Home
Class 9
MATHS
In the following Figure, D and E are ...

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 now answer the equation that you left in the "Introduction " of the capther, whether the filed of Budhia has been actually divided into three parts of equal area ?

Text Solution

Verified by Experts

The correct Answer is:
3
Promotional Banner

Topper's Solved these Questions

  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    CPC CAMBRIDGE PUBLICATION|Exercise EXERCISE 11.3|15 Videos
  • CIRCLES

    CPC CAMBRIDGE PUBLICATION|Exercise EXERCISE 12.6|4 Videos

Similar Questions

Explore conceptually related problems

In, D and E are two points on BC such that BD = De = EC. Show that ar (ABD) = ar (ADE) = ar (AEC). Can you now 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?

In an isosceles triangle ABC with AB = AC, D and E are points on BC such that BE = CD (see figure) Show that AD = AE

D is a point on side BC ΔABC such that AD = AC (see figure). Show that AB > AD.

In Delta ABC , /_ B = /_C , D and E are the points on AB and AC such that BD = CE , prove that DE || BC .

In the figure 11.23, E is any point on median AD of a Delta ABC. Show that ar (ABE) = ar (ACE).

In the following figure, ABCD, DCFR and ABFE are parallelograms. Show that ar (ADE) = ar (BCF)

In a triangle ABC, E is the mid - point of median AD. Show that ar (BED) =1/4 ar (ABC)

In the adjoining figure, D and E are the mid-points of AB and AC respectively. If DE = 4 cm , then BC is equal to :

In the Given figure, E is any point on median AD of a triangle ABC Show that ar (ABE) = ar (ACE)

Diagonals AC and BD of a quadrilateral ABCD each other at P. Show that ar (APB) xx ar (CPD) =ar (APD) xx ar (BPC)