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Question

ABCD is a parallelogram X and Y are the mid points of BC and CD respectively. Then, ar(parallelogram ABCD) is
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A
4×ar(AXY)
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B
2×ar(AXY)
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C
83×ar(AXY)
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D
None of these
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Solution

The correct option is C 83×ar(AXY)
X and Y are the mid-points of sides BC and CD.
In BCD,
XYBD and XY=12BD {From the mid point theorem}

ar(CYX)=14ar(DBC) {property of triangle having mid points}

ar(CYX)=18ar(gmABCD) [ Area of parallelogram is twice the area of triangle made by diagona ] --- ( 1 )

Since, parallelogram ABCD and ABX are between the same parallel lines AD and BC and BX=12BC.

ar(ABX)=14ar(gmABCD) ----- ( 2 )

Similarly, ar(AYD)=14ar(gmABCD) ----- ( 3 )

Now, ar(AXY)=ar(gmABCD)[ar(ABX)+ar(AYD)+ar(CYX)]

=ar(gnABCD)[14ar(gmABCD+14ar(gmABCD)+18ar(gmABCD)] [ From ( 1 ) ( 2 ) and ( 3 ) ]

=ar(gmABCD)58(gmABCD)

ar(AXY)=38ar(gmABCD)

ar(gmABCD)=83×ar(AXY)

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