Part A: Trapeziums & Angle Properties (Pages 35–36)

• (i) In Trapezium PQRS with PQ || RS:
  
  • Side PS acts as a transversal intersecting parallel lines PQ and RS.
  
  Therefore, co-interior angles on the same side of the transversal are supplementary:
  
  ∠P + ∠S = 180° → 105° + ∠S = 180° → ∠S = 75°.
  
  • Similarly, side QR acts as a transversal between parallel lines PQ and RS:
  
  ∠Q + ∠R = 180° → 135° + ∠R = 180° → ∠R = 45°.
• (ii) In Isosceles Trapezium ABCD with AB || DC and AD = BC:
  
  • Since AB || DC, co-interior angles are supplementary:
  
  ∠A + ∠D = 180° → ∠A + 100° = 180° → ∠A = 80°.
  
  • In any isosceles trapezium (where non-parallel legs are equal, AD = BC), base angles are equal:
  
  ∠B = ∠A = 80°.
  
  • Then ∠C = 180° − ∠B = 180° − 80° = 100° (or ∠C = ∠D = 100°).

• Quadrilateral Hierarchy Definitions:
  
  • Kite: A quadrilateral with two distinct pairs of adjacent equal sides.
  
  • Parallelogram: A quadrilateral with both pairs of opposite sides parallel and equal.
  
  • Rhombus: A quadrilateral with all 4 sides equal. (Belongs to BOTH Parallelograms and Kites!).
  
  • Rectangle: A parallelogram with four 90° angles.
  
  • Square: A quadrilateral with 4 equal sides and four 90° angles. (Belongs to ALL sets: Parallelograms, Kites, Rhombuses, and Rectangles!).
• (i) Intersection of Kites and Parallelograms:
  
  Any parallelogram with adjacent sides equal has all four sides equal → this forms a Rhombus (and if angles are 90°, a Square).
• (ii) Can a quadrilateral be both a kite and a rectangle?
  
  A rectangle has opposite sides equal and 90° angles. A kite has adjacent sides equal. A rectangle with adjacent sides equal must have all four sides equal, which is a Square!
• (iii) Is every kite a rhombus?
  
  No. In a standard kite, only adjacent pairs are equal (e.g. sides 3, 3, 5, 5), not all four sides. Every rhombus is a kite, but not every kite is a rhombus.

Part B: Overlapping Rectangles & Geometric Constructions (Pages 36–37)

• In rectangle PAIR, every interior corner angle is 90°. Therefore, ∠PRI = 90°.
• In rectangle RODS, every interior corner angle is 90°. Therefore, ∠SRO = 90°.
• Given: ∠ARO = 30°.
• Notice that ∠ARI + ∠IRO = ∠ARO + … using rotational symmetry of the common corner R:
  
  The angle between corresponding rays of rectangle PAIR and rectangle RODS rotated about R is 30°.
  
  Therefore, the angle of intersection between diagonal/side lines maintains this 30° rotational offset.
• By vertical and alternate angle relations at intersection point O/I: ∠IOD = 30°.

• Geometric Property: In a square, the diagonals are equal in length, bisect each other, and are perpendicular to each other (intersect at 90°).
• Step 1: Draw a straight line segment AB = 6 cm using a ruler. This represents the first diagonal d₁.
• Step 2: With A and B as centres and radius strictly greater than half of AB (e.g., 4 cm), draw arcs on both sides of AB intersecting at points P and Q.
• Step 3: Join PQ. Line PQ is the perpendicular bisector of AB, intersecting AB at its exact mid-point O (where OA = OB = 3 cm).
• Step 4: With O as centre and radius equal to 3 cm (half of 6 cm), draw arcs cutting the perpendicular bisector line at points C (above) and D (below).
  
  Thus, OC = OD = 3 cm, making diagonal CD = 6 cm.
• Step 5: Join AC, CB, BD, and DA.

