Quadrilateral Properties and Reasoning — Solutions

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1. Properties of quadrilateral types

   1. All four angles equal: Rectangle and Square
   2. Opposite sides parallel: Parallelogram, Rectangle, Rhombus, Square
   3. All four sides equal: Rhombus and Square
   4. Diagonals bisect at right angles: Rhombus, Square, and Kite
   5. Exactly one pair of opposite angles equal: Kite
   6. Diagonals equal in length: Rectangle and Square
   7. Exactly one pair of parallel sides: Trapezium
   8. Both diagonals bisect vertex angles: Rhombus and Square

2. Find unknown angles and sides

   1. Parallelogram x + 115 = 180: x = 65° (co-interior angles, parallelogram)
   2. Rhombus 72°: Opposite angle = 72°; adjacent = 180−72=108° each (co-interior angles of parallelogram)
   3. Trapezium ∠A=68°, find ∠D: ∠A + ∠D = 180° (co-interior angles, AB∥CD); ∠D = 112°
   4. Kite ∠B=54°, ∠D=100°: ∠A + ∠C = 360−54−100 = 206°; ∠A = ∠C = 103° (angles of a kite are equal when between unequal pairs)
   5. Rectangle (4x−8)=90: 4x=98; x=24.5 (all angles of rectangle = 90°)
   6. Ratio 3:4:5:6, sum=360: Parts=18; angles=54°, 72°, 90°, 108°; no standard type (not all equal pairs of opposite angles, not a trapezium)

3. Diagonal properties

   1. Rectangle PM=RM: 4x−3 = 2x+7; 2x=10; x=5; PM=17; PR=2×17=34 (diagonals of rectangle are equal and bisect each other)
   2. Rhombus diagonals 16 and 12: Half-diagonals = 8 and 6; side = √(8²+6²) = √(64+36) = √100 = 10 cm (diagonals of rhombus bisect at right angles; Pythagoras)
   3. Parallelogram 3t+1=5t−7: 2t=8; t=4; WM=13; WY=26 (diagonals of parallelogram bisect each other)
   4. Kite ∠PMQ=90° proof: In ▵PQM and ▵PSM: PQ=PS (given); QM=SM (diagonals of kite: PQ=PS, RQ=RS implies P and R lie on perpendicular bisector of QS); PM=PM (common). ∴ ▵PQM≡▵PSM (SSS). ∠PMQ=∠PMS; but ∠PMQ+∠PMS=180° (straight line). ∴ ∠PMQ=90°.

4. Proving quadrilateral types

   1. AB=CD and AB∥CD proves parallelogram: Draw diagonal BD. In ▵ABD and ▵CDB: AB=CD (given); ∠ABD=∠CDB (alternate angles, AB∥CD); BD=BD (common). ∴ ▵ABD≡▵CDB (SAS). Hence AD=CB and AD∥CB. ∴ ABCD is a parallelogram (both pairs of opposite sides equal and parallel).
   2. Diagonals bisect at right angles and are equal: Diagonals bisect each other ⇒ parallelogram. Diagonals equal ⇒ rectangle. Diagonals at right angles ⇒ rhombus. Both properties ⇒ Square.
   3. EFGH with EF=GH=8, EH=FG=5, ∠E=∠G=95°: Opposite sides equal (EF=GH, EH=FG) ⇒ parallelogram. Opposite angles equal (∠E=∠G=95°) confirms parallelogram. Since no right angles and not all sides equal, EFGH is a parallelogram.

5. Problem solving

   1. Parallelogram, BE bisects ∠ABC=64°: ∠ABE=32° (bisected). ∠BCE=32° (alternate angles, AB∥DC). ▵BCE is isosceles (BE=CE); ∠BCF=32°. ∠BAC=∠DCA (alternate angles); use angle sum in ▵ABF for ∠BAF.
   2. AB∥DC, AB=12, DC=7, height=4: Trapezium (one pair of parallel sides). Area = ½(AB+DC)×h = ½(12+7)×4 = ½×19×4 = 38 m²
   3. ABEF is a rectangle: ABCD rectangle: AB∥FE (F midpoint AD, E midpoint BC). AF=BE=½AD (midpoints). ∠A=∠B=90° (rectangle). ∴ ABEF is a rectangle. Area ratio = AB×AE : AB×AD = 1:2 (since AE=½AD).
   4. Rhombus side 10, short diagonal 12: Half short diagonal = 6. Long half = √(10²−6²)=√64=8. Long diagonal = 16. Area = ½×12×16 = 96 cm²

6. Algebraic problems using quadrilateral properties

   1. Parallelogram ∠A=(5x+8)°, ∠B=(3x+12)°; co-interior sum 180°: (5x+8)+(3x+12)=180; 8x+20=180; x=20. ∠A=108°, ∠B=72°, ∠C=108°, ∠D=72° (opposite angles of parallelogram equal)
   2. Trapezium; ∠P=(4y−10)°, ∠Q=(2y+30)°, ∠R=95°; co-interior P+Q=180°: (4y−10)+(2y+30)=180; 6y+20=180; y=26.67. ∠P=96.67°, ∠Q=83.33°, ∠R=95°; ∠S=360−96.67−83.33−95=85°. Verify: 96.67+83.33+95+85=360° ✓
   3. Rhombus ∠WXY=(6t+3)°; all angles of rhombus are either equal (opposite) or supplementary (adjacent): Rhombus opposite angles equal; adjacent supplementary. (6t+3)° must equal its opposite or supplement. In a rhombus with angle θ: 2θ+2(180−θ)=360 always true. Set one angle: if ∠WXY is one interior angle, opposite ∠WZY = (6t+3)°. Adjacent ∠XYZ = 180−(6t+3)°. Without another constraint, let (6t+3)=90 ⇒ t=14.5 for square, or use given that it’s a rhombus angle ≤ 180. ∠WOX=90° (diagonals of rhombus are perpendicular).

