Home
Class 9
MATHS
In the given figure, ABCD is a parallelo...

In the given figure, ABCD is a parallelogram. E and F are any two points on AB and BC, respectively. Prove that `ar (triangleADF)=ar(triangleDCE)`.

Promotional Banner

Topper's Solved these Questions

  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise EXERCISE ( Subjective Problems ) (Long Answer Type )|3 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise EXERCISE ( Integer/Numerical Value Type )|5 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise EXERCISE ( Subjective Problems ) (Very Short Answer Type )|9 Videos
  • CIRCLES

    MTG IIT JEE FOUNDATION|Exercise OLYMPIAD/HOTS CORNER |6 Videos

Similar Questions

Explore conceptually related problems

In the given figure ABCD is a parallelogram. Prove that AB = 2BC.

In the adjoining figure, ABCD is a parallelogram and P is any points on BC. Prove that ar(triangleABP)+ar(triangleDPC)=ar(trianglePDA) .

ABCD is a parallelogram. X and Y are mid-points of BC and CD. Prove that ar(triangleAXY)=3/8ar(||^[gm] ABCD)

In the given figure ABCD is a parallelogram in which AB||CD and E and F are the mid points of AB and BC respectively. What is the area of angleBEF ?

In the adjoining figure, ABCD is a parallelogram. Points P and Q on BC trisect BC. Prove that ar(triangleAPQ)=ar(triangleDPQ)=(1)/(6)ar(triangleABCD) .

In the given figure, ABCD and AEFD a re two parallelograms. ar( Delta PEA) =

In the given figure, ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.

In the given figure ABCD is a trapezium with AB || DC. E and F are respectively midpoint of AD and BC. Prove that, EF=1/(2)(AB+CD)

In the given figure PSDA is a parallelogram in which PQ =QR= RS and AP II BQ II CR II VS.Prove that ar( trianglePQE ) = ar( triangleDCF ).

P is any point on the diagonal BD of the parallelogram ABCD. Prove that ar (triangleAPD) = ar (triangleCPD) .