• Let the side length of square CASE be 2s.
• Since U, V, W, X are midpoints:
  
  CU = UA = AV = VS = SW = WE = EX = XC = s.
• All corner angles ∠C, ∠A, ∠S, ∠E are 90°.
• 1. Equal Sides Proof (Pythagoras Theorem):
  
  In right-angled ΔCUV: UV² = CU² + CV² = s² + s² = 2s² → UV = s√2.
  
  Similarly, in ΔAVW, ΔSWE, ΔEXU: VW = WX = XU = s√2.
  
  All 4 sides of UVWX are equal (UV = VW = WX = XU), which proves UVWX is at least a rhombus.
• 2. Right Angles Proof:
  
  In isosceles right ΔCUV: CU = CV → ∠CUV = ∠CVU = 45°.
  
  In isosceles right ΔEXU: EX = EU → ∠EXU = ∠EUX = 45°.
  
  Along side CE, points X, U lie on a straight line:
  
  ∠CUV + ∠XUV + ∠EUX = 180° (Linear Pair)
  
  45° + ∠XUV + 45° = 180° → ∠XUV = 90°.
  
  By exact symmetry, all four interior angles are 90°.

Part C: Quadrilateral Characterisations & Proofs (Pages 37–39)

• A quadrilateral with four equal sides is by definition a rhombus.
• Every rhombus is a parallelogram, meaning opposite angles are equal, and adjacent co-interior angles are supplementary (sum to 180°).
• Let the given 90° angle be ∠A = 90°.
• Opposite angle ∠C = ∠A = 90°.
• Adjacent angle ∠B = 180° − ∠A = 180° − 90° = 90°.
• Opposite angle ∠D = ∠B = 90°.
• Thus, all four sides are equal, and all four angles equal 90°.

• Let quadrilateral be ABCD with AB = CD and BC = AD.
• Draw diagonal AC. Consider ΔABC and ΔCDA:
  
  • AB = CD (Given)
  
  • BC = DA (Given)
  
  • AC = CA (Common side)
• By SSS Congruence Criterion: ΔABC ≅ ΔCDA.
• By CPCTC (Corresponding Parts of Congruent Triangles):
  
  ∠BAC = ∠DCA → Since these alternate interior angles are equal, AB || CD.
  
  ∠BCA = ∠DAC → Since these alternate interior angles are equal, BC || AD.
• Since both pairs of opposite sides are parallel, ABCD is a parallelogram.

• Let ABCD be a concave (arrowhead/dart) quadrilateral where reflex interior angle occurs at D.
• Join diagonal BD dividing the quadrilateral into two non-overlapping triangles: ΔABD and ΔCBD.
• In ΔABD: Sum of angles = ∠DAB + ∠ABD + ∠ADB = 180°.
• In ΔCBD: Sum of angles = ∠DCB + ∠CBD + ∠CDB = 180°.
• Adding all angles together:
  
  ∠DAB + (∠ABD + ∠CBD) + ∠DCB + (∠ADB + ∠CDB)
  
  = ∠A + ∠B + ∠C + ∠D = 180° + 180° = 360°.

• (i) False: Diagonals equal and bisecting each other defines a rectangle. To be a square, diagonals must also be perpendicular.
• (ii) True: The sum of all four angles is 360°. If three angles are 90°, the fourth angle is 360° − 270° = 90°. A quadrilateral with four right angles is a rectangle.
• (iii) True: By SAS triangle congruence, bisecting diagonals prove opposite sides are parallel and equal, forming a parallelogram.
• (iv) False: A kite also has perpendicular diagonals, but its diagonals do not necessarily bisect each other, and all 4 sides are not equal.
• (v) True: If ∠A = ∠C and ∠B = ∠D, then 2∠A + 2∠B = 360° → ∠A + ∠B = 180°, proving consecutive angles are supplementary, so opposite sides are parallel.
• (vi) True: If all 4 angles are equal, each must be 360° / 4 = 90°, which defines an equiangular quadrilateral (rectangle).
• (vii) False: An isosceles trapezium has only one pair of parallel sides; the other non-parallel pair cannot be parallel.