Perimeter and Area — Solutions

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1. Q1 — Perimeter and area of basic shapes

   Fluency
   • (a) P = 2(9 + 4) = 26 cm. A = 9 × 4 = 36 cm².
   • (b) P = 6 + 7 + 8 = 21 cm. A = ½ × 8 × 5 = 20 cm².
   • (c) C = 2π × 6 = 12π cm ≈ 37.70 cm. A = π × 36 = 36π cm² ≈ 113.10 cm².

2. Q2 — Areas of parallelogram, trapezium, and sector

   Fluency
   • (a) A = bh = 12 × 7 = 84 cm².
   • (b) A = ½(5 + 9) × 6 = ½ × 14 × 6 = 42 cm².
   • (c) A = (135/360) × π × 64 = (3/8) × 64π = 24π ≈ 75.40 cm².

3. Q3 — Circumference and area of circles

   Fluency
   • (a) C = 9π ≈ 28.27 cm. A = 20.25π ≈ 63.62 cm².
   • (b) r = 7 m. C = 14π ≈ 43.98 m. A = 49π ≈ 153.94 m².
   • (c) C = 18.4π ≈ 57.81 mm. A = π × 84.64 ≈ 265.90 mm².

4. Q4 — Composite shape: rectangle with semicircle

   Understanding
   • (a) Semicircle arc = π × 5 = 5π ≈ 15.71 m. Perimeter = 6 + 6 + 10 + 15.71 = 37.71 m.
   • (b) Rectangle = 60 m². Semicircle = ½π(25) = 12.5π ≈ 39.27 m². Total = 99.27 m².

5. Q5 — L-shaped room

   Understanding
   • (a) Sides: 8 + 6 + 5 + 2 + 3 + 4 = 28 m. (Cutting a corner adds two new inner edges equal in total to those removed.)
   • (b) Full rectangle − cut = 48 − 6 = 42 m².

6. Q6 — Sector problems

   Understanding
   • (a) Arc = (72/360) × 2π × 15 = 6π ≈ 18.85 cm.
   • (b) A = (240/360) × π × 100 = (200/3)π ≈ 209.44 cm².
   • (c) 18.85 = (θ/360) × 2π × 15 → θ/360 = 0.200 → θ = 72°.

7. Q7 — Fencing a rectangular paddock

   Understanding
   • (a) P = 2(240 + 185) = 850 m.
   • (b) 850 × $12.50 = $10 625.
   • (c) 850 − 240 = 610 m. Cost = 610 × $12.50 = $7 625.

8. Q8 — Tiling a floor

   Understanding
   • (a) A = 7.2 × 4.8 = 34.56 m².
   • (b) Tile area = 0.09 m². Raw tiles = 384; with 10% waste = 422.4 → 423 tiles.
   • (c) 423 × $4.80 = $2 030.40.

9. Q9 — Running track

   Problem Solving
   • (a) Two semicircles = full circle; circumference = 72π ≈ 226.19 m. Each straight = (400 − 226.19) ÷ 2 ≈ 86.90 m (exact: 200 − 36π).
   • (b) Rectangle: 72 × (200 − 36π) ≈ 6256.3 m². Circle: 1296π ≈ 4071.5 m². Total ≈ 10 327.8 m².

10. Q10 — Stained glass window

    Problem Solving
    • (a) Outer boundary: 80 + 60 + 80 + 60 + 60 = 340 cm.
    • (b) Rectangle = 4800 cm². Triangle = 900√3 ≈ 1558.8 cm². Total ≈ 6358.8 cm².
    • (c) 3.40 m × $8.50 = $28.90.