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Important questions on Parallelograms and Triangles
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Important questions on Parallelograms and Triangles
Last updated at May 29, 2023 by Teachoo
Question 6 Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB) Γ ar (CPD) = ar (APD) Γ ar (BPC). [Hint : From A and C, draw perpendiculars to BD.] Given: A quadrilateral ABCD To prove: A quadrilateral ABCD Construction: Draw AM β₯ BD and CN β₯ BD Proof: Finding Area of Ξ APB and Ξ APD Area of APB Base = BP Height = AM Area Ξ APB = 1/2 Γ Base Γ Height ar (APB) = 1/2 Γ BP Γ AM Area of APD Base = DP Height = AM Area Ξ APD = 1/2 Γ Base Γ Height ar (APD) = 1/2 Γ DP Γ AM Dividing by (1) by (2) (ππ (π΄ππ΅))/(ππ (π΄ππ·)) = (1/2 Γ π΅π Γ π΄π)/(1/2 Γ π·π Γ π΄π) (ππ (π΄ππ΅))/(ππ (π΄ππ·)) = π΅π/π·π β¦(3) Area of BPC Base = BP Height = CN Area Ξ BPC = 1/2 Γ Base Γ Height ar (BPC) = 1/2 Γ BP Γ CN Area of CPD Base = DP Height = CN Area Ξ CPD = 1/2 Γ Base Γ Height ar (CPD) = 1/2 Γ DP Γ CN Dividing by (4) by (5) (ππ (π΅ππΆ))/(ππ (πΆππ·)) = (1/2 Γ π΅π Γ πΆπ)/(1/2 Γ π·π Γ πΆπ) (ππ (π΅ππΆ))/(ππ (πΆππ·)) = π΅π/(π·π ) From (3) and (6) (ππ (π΄ππ΅))/(ππ (π΄ππ·)) = (ππ (π΅ππΆ))/(ππ (πΆππ·)) ar (APB) Γ ar (CPD) = ar (APD) Γ ar (BPC) Hence Proved.