Ex 6.2, 10

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Ex 6.2, 10 Given a square ABCD , let P be a point within it. Join PA, PB, PC, PD (Fig. 6.33). What is the ratio of the areas of the red region (β–³PAB and β–³PCD ) and the green region (β–³PBC and β–³PDA )? Let side of square be a We consider each of the 4 triangles one by one Let 𝒉_𝟏 be height of βˆ† APB 𝒉_𝟐 be height of βˆ† DPC 𝒉_πŸ‘ be height of βˆ† APD 𝒉_πŸ’ be height of βˆ† BPC And since it is a square of side a 𝒉_𝟏+𝒉_𝟐=𝐡𝐢=𝒂 𝒉_πŸ‘+𝒉_πŸ’=𝐴𝐡=𝒂 For βˆ† APB Area βˆ† APB = 1/2 Γ— Base Γ— Height = 1/2 Γ— AB Γ— β„Ž_1 = 𝟏/𝟐 Γ— a Γ— 𝒉_𝟏 For βˆ† DPC Area βˆ† DPC = 1/2 Γ— Base Γ— Height = 1/2 Γ— CD Γ— β„Ž_2 = 𝟏/𝟐 Γ— a Γ— 𝒉_𝟐 For βˆ† APD Area βˆ† APD = 1/2 Γ— Base Γ— Height = 1/2 Γ— AD Γ— β„Ž_4 = 𝟏/𝟐 Γ— a Γ— 𝒉_πŸ’ For βˆ† BPC Area βˆ† BPC = 1/2 Γ— Base Γ— Height = 1/2 Γ— BC Γ— β„Ž_3 = 𝟏/𝟐 Γ— a Γ— 𝒉_πŸ‘ Now, Area of red portion = Area βˆ† APB + Area βˆ† DPC = 1/2 Γ— a Γ— β„Ž_1+1/2 Γ— a Γ— β„Ž_2 = 1/2 Γ— a Γ— (β„Ž_1+β„Ž_2) Since 𝒉_𝟏+𝒉_𝟐 = a = 1/2 Γ— a Γ— π‘Ž = 𝒂^𝟐/𝟐 And, Area of green portion = Area βˆ† APD + Area βˆ† BPC = 1/2 Γ— a Γ— β„Ž_4+1/2 Γ— a Γ— β„Ž_3 = 1/2 Γ— a Γ— (β„Ž_4+β„Ž_3) Since 𝒉_πŸ’+𝒉_πŸ‘ = a = 1/2 Γ— a Γ— π‘Ž = 𝒂^𝟐/𝟐 Thus, Area of red portion = Area of green portion So, their ratio is 1 : 1

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