CBSE Class 9CBSE
Account
It's free!

User


Login
Register


      Forgot password?
Share
Notifications

View all notifications
Books Shortlist
Your shortlist is empty

Solution - In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). - CBSE Class 9 - Mathematics

ConceptTriangles on the Same Base and Between the Same Parallels

Question

In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).

Can you answer the question that you have left in the ’Introduction’ of this chapter, whether the field of Budhia has been actually divided into three parts of equal area?

[Remark: Note that by taking BD = DE = EC, the triangle ABC is divided into three triangles ABD, ADE and AEC of equal areas. In the same way, by dividing BC into n equal parts and joining the points of division so obtained to the opposite vertex of BC, you can divide ΔABC into n triangles of equal areas.]

Solution

You need to to view the solution
Is there an error in this question or solution?

APPEARS IN

R.D. Sharma Mathematics for Class 9 by R D Sharma (2018-19 Session)
Chapter 15: Areas of Parallelograms and Triangles
Q: 2 | Page no. 164

Reference Material

Solution for question: In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). concept: Triangles on the Same Base and Between the Same Parallels. For the course CBSE
S