Home
Class 9
MATHS
In the adjoining figure, MNPQ and ABPQ a...

In the adjoining figure, MNPQ and ABPQ are parallelogram and T is any point on the side BP. Prove that
(i) `ar(MNPQ)=ar(ABPQ)`
(ii)` ar(triangleATQ)=(1)/(2)ar(MNPQ)`.

Promotional Banner

Topper's Solved these Questions

  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise NCERT SECTION ( EXERCISE 9.1)|1 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise NCERT SECTION ( EXERCISE 9.2)|7 Videos
  • AREAS OF PARALLELOGRAMS AND TRIANGLES

    MTG IIT JEE FOUNDATION|Exercise Olympiad/HOTS Corner |14 Videos
  • CIRCLES

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

Similar Questions

Explore conceptually related problems

In the figure, PQRS and ABRS are parallelograms and X is any point on side BR. Show that ar(PQRS) = ar(ABRS)

In Fig.9.17, PQRS and ABRS are parallelograms and X is any point on side BR. Show that (i) ar(PQRS)=ar(ABRS) (ii) ar (AXS)backslash=(1)/(2)ar(PQRS)

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) .

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

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) .

In the given figure LMNO and PMNQ are two parallelograms. R is any point on side MP. If ar( Delta NRQ) = k[ar( I I^(gm) LMNO)] then 2k equals

In the figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then, ar (DeltaDPC) = (1)/(2) ar (EFGD) .

In the adjoing figure, two parallelograms ABCD and AEFB are drawn on opposite sides of AB. Prove that ar("||gm ABCD") +ar("||gm AEFB") =ar("||gm EFCD") .