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Ex 9.4, 6 (Optional) - Diagonals AC and BD of ABCD intersect each

Ex 9.4, 6 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2
Ex 9.4, 6 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3 Ex 9.4, 6 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 4 Ex 9.4, 6 (Optional) - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 5

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Transcript

Question 6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) × ar (CPD) = ar (APD) × ar (BPC). [Hint : From A and C, draw perpendiculars to BD.] Given: A quadrilateral ABCD To prove: A quadrilateral ABCD Construction: Draw AM ⊥ BD and CN ⊥ BD Proof: Finding Area of Δ APB and Δ APD Area of APB Base = BP Height = AM Area Δ APB = 1/2 × Base × Height ar (APB) = 1/2 × BP × AM Area of APD Base = DP Height = AM Area Δ APD = 1/2 × Base × Height ar (APD) = 1/2 × DP × AM Dividing by (1) by (2) (𝑎𝑟 (𝐴𝑃𝐵))/(𝑎𝑟 (𝐴𝑃𝐷)) = (1/2 × 𝐵𝑃 × 𝐴𝑀)/(1/2 × 𝐷𝑃 × 𝐴𝑀) (𝑎𝑟 (𝐴𝑃𝐵))/(𝑎𝑟 (𝐴𝑃𝐷)) = 𝐵𝑃/𝐷𝑃 …(3) Area of BPC Base = BP Height = CN Area Δ BPC = 1/2 × Base × Height ar (BPC) = 1/2 × BP × CN Area of CPD Base = DP Height = CN Area Δ CPD = 1/2 × Base × Height ar (CPD) = 1/2 × DP × CN Dividing by (4) by (5) (𝑎𝑟 (𝐵𝑃𝐶))/(𝑎𝑟 (𝐶𝑃𝐷)) = (1/2 × 𝐵𝑃 × 𝐶𝑁)/(1/2 × 𝐷𝑃 × 𝐶𝑁) (𝑎𝑟 (𝐵𝑃𝐶))/(𝑎𝑟 (𝐶𝑃𝐷)) = 𝐵𝑃/(𝐷𝑃 ) From (3) and (6) (𝑎𝑟 (𝐴𝑃𝐵))/(𝑎𝑟 (𝐴𝑃𝐷)) = (𝑎𝑟 (𝐵𝑃𝐶))/(𝑎𝑟 (𝐶𝑃𝐷)) ar (APB) × ar (CPD) = ar (APD) × ar (BPC) Hence Proved.

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.