Mathematics Standard • Year 11 • Module 2 • Lesson 5
Perimeter and Arc Length
Practise HSC Mathematics Standard 2-style writing on perimeter and arc length — multi-mark short answers and one structured extended response.
1. Short-answer questions
1.1 Find the arc length of a sector with radius 18 cm and central angle 40°. Give your answer correct to 2 d.p. 2 marks Band 3
1.2 Find the perimeter of a sector with radius 14 cm and central angle 225°. Give your answer correct to 2 d.p. 3 marks Band 3-4
1.3 A rectangle (14 cm × 6 cm) has a semicircle of diameter 6 cm removed from one short end. 4 marks Band 4
(a) Sketch and identify which edges form the outer boundary of the remaining shape.
(b) Find the perimeter of the remaining shape correct to 2 d.p.
2. Extended response
2.1 A community garden is being landscaped on flat ground. The garden's footprint consists of a rectangle 12 m long and 8 m wide, with a sector of radius 8 m and angle 90° attached to one short end (the sector forms a quarter-circle "fan" off the end of the rectangle, with the sector's two radii lying along the rectangle's short end and one long side).
Boundary edges to consider:
• Three sides of the rectangle (one long side opposite the sector, the far short end, and the long side that does NOT touch the sector).
• The arc of the sector (replaces the short end where it attaches).
• The radius of the sector that lies along the long side of the rectangle is INTERIOR (one continuous straight edge), so the rectangle's long side adjacent to the sector continues straight without being broken.
(a) Calculate the arc length of the quarter-circle to 2 d.p.
(b) List the boundary edges and calculate the total perimeter to 2 d.p.
(c) Decorative edging costs $32 per metre and is sold in 50 cm increments. Calculate the cost, justifying any rounding direction, and state your conclusion clearly. 6 marks Band 5-6
Explicit marking criteria
Part (a) — 1 mark
• 1 mark — correct arc length using ℓ = (90/360) × 2π × 8 = 4π ≈ 12.57 m.
Part (b) — 2 marks
• 1 mark — correctly identifies the boundary edges (long side that the sector extends + far short end of rectangle + long side opposite the sector + extra radius of sector + arc).
• 1 mark — correct total perimeter to 2 d.p.
Part (c) — 3 marks
• 1 mark — rounds UP the perimeter to the next 50 cm (with justification: edging cannot fall short).
• 1 mark — correct total cost calculation.
• 1 mark — clear concluding sentence stating the cost in dollars and cents with units.
Your response:
Stuck? Edges to count: (i) far long side of rectangle = 12 m; (ii) far short end = 8 m; (iii) near long side = 12 m; (iv) one sector radius (the one not coincident with the rectangle's side) = 8 m; (v) arc = 4π. Total = 12 + 8 + 12 + 8 + 4π = 40 + 4π.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Arc length, r = 18, θ = 40° (2 marks)
Sample response.
ℓ = (40/360) × 2π × 18 = (1/9) × 36π = 4π = 12.57 cm (to 2 d.p.).
Marking notes. 1 mark — correct substitution into the arc length formula. 1 mark — correct evaluation with unit. Common error: using ½πr² (sector area) instead of arc length formula — produces an answer in cm² (wrong unit).
1.2 — Perimeter of sector, r = 14, θ = 225° (3 marks)
Sample response.
Plan: P = 2r + ℓ.
ℓ = (225/360) × 2π × 14 = (5/8) × 28π = 17.5π = 54.978...
2r = 28.
P = 28 + 54.98 = 82.98 cm (to 2 d.p.).
Marking notes. 1 mark — writes plan P = 2r + ℓ (commits to including the radii). 1 mark — correct arc length. 1 mark — correct total perimeter to 2 d.p. Common error: stopping at ℓ = 54.98 cm and calling it the "perimeter" loses 2/3.
1.3 — Rectangle with semicircle removed (4 marks)
Sample response.
(a) Boundary: 2 long sides (14 cm each) + 1 short side (the one NOT cut, 6 cm) + the semicircle arc (replacing the cut short end). The diameter where the semicircle was cut is interior.
(b) Semicircle r = 3 cm. Arc = π × 3 = 3π = 9.425 cm.
P = 2(14) + 6 + 3π = 28 + 6 + 9.425 = 43.42 cm (to 2 d.p.).
Marking notes. (a) 1 mark — correctly identifies which short end remains and that the diameter is NOT on the boundary. (b) 1 mark — correct semicircle arc. 1 mark — correct rectangle straight edges (28 + 6, NOT 28 + 12). 1 mark — correct total with unit. Common error: including both short ends (28 + 12 + 3π = 49.42) loses 2 marks.
2.1 — Community garden composite (6 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Arc length of the quarter-circle.
ℓ = (90/360) × 2π × 8 = (1/4) × 16π = 4π = 12.57 m (to 2 d.p.). [1 mark — arc length.]
(b) Boundary identification and perimeter.
Boundary edges: long side of rectangle that the sector extends (12 m) + far short end of rectangle (8 m) + long side opposite the sector (12 m) + the non-coincident radius of the sector (8 m) + arc (4π m). The radius that lies along the rectangle's side is interior and is not counted. [1 mark — correctly identifies 5 boundary edges, excluding the interior radius.]
P = 12 + 8 + 12 + 8 + 4π = 40 + 4π = 40 + 12.566 = 52.57 m (to 2 d.p.). [1 mark — correct total.]
(c) Edging cost.
52.57 m must be rounded UP to the next 50 cm increment (since edging is sold in half-metre lots). 52.57 → 53.0 m. [1 mark — rounds up, with justification that edging cannot fall short.]
Cost = 53.0 × $32 = $1,696.00. [1 mark — correct cost.]
Conclusion: the gardener should purchase 53.0 m of edging at a total cost of $1,696.00. [1 mark — explicit conclusion sentence with units.]
Total: 6/6.
Band descriptors for marker.
Band 3: Calculates the arc length correctly but counts the interior radius as a boundary edge (gives P = 60.57 m). ≈ 2-3 marks.
Band 4: Perimeter correct, but rounds the edging length DOWN (to 52.5 m) or to the nearest cm rather than the next 50 cm. ≈ 4 marks.
Band 5: Perimeter and cost correct, but conclusion is bare "$1,696" with no units, no rounding justification, or no naming what was being purchased. ≈ 5 marks.
Band 6: Complete, correct rounding direction with justification, conclusion sentence in dollars and cents with units. 6/6.