Review Solutions — Deductive Geometry

1. Congruence tests. Fluency

   • (a) AB=PQ, AC=PR, included ∠A=∠P: SAS ⇒ ▵ABC ≅ ▵PQR.
   • (b) All three sides equal: SSS ⇒ ▵DEF ≅ ▵XYZ.
   • (c) Right angle + hypotenuse + side: RHS ⇒ congruent.
   • (d) Two angles + corresponding side QR=YZ: AAS ⇒ congruent.

2. Scale factors and similarity. Fluency

   • (a) k=3/2; area ratio: 9 : 4.
   • (b) Volume ratio 64:27; length ratio: 4 : 3.
   • (c) k=2/3; QR: 8.
   • (d) Map 1:25 000, 3.6 cm: 0.9 km.

3. Angle facts and parallel lines. Fluency

   • (a) Co-interior sum 180°: x = 22.
   • (b) Corresponding angles equal: y = 20.
   • (c) Exterior angle − known interior: 53°.
   • (d) Opposite angles of parallelogram equal: x=30, ∠A = 70°.

4. Quadrilateral identification. Fluency

   • (a) All sides equal + one right angle: Square.
   • (b) Parallelogram + perpendicular diagonals: Rhombus.
   • (c) One pair parallel + equal diagonals: Isosceles trapezium.
   • (d) Two pairs adjacent equal sides + one diagonal bisects other at 90°: Kite.

5. Similar triangles (diagram, k=2/3). Understanding

   • (a) Ratios AD/AB, AE/AC, DE/BC: All equal 2/3.
   • (b) Similarity test: Common angle + sides in same ratio about it. SAS similarity.
   • (c) Prove DE ∥ BC: ∠ADE = ∠ABC (corresponding, similar triangles) ⇒ DE ∥ BC. □
   • (d) Scale factor and area ratio: k = 2/3. Area ratio = 4/9.

6. Parallelogram PQRS, diagonals meet at T. Understanding

   • (a) ▵PQT ≅ ▵RST: PQ=RS (opp. sides); alternate angles at P and R; alternate at Q and S. AAS ⇒ congruent. □
   • (b) Diagonals bisect each other: PT=RT, QT=ST (corresponding sides of congruent triangles). □
   • (c) ∠QSR when ∠PQS=35°: 35° (alternate angles, PQ∥SR).
   • (d) Confirm parallelogram from angles 62° and 118°: 62+118=180° (co-interior). PQ∥SR confirmed. Parallelogram.

7. Intercept theorem applications. Understanding

   • (a) AD=5, DB=3, AE=10; find EC: 5/3 = 10/EC ⇒ EC = 6.
   • (b) Intercepts 4,6 and 5,x: x = 7.5.
   • (c) Shelf 90 cm in ratio 2:3:4: 20 cm, 30 cm, 40 cm.
   • (d) Trapezium diagonals, AB=8, DC=12; ratio AE/EC: 2 : 3.

8. Coordinate proof (A(1,2), B(5,2), C(7,6), D(3,6)). Understanding

   • (a) Side lengths: AB=CD=4, BC=DA=√20.
   • (b) Slopes, quadrilateral type: AB & CD slope 0; BC & DA slope 2. Parallelogram.
   • (c) Diagonal lengths: AC=√52, BD=√20. Unequal ⇒ not a rectangle.
   • (d) Rhombus check: AB=4 ≠ BC=√20. Not a rhombus.

9. Shadow and height (person 1.8 m, shadow 2.4 m). Problem Solving

   • (a) Ratio setup: h/1.8 = 32/2.4.
   • (b) Building height: 24 m.
   • (c) Building shadow when person shadow = 4.2 m: 56 m.
   • (d) Tree 15 m shadow (same time as b): 20 m.

10. Trapezium (AB=10, DC=6, h=8). Problem Solving

    • (a) Midsegment MN: 8 cm.
    • (b) Prove MN ∥ AB: M and N both at y=4 in coordinate setup ⇒ MN horizontal ⇒ MN ∥ AB. □
    • (c) Area of trapezium: 64 cm².
    • (d) Triangle area ratio = AB:DC: ▵ABC=40, ▵ACD=24. 40:24 = 5:3 = AB:DC. □

11. Scale model (1:200 building). Problem Solving

    • (a) Model height: 24 cm.
    • (b) Model floor area: 600 cm².
    • (c) Model volume: 15 000 cm³.
    • (d) Model mass (density 0.05 g/cm³): 0.75 kg.

12. Ladder and shelf (L=5 m). Problem Solving

    • (a) Height h in terms of x: h = √(25 − x²).
    • (b) Similar triangle ratio: y/h = (x−1)/x ⇒ y = h(x−1)/x.
    • (c) x=3: h and y: h = 4. y = 8/3 ≈ 2.67 m.
    • (d) x=4: h, y, and comparison: h=3. y=9/4=2.25. x=3 gives greater y.