Name: Memorandum Gr 10 CAPS Reference Topic 2.1 πΊπΊ Time: Euclidean Geometry Properties of Quadrilaterals π½π½ 2 1 Date: ππ π»π» 60° 1 Statement Reason 2.1.1 ππππ = 4ππππ πΊπΊπΊπΊ ππππππππππππ π»π»π»π» ππππ ππππππππ πΊπΊπΊπΊπΊπΊπΊπΊ 2.1.2 πΌπΌοΏ½1 = 60° πΊπΊπΊπΊ ππππππππππππ πΌπΌΜ ππππ ππππππππ πΊπΊπΊπΊπΊπΊπΊπΊ 2.1.3 In βπ»π»π»π»π»π» and βπ½π½π½π½π½π½ 1) π»π»π»π» = ππππ = 4ππππ 2) π»π»π»π» = π½π½π½π½ 3) ππππ = ππππ ∴ βπ»π»π»π»π»π» ≡ βπ½π½π½π½π½π½ Proven at 2.1.1 Given Common π π , π π , π π πΌπΌ ππ ππ π―π―π―π― ππ 2.2 ππ π΄π΄ π΅π΅ π·π· ππ Statement Reason 2.2 π΄π΄π΄π΄ = π΅π΅π΅π΅ Given ABCD is a parallelogram But π΄π΄π΄π΄ = ππππ ∴ π΄π΄π΄π΄ + ππππ = π΄π΄π΄π΄ And π΅π΅π΅π΅ = ππππ ∴ π΅π΅π΅π΅ + ππππ = π΅π΅π΅π΅ ∴ ππππ = ππππ And OD//PC ∴ ππππππππ is a parallelogram πΆπΆ Given Given Given One pair of opposite sides are equal and parallel. 2.3 π΄π΄ π΅π΅ 1 1 2 2 2 1 1 2 π·π· πΆπΆ Statement Reason In βπ΄π΄π΄π΄π΄π΄ and βπΆπΆπΆπΆπΆπΆ 1) π΄π΄π΄π΄ = πΆπΆπΆπΆ 2) π΅π΅π΅π΅ = π΅π΅π΅π΅ 3) π΄π΄π΄π΄ = π΅π΅π΅π΅ Opposite sides of a parallelogram π΄π΄π΄π΄π΄π΄π΄π΄ are equal Common Given ∴ βπ΄π΄π΄π΄π΄π΄ ≡ βπΆπΆπΆπΆπΆπΆ π π , π π , π π οΏ½ = 90° ∴ πΆπΆΜ = π΅π΅οΏ½ = π΄π΄Μ = π·π· Interior <s of a quadrilaterals is equal to 360° οΏ½ ∴ πΆπΆΜ = π΅π΅οΏ½ and π΄π΄Μ = π·π· ∴ π΄π΄π΄π΄π΄π΄π΄π΄ a rectangle 2.4 π»π» π·π· πΊπΊ 2 1 ππ πΈπΈ Statement Reason 2.4.1 π·π·πποΏ½πΊπΊ = 90° π·π·π·π· ⊥ πΈπΈπΈπΈ ππππ ππππππππ π·π·π·π·π·π·π·π· 2.4.3 π·π·π·π· = π»π»π»π» ππππ = π·π·π·π· ∴ π·π·π·π·π·π·π·π· is a rectangle Given Given 2.4.2 π·π·π·π· 2 = 42 + 32 = 16 + 9 π·π·π·π· = √25 π·π·π·π· = 5 ππππ (Pyth) Opposite of rectangle are equal πΉπΉ 2.5 ππ 1 ππ 2 ππ 1 2 2 ππ Statement 2.5 In βππππππ and βππππππ 1) ππππ = ππππ οΏ½2 = π π οΏ½2 2) ππ οΏ½1 + ππ οΏ½2 = ππ οΏ½2 + ππ οΏ½2 = 180° 3) ππ οΏ½ οΏ½ ∴ ππ1 = ππ1 ∴ βππππππ ≡ βππππππ ∴ βππππππ ≡ βππππππ ∴ ππππππππ is a parallelogram 1 ππ 2 Reason Given Alternating <s PS//QR <s on a straight line ππ, π΄π΄, π΄π΄ 1 π π