Home
Class 9
MATHS
In a triangle ABC, E is the mid - point...

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

Text Solution

Verified by Experts

The correct Answer is:
`triangle ABC`
Promotional Banner

Topper's Solved these Questions

  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    CPC CAMBRIDGE PUBLICATION|Exercise Exercise 11.4|8 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    CPC CAMBRIDGE PUBLICATION|Exercise EXERCISE 11.2|6 Videos
  • CIRCLES

    CPC CAMBRIDGE PUBLICATION|Exercise EXERCISE 12.6|4 Videos

Similar Questions

Explore conceptually related problems

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

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

D,E and F are respectively the mid-points of the sides BC, CA and AB of triangle ABC show that (i) BDEF is a parallelogram. (ii) ar (DEF) = 1/4 ar (ABC) (iii) ar (BDEF) = 1/2 ar (ABC)

If E,F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = (1)/(2) ar (ABCD)

In P is a point in the interior of a parallelogram ABCD. Show that (ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

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 a triangle ABC, AD is the median drawn on the side BC is produced to E such that AD = ED prove that ABEC is a parallelogram.

IN the figure, P is a point in the interior of a parallelogram ABCD. Show that (i) ar(APB) + ar(PCD) = 1/2 ar (ABCD) (ii) ar (APD) + ar(PBC) = ar(APB) + ar (PCD)