Similar Figures and Scale Factors — Solutions

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1. State whether each pair of shapes must be similar

   1. Yes, always similar. All squares have equal angles (90°) and sides in the same ratio (the ratio of their sides is constant). Scale factor = larger side ÷ smaller side.
   2. Check ratios: 6/4 = 1.5 and 9/6 = 1.5. Ratios are equal. Yes, similar. Scale factor = 1.5.
   3. Yes, always similar. All equilateral triangles have three 60° angles. By AA similarity, any two equilateral triangles are similar.
   4. Yes, similar. Both have a 90° angle and a 35° angle, so the third angle must also be 55°. By AA, they are similar.
   5. Yes, similar. Both have an apex angle of 50°. The base angles are equal (isosceles), so each base angle = (180°−50°)/2 = 65°. All angles match. By AA, they are similar.
   6. Yes, always similar. All regular hexagons have interior angles of 120° and equal side lengths, so all side ratios are constant.

2. Calculate scale factors

   1. k = 10/4 = 2.5 (enlargement)
   2. k = 5/15 = 1/3 (reduction)
   3. k = 10.5/7 = 1.5 (enlargement)
   4. k = 8/20 = 0.4 (reduction)
   5. Check: 7.5/3 = 2.5 and 12.5/5 = 2.5. k = 2.5 (enlargement)
   6. 9/12 = 0.75, 12/16 = 0.75, 15/20 = 0.75. k = 0.75 (reduction)
   7. k = 4/6 = 2/3 (reduction)
   8. k = 12/9 = 4/3 ≈ 1.333 (enlargement)
   9. 10/5=2, 24/12=2, 26/13=2. k = 2 (enlargement)
   10. Perimeter scale factor = k. k = 36/48 = 0.75 (reduction)

3. Similarity tests for triangles

   1. ▵ABC: ∠A=50°, ∠B=70°, ∠C=60°. ▵PQR: ∠P=50°, ∠R=60°, ∠Q=70°. Two pairs of angles match. ▵ABC ~ ▵PQR by AA (A↔P, B↔Q, C↔R).
   2. DE/GH = 4/6 = 2/3, EF/HI = 6/9 = 2/3, and included ∠E = ∠H = 80°. ▵DEF ~ ▵GHI by SAS similarity.
   3. 6/3=2, 10/5=2, 14/7=2. All three ratios equal 2. ▵JKL ~ ▵MNO by SSS similarity. Scale factor = 2.
   4. PQ/ST = 8/6 = 4/3, QR/TU = 12/9 = 4/3, and ∠Q = ∠T = 55°. ▵PQR ~ ▵STU by SAS similarity.
   5. AB/XY = 6/9 = 2/3, BC/YZ = 8/12 = 2/3, and ∠B = ∠Y = 90°. ▵ABC ~ ▵XYZ by SAS similarity.

4. Formal proofs of similarity

   1. ∠BAC = ∠DEC (alternate angles, AB ∥ DE). ∠ABC = ∠EDC (alternate angles, AB ∥ DE). Therefore ▵ABC ~ ▵EDC by AA.
   2. ∠P is common to both triangles. ST ∥ QR, so ∠PST = ∠PQR (corresponding angles). Therefore ▵PST ~ ▵PQR by AA.
   3. ∠A is common. AB/AD = 4/8 = 1/2, AC/AE = 6/12 = 1/2. Two sides in same ratio with included angle equal. Therefore ▵ABC ~ ▵ADE by SAS similarity.

5. Real-world similarity and scale

   1. Scale factor k = 25/10 = 2.5. New height = 15 × 2.5 = 37.5 cm. Reason: similar figures have sides in the same ratio.
   2. The sun's rays are parallel, so both the person and tree cast shadows from effectively parallel light. The angles of elevation are equal, forming similar right triangles. Person: 1.5/2 = 0.75. Tree height: h/8 = 0.75, h = 6 m.
   3. Scale 1:24. Real length = 4.32 m = 432 cm. Model = 432/24 = 18 cm.
   4. Let the right angle be at C, with altitude CD to hypotenuse AB. In ▵ACD and ▵ABC: ∠A is common, ∠ACD = ∠ABC = 90° − ∠A. By AA, ▵ACD ~ ▵ABC. Similarly for ▵CBD ~ ▵ABC. Similarity test: AA.

6. Find the unknown side

   1. Scale factor k = 15/10 = 1.5. x = 6 × 1.5 = 9 cm.
   2. Scale factor k = 6/8 = 0.75. Other sides: 12 × 0.75 = 9 cm and 16 × 0.75 = 12 cm.
   3. Scale factor k = 9/12 = 0.75. New shorter side = 8 × 0.75 = 6 cm.
   4. Scale factor k = 8/5 = 1.6. Perimeter of larger pentagon = 35 × 1.6 = 56 cm.

7. Area and perimeter ratios

   1. Perimeter scale factor = k = 3. Perimeter of larger = 24 × 3 = 72 cm.
   2. Scale factor k = 10/4 = 2.5. Area ratio = k² = 6.25. Area of larger = 100 cm². Alternatively: 10 × 10 = 100 cm² directly.
   3. Real dimensions: 6 × 50 = 300 cm = 3 m and 4 × 50 = 200 cm = 2 m. Actual area = 3 × 2 = 6 m².

8. Maps and models

   1. Actual distance = 8 cm × 250 000 = 2 000 000 cm = 20 km.
   2. Actual dimensions: 5.2 × 100 = 520 cm = 5.2 m and 3.8 × 100 = 380 cm = 3.8 m. Floor area = 5.2 × 3.8 = 19.76 m².
   3. Actual wingspan = 18.5 × 72 = 1332 cm = 13.32 m.
   4. Area scale factor = (20 000)² = 4 × 10&sup8;. Actual area = 12 × 4 × 10&sup8; cm² = 48 × 10&sup8; cm² = 480 000 m². (Converting: 48 × 10&sup8; cm² ÷ 10 000 = 480 000 m².)

9. Indirect measurement

   1. Scale factor from small to large = 4. River width = 6.5 × 4 = 26 m.
   2. Similar triangles (same sun angle). Building height / 18 = 2 / 3 → building height = 18 × (2/3) = 12 m.
   3. The lamp post and the person form similar triangles with a common apex at the light source. The shadow tip is 2 m behind the person, so the total distance from the post to shadow tip = 5 + 2 = 7 m. Let lamp post height = H. H/7 = 1.8/2 (similar triangles). H = 7 × 0.9 = 6.3 m.

10. Multi-step reasoning

    1. (a) ∠P is common to both triangles. XY ∥ QR, so ∠PXY = ∠PQR (corresponding angles). By AA, ▵PXY ~ ▵PQR. (b) Scale factor k = PQ/PX = 10/4 = 2.5. PR = PY × k = 5 × 2.5 = 12.5 cm. YR = 12.5 − 5 = 7.5 cm. (c) Perimeter ratio = 1 : 2.5 = 2 : 5.
    2. (a) Scale factor k = 5/2 = 2.5. (b) Volume scales as k³ = 2.5³ = 15.625. Volume of larger = 32π × 15.625 = 500π cm³.
    3. Scale factor k = 12/8 = 1.5. Volume ratio = k³ = 1.5³ = 3.375. Larger tin capacity = 400 × 3.375 = 1350 mL.