by pravin3103

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Given that ABCD ia a //gm. X and Y are the mid points of BC and CD

Construction: Join BD

Since X and Y are the mid points of sides BC and CD respectively, therefore in triangle BCD, XY//BD and XY=1/2 BD

implies area of triangle CYX= 1/4 area of triangle DBC

{ In triangle BCD, if X is the mid point of BC and Y is the mid pt of CD then area triangle CYX=1/4 area triangle DBC}

IMPLIES AREA TRIANGLE CYX= 1/8 area //gm ABCD

[Area of//gm is twice the area of triangle made by the diagonal]

Since //gm ABCD and triangle ABX are between same // lines AB and BC and BX=1/2BC

Therefore, area triangle ABX= 1/4 area //gm ABCD

Similarly, area triangle AYD= 1/4 area //gm ABCD

Now, area triangle AXY= area//gm ABCD- {ar triangleABX + ar AYD + ar CYX}

= ar//gmABCD - {1/4 + 1/4 + 1/8} area //gmABCD

=area//gmABCD- 5/8 area //gmABCD

=3/8 area //gm ABCD.

. X and Y are the midpointsof BC and CD.TPT:

area(ΔAXY)=3/8 area(parallelogram ABCD)

Proof:

construction: join AC and BD.in ΔADC , AY is the median. since median divides the triangle into two equal areas.

area(ΔAYC)=1/2area(ΔADC)=1/4 area(parallelogram ABCD) .........(1

)since diagonals of a parallelogram divides it into two equal areas

similarly in ΔABC, area(ΔAXC)=1/2 area(ΔABC)=1/4 area(parallelogram ABCD) .............(2)from (1) and (2) area(quadrilateral AXCY) = area(ΔAYC)+area(ΔAXC)=(1/4+1/4) area(parallelogram ABCD)therefore area(quadrilateral AXCY)=1/2area(parallelogram ABCD) .........(3)area(ΔAXY)=area(quadrilateral AXCY)area(ΔXCY)[since area(ΔXCY)=1/4area(ΔBCD)= area(parallelogram ABCD)]area(ΔAXY)=1/2 area(parallelogram ABCD)1/8area(parallelogram ABCD)=3/8 area(parallelogram ABCD)