1. Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)
  2. Serial order wise
  3. Ex 9.2

Ex 9.2, 3 - Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)

Last updated at May 29, 2018 by

Ex 9.2, 3 - P and Q are any two points lying on sides DC - Paralleograms & triangles with same base & same parallel lines

Ex 9.2, 3 - Chapter 9 Class 9 Areas of Parallelograms and Triangles - Part 2

  1. Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)
  2. Serial order wise

    3. Ex 9.2

    Ex 9.3 →

Transcript

Ex 9.2, 3 P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar (APB) = ar (BQC). Given: A parallelogram ABCD where P & Q are any points on DC and AD resp. To prove: ar (APB) = ar (BQC) Proof: ABCD is a parallelogram So, AB CD & AD BC From (1) & (2) Area ( APB) = Area (BQC) Hence proved

Chapter 9 Class 9 - Areas of Parallelograms and Triangles (Deleted)
Serial order wise

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.