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ABCD is a parallelogram. X and Y are mid...

ABCD is a parallelogram. X and Y are mid-points of BC and CD. Prove that `ar(triangleAXY)=3/8ar(||^[gm] ABCD)`

Text Solution

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area (ABCD)=base*height=BC*h
Area (ABX)=`1/2*(BX)*h=(BC)/4*h`
Area(XCY)=`1/2*(BC)/2*h/2=((BC)*h)/8`
Area(AXY)=`(BC)*h-((BC)*h)/4-((BC)*h)/8-((BC)*h)/4`
`=3/8(arABCD)`.
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