1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
  3. Ex 9.3

Ex 9.3, 14 - Class 9

Last updated at Dec. 8, 2016 by

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

  1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
    3. Ex 9.3
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Ex 9.3, 14 In figure, AP ∥ BQ ∥ CR. Prove that ar(AQC) = ar(PBR). Given: AP ∥ BQ ∥ CR To prove: ar(AQC) = ar(PBR) Proof: Since ΔABQ and ΔPBQ lie on the same base BQ and are between parallel lines BQ & AP ∴ ar(ABQ) = ar(PBQ) Also, ΔCBQ and ΔRBQ 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

Chapter 9 Class 9 Areas of parallelograms and Triangles
Serial order wise

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