Solutions — Applications of Measurement

1. Unit conversions. Fluency

   • (a) 3.5 m² to cm²: 3.5 × 10 000 = 35 000 cm².
   • (b) 85 000 cm² to m²: 85 000 ÷ 10 000 = 8.5 m².
   • (c) 0.6 m³ to litres: 0.6 × 1000 = 600 L.
   • (d) 4.2 km² to hectares: 4.2 × 100 = 420 ha.

2. Composite surface areas. Fluency

   • (a) L-shaped cross-section: Big rectangle: 8×5 = 40 m². Remove notch: 5×3 = 15 m². Area = 40 − 15 = 25 m².
   • (b) Circle with square hole: π(16) − 9 = 16π − 9 ≈ 50.27 − 9 = 41.27 cm².
   • (c) Walls minus windows/door: Total wall area: 2(6+4)×2.8 = 56 m². Windows: 2×1.2×1.0 = 2.4 m². Door: 0.9×2.1 = 1.89 m². Paintable = 56 − 2.4 − 1.89 = 51.71 m².
   • (d) Annulus: π(100) − π(36) = 64π ≈ 201.06 cm².

3. Composite volumes. Fluency

   • (a) Cylinder minus cone (same r=5, h=12): V_cyl = π(25)(12) = 300π. V_cone = ⅓π(25)(12) = 100π. Remaining = 200π ≈ 628.32 cm³.
   • (b) Hemisphere (r=6) + cone (r=6, h=8): V_hemi = ⅔π(216) = 144π. V_cone = ⅓π(36)(8) = 96π. Total = 240π ≈ 753.98 cm³.
   • (c) L-prism (10×8 minus 6×5, depth 4): A = 80 − 30 = 50 cm². V = 50 × 4 = 200 cm³.
   • (d) Tank 50×30×40 half full, sphere r=6 dropped in: V_sphere = &frac43;π(216) = 288π ≈ 904.78 cm³. Base area of tank = 50×30 = 1500 cm². Rise = 904.78/1500 ≈ 0.60 cm.

4. Real-world contexts. Fluency

   • (a) Lawn 12×8=96 m² at 40 g/m²: 96 × 40 = 3840 g = 3.84 kg.
   • (b) Tiles 0.3×0.3 m, floor 3.6×2.4=8.64 m²: Tiles needed (exact): 8.64/0.09 = 96. With 10% extra: 96×1.1 = 105.6 ⇒ 106 tiles.
   • (c) Garden bed 4×3×0.15 m: V = 1.8 m³. Cost = 1.8×$65 = $117.
   • (d) Tank r=0.8, h=1.5 m. Rain 12 mm/h on 80 m²: V_tank = π(0.64)(1.5) = 0.96π ≈ 3015.9 L. Collection rate: 80×0.012 m³/h = 0.96 m³/h = 960 L/h. Time = 3015.9/960 ≈ 3.1 hours (about 3 h 9 min).

5. Swimming pool cross-section. Understanding

   • (a) Shape: Trapezium (trapezoid). The two parallel sides are the depths at each end: 1.2 m (shallow) and 3.0 m (deep). The non-parallel sides are the pool floor (sloped) and the water surface (25 m horizontal).
   • (b) Cross-section area: A = ½(1.2 + 3.0) × 25 = ½ × 4.2 × 25 = 52.5 m².
   • (c) Volume: V = 52.5 × 10 = 525 m³.
   • (d) Litres and pump time: 525 m³ = 525 000 L. Time = 525 000 / 500 = 1050 min = 17 hours 30 minutes.

6. Scaling. Understanding

   • (a) Dimensions doubled: Volume scales by k³ = 2³ = 8 times.
   • (b) Radius tripled: Surface area scales by k² = 3² = 9. New TSA is 9 times the original.
   • (c) Model roof area: Scale 1:50 ⇒ area scale 1:50² = 1:2500. 600 m² = 600 × 10 000 cm² = 6 000 000 cm². Model = 6 000 000 / 2500 = 2400 cm².
   • (d) Smaller cylinder r=2, larger V=500, r=5: Scale factor k = 2/5 = 0.4. Volume scales by k³ = 0.064. V_small = 500 × 0.064 = 32 cm³.

