L25 Solutions — Congruence and Similarity

1. Congruence tests. Fluency

   • (a) SSS (Side-Side-Side).
   • (b) SAS (Side-Angle-Side).
   • (c) RHS (Right angle-Hypotenuse-Side).
   • (d) No. SSA (two sides + non-included angle) can produce two different triangles (the ambiguous case), so it does not guarantee congruence.

2. Scale factors. Fluency

   • (a) 9/6 = 12/8 = 15/10 = 1.5.
   • (b) k = 35/20 = 1.75.
   • (c) k² = 16. Area ratio = 16 : 1.
   • (d) Volume ratio = k³ = 27/8 ⇒ k = ³√(27/8) = 3/2 = 1.5. Length ratio = 3 : 2.

3. Finding missing lengths. Fluency

   • (a) k = 8/5. EF = BC×k = 7×(8/5) = 11.2 cm.
   • (b) k = 9/12 = 0.75. YZ = 16×0.75 = 12.
   • (c) 6×1.5 = 9 cm; 9×1.5 = 13.5 cm. Dimensions: 9 cm × 13.5 cm.
   • (d) k = 24/15 = 16/10 = 1.6.

4. Identify the similarity test. Fluency

   • (a) AA (Angle-Angle).
   • (b) SSS similarity (all sides in equal ratio).
   • (c) SAS similarity (two sides in same ratio, included angle equal).
   • (d) ∠R = 48°, ∠S = 75° ⇒ ∠T = 57°. Corresponding angles: ∠W = ∠T = 57°.

5. Similar triangles in a diagram (DE ∥ BC). Understanding

   • (a) ∠DAE = ∠BAC (common). ∠ADE = ∠ABC (corresponding angles, DE∥BC). ∴ ▵ADE ~ ▵ABC (AA).
   • (b) AB = AD + DB = 6 + 4 = 10. k = AB/AD = 10/6 = 5/3.
   • (c) BC = DE × k = 9 × (5/3) = 15.
   • (d) AE/AC = AD/AB = 6/10. AE = 15 × (6/10) = 9.

6. Congruence proof (parallelogram ABCD). Understanding

   • (a) In a parallelogram, opposite sides are parallel and equal: AB ∥ DC and AB = DC (definition/property of parallelogram).
   • (b) ∠BAC = ∠DCA because AB ∥ DC and AC is a transversal ⇒ alternate interior angles are equal.
   • (c) ∠BCA = ∠DAC because AD ∥ BC (opposite sides of parallelogram) and AC is a transversal ⇒ alternate interior angles are equal.
   • (d) In ▵ABC and ▵CDA: AC = CA (common side); ∠BAC = ∠DCA (alternate angles, AB∥DC); ∠BCA = ∠DAC (alternate angles, AD∥BC). ∴ ▵ABC ≅ ▵CDA (AAS).

7. Area and volume ratios. Understanding

   • (a) k = 3/5. Area ratio = k² = 9 : 25.
   • (b) k² = 128/72 = 16/9. k = 4/3.
   • (c) k = 1/24. Volume ratio = k³ = 1/13824. Model = 3 m³ × 10&sup6; cm³/m³ / 13824 ≈ 217 cm³.
   • (d) k³ = 250/432 = 125/216. k = 5/6. Radii ratio = 5 : 6.

8. Similar triangles in context. Understanding

   • (a) h/2 = 12/3 ⇒ h = 8 m.
   • (b) 4.6 × 50000 = 230000 cm = 2.3 km.
   • (c) k = 2.1/1.2 = 1.75. Run = 4.8 × 1.75 = 8.4 m.
   • (d) Midpoint theorem: DE = BC/2 = 9 cm.

9. Height of a building. Problem Solving

   • (a) Two right triangles sharing a common angle at the surveyor’s eye. Small triangle: base 6 m (eye to rod), height 1.5 m (rod). Large triangle: base 20 m (eye to building), height h.
   • (b) Small: base = 6 m, height = 1.5 m. Large: base = 20 m, height = h.
   • (c) h/1.5 = 20/6 ⇒ h = 1.5 × (20/6) = 5 m.
   • (d) The similar triangles measure from eye level. h_above_eye = 5 m. Actual building height = 5 + 1.6 = 6.6 m.

10. Nested similar triangles. Problem Solving

    • (a) ∠DAE = ∠BAC (common angle). ∠ADE = ∠ABC (corresponding angles, DE∥BC). ∴ ▵ADE ~ ▵ABC (AA).
    • (b) AB = AD + DB = 4 + 6 = 10. Scale factor = AD/AB = 4/10 = 2/5.
    • (c) x/y = 2/5 ⇒ x = (2/5)y.
    • (d) Area(▵ADE)/Area(▵ABC) = k² = (2/5)² = 4/25. Area(▵ABC) = 16×(25/4) = 100 cm². Area(BCED) = 100 − 16 = 84 cm².