c) unfinished or uses calculus prematurely without setting up the function. ≈ 2-3 marks.
Band 4: Completes (a) and (b); attempts (c) but makes algebra errors in completing the square OR forgets to verify the domain. ≈ 4-5 marks.
Band 5: Full derivation, correct (b), correct optimisation; missing the angle verification in (c). ≈ 6 marks.
Band 6: All parts complete with explicit reasoning; the maximisation uses completing the square cleanly (or calculus equivalently); the θ = 2 rad verification is given and shown to lie in (0, 2π). 7/7. Top scripts also note that r = 6 makes the bent wire's "two radii" segment 12 cm and the arc l = rθ = 12 cm — an elegant equal-split of the 24 cm wire.