7. Proving a quadrilateral is a specific type

   1. ∠A=∠C=80°, ∠B=∠D=100°; student claims parallelogram: Having opposite angles equal is a necessary condition for a parallelogram, but it is also satisfied by other shapes. However, if opposite angles of a quadrilateral are equal (both pairs), then the quadrilateral IS a parallelogram. So the student is correct — equal opposite angles imply a parallelogram (this can be proved by showing co-interior angles sum to 180° on each pair of sides, hence both pairs of sides are parallel).
   2. EFGH with EF=FG=GH=HE (all sides equal): Step 1: Draw diagonal EG. ▵EFG and ▵EHG: EF=EH (given); FG=HG (given); EG=EG (common). ∴ ▵EFG≡▵EHG (SSS). Hence ∠FEG=∠HEG and ∠FGE=∠HGE. Similarly using diagonal FH: ▵EFH≡▵GFH (SSS) ⇒ EF∥GH and EH∥FG. So EFGH is a parallelogram. Since all sides are equal, EFGH is a rhombus.
   3. Diagonals bisect each other and ∠ABC=90°: Diagonals bisect each other ⇒ ABCD is a parallelogram (property of parallelogram). In a parallelogram, ∠ABC=90° ⇒ all angles = 90° (co-interior angles, parallel sides; opposite angles equal). A parallelogram with all angles 90° is a rectangle. ∴ ABCD is a rectangle.

8. Combining quadrilateral and triangle reasoning

   1. Rectangle ABCD, ∠ACD=34°: ∠CAB=34° (alternate angles, AB∥DC); ∠ABC=90° (rectangle); ∠ADB=34° (alternate angles, AB∥DC); ∠ACB=90−34=56° (angle sum in ▵ABC or complementary in right angle)
   2. Parallelogram PQRS, diagonal PR; ∠QPR=28°, ∠PRS=41°: ∠QPR=28° (given); ∠SPR=∠PRS=41° (alternate angles, PS∥QR... wait: ∠PRS is angle at R in ▵PRS). ∠QRP=∠QPR=28° (alternate, PQ∥SR); ∠PQR=180−28−(28+41)=83° (angle sum ▵PQR); opposite angles: ∠PSR=∠PQR=83°... use: ∠QRS=∠QPR+∠PRS via alternate angles: ∠QPR=28 corresponds to alternate at R =28; ∠QRS=28+41=69°; ∠PQR=180−69=111° (co-interior); ∠PSR=69°; ∠QPS=111°
   3. Kite ABCD, AB=AD, CB=CD; ∠ABC=70°, ∠ADC=130°: ∠BAD+∠BCD=360−70−130=160°; kite: ∠ABC=∠ADC only if it’s a different pair — actually kite has ∠B≠∠D here (70≠130). The two non-vertex angles are ∠BAD and ∠BCD. ∠BAD+∠BCD=160°. The kite’s axis of symmetry means ∠BAD is the angle between the two equal-side pairs; by symmetry ∠BAD is split equally by AC. Without extra info, ∠BAD+∠BCD=160°; no constraint forces them equal. Verify: 70+130+∠A+∠C=360; ∠A+∠C=160°

9. Diagonal properties and area in quadrilaterals

   1. Rhombus diagonals 24 cm and 10 cm: (a) Half-diagonals = 12 and 5; side = √(12²+5²) = √169 = 13 cm. (b) Area = ½×24×10 = 120 cm². (c) Acute angle: tan(θ) = 12/5 ⇒ θ = tan⊃⁻¹(2.4) = 67.4°; acute angle of rhombus = 2×tan⊃⁻¹(5/12) = 2×22.6° = 45.2°; obtuse = 134.8°
   2. Square with diagonal 10√2 cm: Diagonal = side×√2; side = 10√2÷√2 = 10 cm. Perimeter = 4×10 = 40 cm. Area = 10² = 100 cm²
   3. Rectangle AB=12, BC=5: (a) AC = √(12²+5²) = √169 = 13 cm. (b) ∠ with AB = tan⊃⁻¹(5/12) = tan⊃⁻¹(0.4167) = 22.6°. (c) Area = 12×5 = 60 cm²; Perimeter = 2(12+5) = 34 cm

10. Real-world quadrilateral reasoning

    1. Parallelogram AB=15 m, BC=9 m, ∠ABC=70°: (a) Area = AB×BC×sin(70°) = 15×9×0.9397 = 126.9 m². (b) Perimeter = 2(15+9) = 48 m. (c) ∠A=∠C=110° (co-interior with 70°); ∠B=∠D=70°
    2. Trapezium: top 60 cm, bottom 90 cm, non-parallel sides 40 cm each: Overhang each side = (90−60)/2 = 15 cm. Height = √(40²−15²) = √(1600−225) = √1375 ≈ 37.1 cm. Area = ½(60+90)×37.1 = ½×150×37.1 = 2782.1 cm²
    3. Builder claims rectangle from all four sides + ∠A=90°: The second builder is correct — knowing four sides and one angle is not sufficient (a quadrilateral with given sides is not rigid; it can be ‘squashed’). Minimum to prove a rectangle: (1) show it is a parallelogram AND one angle is 90°; or (2) show both diagonals are equal AND bisect each other; or (3) show all four angles are 90°. Just four sides + one angle does not guarantee all angles are right angles.