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

Ex 9.3, 1 - Chapter 9 Class 9 Areas of parallelograms and Triangles

Last updated at May 29, 2018 by

Ex 9.3, 1 - In fig, E is any point on median AD of - Ex 9.3

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  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, 1 In Fig, E is any point on median AD of a ABC. Show that ar (ABE) = ar (ACE). Given : ABC is a triangle with AD as medium i.e. BD = CD E is any point on AD. To Prove : ar ( ABE) = ar ( ACE) Proof: Join EB and EC Subtracting (1) and (2) ar ( ABD) ar ( EBD) = ar ( ADC) ar ( EDC) ar ( ABE) = ar ( ACE) Hence, Proved

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

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