Home
Class 9
MATHS
If E, F G and H are respectively the mi...

If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH) `=1/2` ar (ABCD) .

Answer

Step by step text solution for If E, F G and H are respectively the midpoints of the sides AB, BC, CD and AD of a parallelogram ABCD, show that ar(EFGH) =1/2 ar (ABCD) . by MATHS experts to help you in doubts & scoring excellent marks in Class 9 exams.

Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

Similar Questions

Explore conceptually related problems

E and F are respectively the mid-points of equal sides AB and AC of DeltaABC (see figure) Show that BF = CE.

P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD show that ar(DeltaAPB) = ar Delta(BQC)

Knowledge Check

  • In DeltaABC the midpoints are D,E and F of the sides AB,BC and CA, then DeltaDEF:DeltaABC is

    A
    `1:1`
    B
    `1:3`
    C
    `1:2`
    D
    `1:4`
  • Similar Questions

    Explore conceptually related problems

    If M and N are the mid - points of the sides BC and CD respectively of a parallelogram ABCD, then AM + AN equals

    ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH . Prove that EFGH is a square.

    In the figure, ΔABC, D, E, F are the midpoints of sides BC, CA and AB respectively. Show that (i) BDEF is a parallelogram (ii) ar(DeltaDEF)=1/4ar(DeltaABC) (iii) ar(BDEF)=1/2ar(DeltaABC)

    ABCD is quadrilateral E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.

    In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB .

    In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF=2AB .

    In the figure D, E are points on the sides AB and AC respectively of DeltaABC such that ar(DeltaDBC) = ar(DeltaEBC) . Prove that DE || BC.