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Question

ABCD is a trapezium in which ABDC,AB=50 cm, DC=30 cm. If X and Y are respectively the mid-points of AD and BC, prove that
ar(trap.DCYX)=79ar(trap.XYBA).

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Solution


ABCD is a trapezium.
Join DY and produce it to meet AB produced at P.
In BYP and CYD
BYD=CYD [ Vertically opposite angles ]
DCAP and BC is the transversal.
DCY=PBY [ Alternate angles ]
BYPCYD [ By ASA congruence rule ]
DY=YP and DC=BP [ CPCT ]
Y is the mid point of AD
XYAP and XY=12×AP [ Mid -point theorem ]

XY=12×(AB+BP)

XY=12×(AB+DC)

XY=12×(50+30)

XY=12×80

XY=40cm
Since X and Y are the mid points of AD and BC respectively .
Trapezium DCYX and XYBA are of same height, say hcm
Now,
ar(trap.DCYX)ar(trap.XYBA)=12(DC+XY)×h12(AB+XY)×h=30+4050+40=7090=79

9ar(trap.DCXY)=7ar(trap.XYBA)

ar(trap.DCYX)=79ar(trap.XYBA)


1264009_693954_ans_618453e205d74c879d30ab461a6b5906.png

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