Home
Class 9
MATHS
In the figure, P is a point in the inter...

In the figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar`(triangleAPB)+ar(trianglePCD=1/2ar(ABCD)`
(ii) ar`(triangleAPD)+ar(trianglePBC)=ar(triangleAPB)+ar(trianglePCD)`

Promotional Banner

Topper's Solved these Questions

  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise NCERT SECTION ( EXERCISE 9.3)|19 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise NCERT SECTION ( EXERCISE 9.4)|19 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise NCERT SECTION ( EXERCISE 9.1)|1 Videos
  • CIRCLES

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

Similar Questions

Explore conceptually related problems

In the adjoining figure, P is the point in the interior of a parallelogram ABCD.Show that ar(DeltaAPB) ar(DeltaPCD) =1/2ar(||gm ABCD)

In Fig.9.16,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)

ABCD is a parallelogram and O is a point in its interior. Prove that (i) ar(triangleAOB)+ar(triangleCOD) =(1)/(2)ar("||gm ABCD"). (ii) ar(triangleAOB)+ar(triangleCOD)=ar(triangleAOD)+ar(triangleBOC) .

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

In the figure, E is any point on median AD of a triangleABC . Show that ar( triangleABE )= ar( triangleACE ).

In the adjacent figure P is a point inside the parallelogram ABCD. Prove that ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

In the adjacent figure P is a point inside the parallelogram ABCD. Prove that ar (APB) + ar (PCD) = 1/(2) ar (ABCD)

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

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) .