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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

Let us draw a line segment AM ⊥ BC.

We know that,

Area of a triangle = 1/2 × Base × Altitude

`"Area "(triangleADE)=1/2xxDExxAM`

`"Area "(triangleABD)=1/2xxBDxxAM`

`"Area "(triangleAEC)=1/2xxECxxAM`

It is given that DE = BD = EC

`⇒ 1/2xxDExxAM=1/2xxBDxxAM=1/2xxECxxAM`

⇒ Area (ΔADE) = Area (ΔABD) = Area (ΔAEC)

It can be observed that Budhia has divided her field into 3 equal parts.

Is there an error in this question or solution?

APPEARS IN

 NCERT Mathematics Textbook for Class 9 (with solutions)
Chapter 9: 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: null - Triangles on the Same Base and Between the Same Parallels. For the course CBSE
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