Topic Review — Similarity and Scale — Solutions

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1. Similar figures — identifying and scale factors

   1. k = 12/8 = 1.5
   2. k = 15/10 = 1.5. EF = 14 × 1.5 = 21
   3. Perimeter scale factor = k = 2. Larger perimeter = 30 × 2 = 60 cm
   4. k² = 144/36 = 4. k = 2
   5. 1. AA (two pairs of equal angles)
      2. 4:6:8 = 2:3:4 and 6:9:12 = 2:3:4. SSS similarity, k = 1.5

2. Finding unknown sides

   1. k = AB/AD = (AD+DB)/AD = 10/6 = 5/3. BC = DE × k = 9 × (5/3) = 15
   2. k = 7.5/5 = 1.5. Longer side = 8 × 1.5 = 12 cm
   3. h/2 = 20/1.6 = 12.5. h = 2 × 12.5 = 25 m
   4. k = UV/RS = 21/7 = 3. VW = ST × 3 = 11 × 3 = 33. WU = TR × 3 = 9 × 3 = 27

3. Scale drawings and maps

   1. 6.5 × 20 000 = 130 000 cm = 1300 m
   2. 8.4 km = 840 000 cm. 840 000 ÷ 200 000 = 4.2 cm
   3. Actual: 6 × 50 = 300 cm = 3 m; 4.5 × 50 = 225 cm = 2.25 m. Area = 3 × 2.25 = 6.75 m²
   4. 3 cm : 45 m = 3 cm : 4500 cm = 1 : 1500. Scale = 1 : 1500
   5. Actual: 8 × 5000 = 40 000 cm = 400 m; 5 × 5000 = 25 000 cm = 250 m. Area = 400 × 250 = 100 000 m² = 10 ha

4. Combined similarity and scale

   1. Bridge model (1 : 200):
      1. Length = 45 × 200 = 9000 cm = 90 m; Width = 6 × 200 = 1200 cm = 12 m
      2. Area scale factor = 200² = 40 000. Model area = 1296 m² ÷ 40 000 = 0.0324 m² = 324 cm²
      3. Model area = 45 × 6 = 270 cm² — wait, this differs from part (ii). Checking: real area = 90 × 12 = 1080 m² ≠ 1296 m². Using the given 1296 m²: model area = 1296 × 10 000 (cm²/m²) ÷ 40 000 = 12 960 000 ÷ 40 000 = 324 cm². Model dimensions from 324 cm²: e.g. 45 cm × 7.2 cm = 324 cm² would match the given area. Note: 45×6=270, so the given area of 1296 m² corresponds to actual dimensions of 90 m × 14.4 m.
   2. Right triangle altitude:
      1. Altitude² = 6² = 36. Product of segments = 4 × 9 = 36. ✓
      2. Let the original triangle be ABC with right angle at C and altitude CD to AB. ▵ACD: ∠A is common, ∠ADC = ∠ACB = 90°. By AA, ▵ACD ~ ▵ABC. Similarly ▵CBD ~ ▵ABC by AA (shared angle at B, right angle).
   3. Photo enlargements:
      1. 150%: k=1.5. New dims: 15×1.5=22.5 cm, 10×1.5=15 cm. 250%: k=2.5. New dims: 15×2.5=37.5 cm, 10×2.5=25 cm.
      2. Original area = 150 cm². 150% enlargement: 22.5×15 = 337.5 cm². 250% enlargement: 37.5×25 = 937.5 cm².
      3. 150%: area factor = 1.5² = 2.25 (337.5/150 = 2.25 ✓). 250%: area factor = 2.5² = 6.25 (937.5/150 = 6.25 ✓).

5. Mixed extended response — paddock survey

   1. 1. 4.5 × 10 000 = 45 000 cm = 450 m; 6.2 × 10 000 = 62 000 cm = 620 m; 7.8 × 10 000 = 78 000 cm = 780 m
      2. Perimeter = 450 + 620 + 780 = 1850 m
      3. Check if right-angled: 450²+620² = 202 500+384 400 = 586 900. 780² = 608 400. Not a perfect right triangle, but close. Using Heron’s formula: s = 925. Area = √(925 × 475 × 305 × 145) = √(925×475×305×145). 925×475 = 439 375; 305×145 = 44 225; product = 439 375×44 225 ≈ 19 428 109 375. √19 428 109 375 ≈ 139 386 m² ≈ 13.94 ha. (Alternatively, treating as right triangle with legs 450 and 620: Area = ½×450×620 = 139 500 m² = 13.95 ha — very close.)
      4. Fencing cost = 1850 × $28 = $51 800