Question 14 - Area of Triangles - Areas of Parallelograms and Triangles

Last updated at Dec. 13, 2024 by

Ex 9.3, 14 - In figure, AP || BQ || CR. Prove ar(AQC) - Triangles with same base & same parallel lines

Ex 9.3, 14 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

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Question 14 In figure, AP BQ CR. Prove that ar(AQC) = ar(PBR). Given: AP BQ CR To prove: ar(AQC) = ar(PBR) Proof: Since ABQ and PBQ lie on the same base BQ and are between parallel lines BQ & AP ar(ABQ) = ar(PBQ) Also, CBQ and RBQ lie on the same base BQ and are between parallel lines BQ & CR ar(CBQ) = ar(RBQ) Adding (1) & (2) ar(AQB) + ar(BQC) = ar(PBQ) + ar(QBR) ar(AQC) = ar(PBR). 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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo