Mathematics Standard • Year 11 • Module 2 • Lesson 6
Area of Sectors, Annuli, and Composite Shapes
Practise HSC-style writing on this lesson: multi-mark short answers plus one structured extended response with explicit marking criteria.
1. Short-answer questions
1.1 Find the area of a sector with radius 9 m and central angle 200°. Give your answer correct to 2 decimal places. 3 marks Band 3
1.2 A circular table has diameter 1.6 m. A circular lazy Susan with diameter 0.8 m sits centred on the table. Find the area of the table not covered by the lazy Susan, correct to 2 decimal places. 3 marks Band 3-4
1.3 A triangular block of land has two boundary fences of 95 m and 140 m meeting at an angle of 110°.
(a) Calculate the area of the block to the nearest m².
(b) The owner claims the block is "over 1 hectare". Test this claim and justify your answer in one sentence. 4 marks Band 4
2. Extended response
2.1 A logo design is shown below.
The logo is built from an equilateral triangle of side length 10 cm.
A sector of radius 3 cm and central angle 60° is drawn at each of the three vertices (using the included angle of the triangle).
An annulus with outer radius 4 cm and inner radius 2 cm sits on top of the triangle as a decorative ring (it does not overlap any sector).
(a) Use the sine area rule to find the area of the equilateral triangle in exact form (in terms of √3) and to 2 d.p.
(b) Find the total area of the three corner sectors in exact form (in terms of π).
(c) Find the area of the annulus in exact form (in terms of π).
(d) Find the area of the logo (triangle − sectors + annulus), correct to 2 d.p. State your conclusion. 7 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct substitution ½ × 10 × 10 × sin 60°.
• 1 mark — correct exact answer 25√3 cm² (or equivalent) AND decimal to 2 d.p.
Part (b) — 1 mark
• 1 mark — correct total (the three 60° sectors make half a circle of radius 3, giving 9π/2 cm²).
Part (c) — 1 mark
• 1 mark — correct annulus area π(16 − 4) = 12π cm² with R and r identified.
Part (d) — 3 marks
• 1 mark — sets up the correct expression: triangle − sectors + annulus.
• 1 mark — numerical result to 2 d.p. using unrounded intermediates.
• 1 mark — explicit conclusion sentence stating the logo area with units.
Your response:
Stuck on (d)? Combine your exact answers from (a), (b), (c): 25√3 − 9π/2 + 12π, then evaluate once.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Sector with r = 9, θ = 200° (3 marks)
Sample response.
A = (200/360) × π × 9² = (5/9) × 81π = 45π.
A = 45π m² (exact) ≈ 141.37 m² (2 d.p.).
Marking notes. 1 mark — correct substitution (fraction and radius squared shown). 1 mark — simplified to 45π or correct unrounded form. 1 mark — final decimal to 2 d.p. with units m². A bare "141.37" with no working scores 1/3.
1.2 — Table minus lazy Susan (3 marks)
Sample response.
Outer radius R = 1.6 ÷ 2 = 0.8 m. Inner radius r = 0.8 ÷ 2 = 0.4 m.
A = π(0.8² − 0.4²) = π(0.64 − 0.16) = 0.48π.
A ≈ 1.51 m² (2 d.p.).
Marking notes. 1 mark — both radii correctly found from diameters (this is the standard "diameter trap" mark). 1 mark — correct annulus substitution and simplification. 1 mark — decimal answer to 2 d.p. with units. Common error: substituting diameter directly into π(R² − r²) gives 4× the correct answer.
1.3 — Triangular block (4 marks)
(a) Sample response. A = ½ × 95 × 140 × sin 110° = 6 650 × 0.93969... ≈ 6 249 m².
(b) Sample response. The claim is incorrect. 6 249 m² is well below 10 000 m² (= 1 hectare), so the block is only about 0.62 ha — not "over 1 hectare".
Marking notes. (a) 1 mark — correct substitution into A = ½ab sin C with the 110° identified as the included angle. 1 mark — correct evaluation to nearest m². (b) 1 mark — correct comparison with 10 000 m². 1 mark — conclusion sentence stating the claim is incorrect with reasoning. A response of just "no" with no justification = 0 on part (b).
2.1 — Logo design (7 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Equilateral triangle area.
Sine area rule with a = b = 10 cm, included angle C = 60° (equilateral).
A = ½ × 10 × 10 × sin 60° = 50 × (√3/2) = 25√3 cm² ≈ 43.30 cm². [2 marks — substitution + exact & decimal answer.]
(b) Three corner sectors total.
Each sector: (60/360) × π × 3² = ⅙ × 9π = 3π/2.
Three sectors: 3 × 3π/2 = 9π/2 cm². (The three 60° angles sum to 180° = half a circle of radius 3, area ½ × 9π = 9π/2 — same answer.) [1 mark.]
(c) Annulus area.
R = 4 cm (outer), r = 2 cm (inner). A = π(16 − 4) = 12π cm². [1 mark.]
(d) Logo area.
Logo = triangle − sectors + annulus = 25√3 − 9π/2 + 12π = 25√3 + 15π/2. [1 mark — correct combination.]
≈ 43.3013 + 23.5619 ≈ 66.86 cm². [1 mark — evaluation to 2 d.p. using unrounded intermediates.]
Conclusion: the total area of the logo is approximately 66.86 cm². [1 mark — explicit conclusion sentence with units.]
Total: 7/7.
Band descriptors for marker.
Band 3: Calculates one of the three component areas correctly but does not combine them or omits the annulus. ≈ 2-3 marks.
Band 4: Two of three components correct (commonly triangle + sectors), uses the wrong sign for the annulus (subtracted instead of added), or stops at exact form without a decimal. ≈ 4-5 marks.
Band 5: All three component areas correct, combined with the right signs, but the final answer is given only as an exact expression without a decimal, or the conclusion sentence is missing. ≈ 6 marks.
Band 6: Full numerical and exact-form solution including correct combination 25√3 − 9π/2 + 12π AND a clear conclusion sentence with units. 7/7.