Mathematics Standard • Year 11 • Module 2 • Lesson 6
Area of Sectors, Annuli, and Composite Shapes
Apply the three formulas to real Australian contexts: garden beds, sports field markings, signage and triangular surveys.
Problem 1 — Circular garden bed with pond
A circular garden bed in a Newcastle park has radius 4.5 m, with a circular pond of radius 1.2 m in the centre. The gardener needs to turf only the annular region between the pond and the garden's edge.
Set up: What are we solving for?
(i) Calculate the area of the entire circular garden bed in exact form (in terms of π). 1 mark
(ii) Calculate the annulus area (turf required), in exact form and to 2 d.p. 2 marks
(iii) Turf costs $14.50 per m². Find the total turfing cost to the nearest dollar. 2 marks
Stuck? Revisit lesson § Card 2 — Annulus Area. Use the factored form A = π(R² − r²) and multiply by π once.Problem 2 — Cricket boundary "wedge" for fielding drill
A coach marks a sector on the cricket field for a fielding drill. The sector has radius 25 m (from the wicket) and an angle of 60°.
Set up: What are we solving for?
(i) Find the area of the sector in exact form (in terms of π). 2 marks
(ii) Convert your exact answer to a decimal correct to the nearest m². 1 mark
(iii) A second sector is added with the same radius but an angle of 120°. Without recalculating from scratch, state by what factor the second sector's area is greater than the first, and justify in one sentence. 2 marks
Stuck on (iii)? The radius is the same, so only the fraction θ/360 changes — compare 60/360 with 120/360.Problem 3 — Triangular paddock survey
A farmer surveys a paddock and finds it forms a triangle with two boundary fences of length 180 m and 220 m meeting at an angle of 75°.
Set up: What are we solving for?
(i) Calculate the paddock area in square metres to the nearest m². 2 marks
(ii) Convert your answer to hectares (1 hectare = 10 000 m²), correct to 2 decimal places. 1 mark
(iii) Pasture seed costs $45 per hectare. Find the total seed cost to the nearest dollar. 2 marks
Stuck? Revisit lesson § Card 3 — Sine Area Rule. The 75° sits between the two given sides — that is the included angle.Problem 4 — Equilateral logo with corner sectors
A community group designs a logo as an equilateral triangle with side length 12 cm. A sector of radius 4 cm and central angle 60° is drawn at each vertex (using the included angle of the triangle).
Set up: What are we solving for?
(i) Find the area of the equilateral triangle using the sine area rule, in exact form and to 2 d.p. 2 marks
(ii) Find the total area of the three corner sectors, in exact form. 2 marks
(iii) Find the shaded logo area (triangle minus the three sectors), to 2 d.p. State your conclusion. 2 marks
Stuck? Revisit lesson § Worked Example 4 — the three 60° sectors together make exactly ½ of a full circle of radius 4.Problem 5 — Shade sail with cut-out arc
A school orders a square shade sail of side 6 m. A quarter-circle sector of radius 6 m (centred at one corner of the square) is cut out so the sail clears a tree. The shaded sail is the square minus the sector.
Set up: What are we solving for?
(i) Find the area of the original square. 1 mark
(ii) Find the area of the quarter-circle sector cut out, in exact form and to 2 d.p. 2 marks
(iii) Find the remaining sail area to 2 d.p. The fabric supplier rounds up to the next whole m². State the rounded purchase figure and explain in one sentence why you round up. 3 marks
Stuck? Revisit lesson § Composite Areas — Remaining = Square − Sector. Rounding down would leave the sail short of fabric.How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Garden bed with pond
Set up. We need the annular area between the pond (r = 1.2 m) and the garden edge (R = 4.5 m), then multiply by the turf rate.
(i) Total garden area = π(4.5)² = 20.25π m².
(ii) A = π(4.5² − 1.2²) = π(20.25 − 1.44) = 18.81π m² ≈ 59.10 m².
(iii) Cost = 59.10 × $14.50 = $856.95 ≈ $857. (Round to the nearest dollar as the question states.)
Problem 2 — Cricket fielding wedge
Set up. Sector area uses the fraction θ/360 of πr² with r = 25 m and θ = 60°.
(i) A = (60/360) × π × 625 = ⅙ × 625π = 625π/6 m².
(ii) 625π/6 = 327.249... ≈ 327 m².
(iii) The second sector is twice the first. The radius is unchanged, so only the fraction θ/360 matters: 120/360 = 2 × 60/360.
Problem 3 — Triangular paddock
Set up. Two sides and the included angle — use A = ½ab sin C.
(i) A = ½ × 180 × 220 × sin 75° = 19 800 × 0.96593... = 19 125.4... ≈ 19 125 m².
(ii) 19 125 / 10 000 = 1.91 hectares (2 d.p.).
(iii) Seed cost = 1.91 × $45 = $85.97... ≈ $86. (Use the unrounded 1.9125... for accuracy: 1.9125 × 45 = $86.06; either $86 accepted.)
Problem 4 — Equilateral logo
Set up. Triangle area uses sine area rule with sides 12, 12 and included angle 60°. Three sectors at the vertices share radius 4 cm and angle 60°.
(i) Triangle: A = ½ × 12 × 12 × sin 60° = 72 × (√3/2) = 36√3 cm² ≈ 62.35 cm².
(ii) One sector = (60/360) × π × 16 = ⅙ × 16π = 8π/3. Three sectors = 8π cm². (The three 60° angles total 180° = half a circle.)
(iii) Shaded = 36√3 − 8π ≈ 62.3538 − 25.1327 ≈ 37.22 cm². The shaded logo area is approximately 37.22 cm².
Problem 5 — Shade sail
Set up. Square area minus quarter-circle area, then round up because the supplier won't sell partial m² and short fabric leaves a hole.
(i) Square area = 6² = 36 m².
(ii) Sector (θ = 90°, r = 6): A = ¼ × π × 36 = 9π m² ≈ 28.27 m².
(iii) Remaining = 36 − 9π ≈ 36 − 28.27 = 7.73 m². Rounded up to next whole m² = 8 m². You round up because rounding down to 7 m² would mean ordering less fabric than the sail needs — the sail wouldn't fit.