**Download NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3 – Areas of Parallelograms And Triangles. This Exercise contains 16 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 10 or other Chapters, you can click the link at the end of this Note.**

### NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3

**NCERT Solutions for Class 9 Maths Chapter 9 Exercise 9.3**

**1. In given figure, E is any point on median AD of a ∆ ABC. Show that **

**ar (ABE) = ar (ACE).**

**Solution :**

Given that, AD is the median of ∆ ABC, BD = DC

=> ar (∆ ABD ) = ar (∆ ADC )

=> ar (∆ ABE ) + ar (∆ EBD ) = ar (∆ ACE ) + ar (∆ ECD ) ….(1)

It can be clearly observed that, in the ∆ EBC, ED is the median.

=> ar ( ∆ EBD ) = ar ( ∆ ECD ) ….(2)

On subtracting equation 2 from equation 1, we get

ar ( ∆ ABE ) = ar ( ∆ ACE )

**Hence Proved.**

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