7. Cost and material problems. Understanding

   • (a) Wall 8×2.5 minus window 1.5×1: Area = 8×2.5 − 1.5×1 = 20 − 1.5 = 18.5 m². Cost = 18.5×$15 = $277.50.
   • (b) Cyl paint area (curved + 1 base, r=1, h=2): Area = 2π(1)(2) + π(1)² = 4π + π = 5π ≈ 15.71 m². Litres = 15.71/12 ≈ 1.3 L.
   • (c) Concrete path 1.5 m wide around 20×8 pool, 10 cm deep: Outer dimensions: (20+3)×(8+3) = 23×11 = 253 m². Pool area: 20×8 = 160 m². Path area: 253−160 = 93 m². V = 93×0.1 = 9.3 m³.
   • (d) Gold sphere r=1.5 cm reshaped to wire r=0.05 cm: V_sphere = &frac43;π(1.5)³ = &frac43;π(3.375) = 4.5π cm³. Wire: π(0.05)²×L = 4.5π ⇒ 0.0025L = 4.5 ⇒ L = 1800 cm = 18 m.

8. Rates and flow. Understanding

   • (a) Pipe r=2 cm, speed 0.5 m/s = 50 cm/s: A = π(4) = 4π cm². Volume/s = 4π×50 = 200π ≈ 628.32 cm³/s.
   • (b) Conical tank r=3, h=4, fill at 0.5 m³/min: V = ⅓π(9)(4) = 12π ≈ 37.70 m³. Time = 37.70/0.5 ≈ 75.4 min (about 1 h 15 min).
   • (c) Sphere r=10 cm, radius halves to 5 cm: TSA = 4π(5)² = 100π ≈ 314.16 cm².
   • (d) Conical pile r=h=60 cm: V = ⅓π(60)²(60) = ⅓π(216000) = 72000π ≈ 226 195 cm³. Slant l = 60√2. Curved SA = πrl = π(60)(60√2) = 3600π√2 ≈ 16 013 cm². (Base: πr² = 3600π ≈ 11 310 cm²; total if closed ≈ 27 323 cm².)

9. Grain silo. Problem Solving

   • (a) Total volume (cylinder r=3, h=8 + hemisphere r=3): V_cyl = π(9)(8) = 72π. V_hemi = ⅔π(27) = 18π. Total = 90π ≈ 282.74 m³.
   • (b) Total outer surface area: Cylinder curved: 2π(3)(8) = 48π. Cylinder bottom base: π(9) = 9π. Hemisphere dome: 2π(9) = 18π. Total = 75π ≈ 235.62 m². (No top circle of cylinder — the hemisphere sits there.)
   • (c) 90% full, density 780 kg/m³: V_grain = 0.9 × 90π = 81π ≈ 254.47 m³. Mass = 254.47 × 780 ≈ 198 487 kg ≈ 198.5 tonnes.
   • (d) Pure cylinder, same volume (90π), total height 11 m: πr²(11) = 90π ⇒ r² = 90/11 ≈ 8.182 ⇒ r ≈ 2.86 m.

10. Olympic athletics track. Problem Solving

    • (a) Inner edge of lane 1: 2×84.39 + 2π×36.8 = 168.78 + 231.22 ≈ 400.00 m ✓.
    • (b) Area enclosed by inner edge: Two semicircles = one full circle: π(36.8)² ≈ 4254 m². Two rectangles (straight sections): 2×84.39×2×36.8... Actually the enclosed area = rectangle + circle: 84.39×(2×36.8) + π(36.8)² = 84.39×73.6 + π(1354.24) = 6211.1 + 4254.0 ≈ 10 465 m².
    • (c) Area of 8 lanes (each 1.22 m wide): Total width of 8 lanes = 8×1.22 = 9.76 m. Outer edge radius = 36.8 + 9.76 = 46.56 m. Total area enclosed by outer edge: 84.39×(2×46.56) + π(46.56)² = 7843.9 + 6810.6 = 14 654.5 m². Track area = 14 654.5 − 10 465 = 4190 m².
    • (d) Rubber volume (12 mm = 0.012 m deep): V = 4190 × 0.012 ≈ 50.3 m³.