L27 Solutions — Applications of Deductive Geometry

1. Quadrilateral properties. Fluency

   • (a) Square.
   • (b) Opposite angles equal: 70° and 70°. Co-interior sum to 180°: 70°, 110°, 70°, 110°.
   • (c) Diagonals bisect at 90°. Half-diagonals: 5 and 12. Side = √(25+144) = √169 = 13 cm.
   • (d) Diagonal = 13, side = 5. Other side = √(169−25) = √144 = 12 cm.

2. Coordinate geometry and quadrilaterals. Fluency

   • (a) All sides at right angles, all 90° angles. Rectangle (in fact a square since 5≠4, so rectangle).
   • (b) Diagonal AC: midpoint = (2.5, 2). Diagonal BD: midpoint = (2.5, 2). Same midpoint — diagonals bisect each other.
   • (c) Slope EF = (1−1)/(4−1) = 0. Slope GH = (4−4)/(5−2) = 0. EF ∥ HG. Slope EH = (4−1)/(2−1) = 3. Slope FG = (4−1)/(5−4) = 3. EH ∥ FG. Both pairs parallel ⇒ parallelogram. □
   • (d) Slope EF = 0 (horizontal). Slope EH = 3 (not undefined/0). Since slope EH ≠ undefined, angle HEF ≠ 90°. Not a rectangle. (tanθ = 3 ⇒ θ ≈ 71.6°, not 90°.) □

3. Intercept theorem. Fluency

   • (a) 6/9 = 8/y ⇒ y = 8×9/6 = 12.
   • (b) AD/DB = AE/EC ⇒ 4/6 = 5/EC ⇒ EC = 5×6/4 = 7.5.
   • (c) Scale factor = AD/AB = 4/10 = 2/5. DE = (2/5)×15 = 6.
   • (d) 3/5 = x/10 ⇒ x = 6.

4. Identify quadrilateral from properties. Fluency

   • (a) Equal diagonals that bisect each other ⇒ Rectangle.
   • (b) Diagonals bisect at right angles ⇒ Rhombus or square.
   • (c) One pair parallel, unequal diagonals ⇒ Trapezium.
   • (d) Two pairs adjacent equal sides, one diagonal bisects the other at 90° ⇒ Kite.

5. Parallel intercept theorem (diagram). Understanding

   • (a) When parallel lines cut two transversals, the ratios of corresponding intercepts are equal: AB/BC = PQ/QR.
   • (b) 6/9 = x/12.
   • (c) x = 6×12/9 = 72/9 = 8.
   • (d) ST/TU = AB/BC = 6/9 = 2/3. 10/TU = 2/3 ⇒ TU = 15.

6. Proving a rectangle (parallelogram with equal diagonals). Understanding

   • (a) In ▵ABC and ▵DCB: AC = DB (equal diagonals, given); AB = DC (parallelogram, opposite sides); BC = CB (common). ∴ ▵ABC ≅ ▵DCB (SSS).
   • (b) ∠ABC = ∠DCB (corresponding angles, congruent triangles). □
   • (c) ∠ABC + ∠DCB = 180° (co-interior, AB∥DC). Since equal: 2∠ABC = 180° ⇒ ∠ABC = 90°.
   • (d) ABCD is a parallelogram with ∠ABC = 90°. All angles are therefore 90°. ∴ ABCD is a rectangle. □

7. Kite properties (AB=AD, CB=CD). Understanding

   • (a) In ▵ABD and ▵CBD: AB = CB (given); AD = CD (given); BD = BD (common). ∴ ▵ABD ≅ ▵CBD (SSS).
   • (b) ∠ABD = ∠CBD (corresponding angles, ▵ABD ≅ ▵CBD). □
   • (c) From congruence, ∠BAM = ∠BCM (where M is the intersection of diagonals). Also AM = CM (corresponding sides). BD ⊥ AC because ∠AMB = ∠CMB and they sum to 180° ⇒ each = 90°. So BD is the perpendicular bisector of AC. □
   • (d) ∠ABD = ∠CBD = 34° ⇒ ∠ABC = 68°. ∠ADB = ∠CDB = 62° ⇒ ∠ADC = 124°. ∠BAD: in ▵ABD, ∠BAD = 180−34−62 = 84°. ∠BCD = ∠BAD = 84°. Angles: ABC = 68°, ADC = 124°, BAD = BCD = 84°.

8. Coordinate proof (P(0,0), Q(6,0), R(8,4), S(2,4)). Understanding

   • (a) PQ = 6. QR = √(4+16) = √20 = 2√5. RS = 6. SP = √(4+16) = 2√5. PQ=RS=6, QR=SP=2√5.
   • (b) Slope PQ = 0. Slope RS = 0. PQ∥RS. Slope QR = (4−0)/(8−6) = 2. Slope SP = (4−0)/(2−0) = 2. QR∥SP. Both pairs parallel with opposite sides equal ⇒ Parallelogram.
   • (c) Midpoint PR = (4, 2). Midpoint QS = (4, 2). Same midpoint ⇒ confirms parallelogram.
   • (d) Slope PQ = 0 (horizontal). Slope PS = 2. Product = 0×2 = 0 ≠ −1. PQ ⊥ PS would need product = −1. Not a rectangle. □

9. Tiling with equilateral triangles. Problem Solving

   • (a) 60° (all angles of an equilateral triangle).
   • (b) Two triangles share a side to form a rhombus. The angles at the shared edge: one triangle contributes 60° each side, the other contributes 60°. Rhombus angles: 60°, 120°, 60°, 120°.
   • (c) Height of equilateral triangle (side 4): h = 4×(√3/2) = 2√3 ≈ 3.46 cm. Strip height = 2√3 ≈ 3.46 cm.
   • (d) Area of 1 equilateral triangle = (√3/4)×16 = 4√3 cm². 8 triangles per row × 6 rows = 48 triangles. Total area = 48×4√3 = 192√3 ≈ 332.6 cm².

10. Bridge truss (isosceles ▵ABC: AB=AC=25, BC=30). Problem Solving

    • (a) M is midpoint of BC, so BM = 15. AM = √(25²−15²) = √(625−225) = √400 = 20 m.
    • (b) D, E, F are midpoints of AB, AC, BC. By midpoint theorem, DE ∥ BC and DE = BC/2 = 15; EF ∥ AB, DF ∥ AC. Scale factor = 1/2. ∴ ▵DEF ~ ▵ABC (AA, or midpoint theorem) with k = 1/2.
    • (c) Perimeter ▵ABC = 25+25+30 = 80 m. Perimeter ▵DEF = 80×(1/2) = 40 m.
    • (d) Total = 80 + 40 + 20 = 140 m.