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

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

Last updated at May 29, 2018 by

Subscribe to our Youtube Channel - https://www.youtube.com/channel/UCZBx269Tl5Os5NHlSbVX4Kg

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

  1. Chapter 9 Class 9 Areas of Parallelograms and Triangles
  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
Serial order wise

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.