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

Ex 9.2, 3 - Class 9

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

  1. Chapter 9 Class 9 Areas of parallelograms and Triangles
  2. Serial order wise
    3. Ex 9.2
    Ex 9.3 →
Ask Download

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 (ΔAPB) = 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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.