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Ex 9.2, 4 - In figure, P is a point in the interior of - Paralleograms & triangles with same base & same parallel lines

Ex 9.2, 4 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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Ex 9.2, 4 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 3 Ex 9.2, 4 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 4 Ex 9.2, 4 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 5

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Ex 9.2, 4 In the given figure, P is a point in the interior of a parallelogram ABCD. Show that ar (APB) + ar (PCD) = 1/2 ar (ABCD) Given: A parallelogram ABCD To prove: ar (APB) + ar(PCD) = 1/2 ar (ABCD) Proof: Since ABCD is a parallelogram AB ∥ CD & AD ∥ BC We draw line EF passing through P, parallel to AB & DC i.e. EF ∥ AB ∥ CD Here, AE ∥ BF & AB ∥ EF ∴ EFBA is a parallelogram Similarly, EFCD is a parallelogram Now, ΔAPB and parallelogram EFBA are on the same base AB and between the same parallel lines AB and EF, ∴ Area (ΔAPB) = 1/2 Area (EFBA) Similarly, ΔPCD and parallelogram EFCD are on the same base CD and between the same parallel lines CD and EF, ∴ Area (ΔPCD) = 1/2 Area (EFCD) Adding (1) & (2), we obtain Area (∆APB) + Area (∆PCD) = 1/2 Area (EFAB) + 1/2 Area (EFCD) Area (ΔAPB) + Area (ΔPCD) = 1/2 Area (ABCD) Ex 9.2, 4 In the given figure, P is a point in the interior of a parallelogram ABCD. Show that (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD) Given: A parallelogram ABCD To prove: ar (APD) + ar(PBC) =ar (APB) + ar (PCD) Proof: Since ABCD is a parallelogram AB ∥ CD & AD ∥ BC We draw line GH passing through P, parallel to AD & BC i.e. GH ∥ AD ∥ BC Here, GH ∥ AD & AG ∥ DH ∴ AGHD is a parallelogram Similarly, GBCH is a parallelogram Now, ΔAPD and parallelogram AGHD are on the same base AD and between the same parallel lines AD and GH, ∴ Area (ΔAPD) = 1/2 Area (AGHD) Similarly, ΔPBC and parallelogram GBCH are on the same base BC and between the same parallel lines BC and GH, ∴ Area (ΔPCB) = 1/2 Area (GBCH) Adding (1) & (2), we obtain Area (∆APD) + Area (∆PBC) = 1/2 Area (AGHD) + 1/2 Area (GBCH) Area (∆APD) + Area (∆PBC) = 1/2 Area (ABCD) Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD) Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.