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ABCD is a trapezium in which AB parallel to DC, DC=30cm and AB=50cm, if X and Y are respectively the mid points of AD and BC, prove that ar[DCYX]= 7/9ar[XYBA].


Given: ABCD is a trapezium with AB || DC.
Construction: Join DY and produce it to meet AB produced at P.

In triangle BYP and triangle CYD
∠BYD = ∠CYD (vertically opposite angles)
∠DCY = ∠PBY (alternate opposite angles as DC || AP and BC is the transversal)

and BY = CY (Y is the mid point of BC)

Thus △ BYP ≅ △CYD (By ASA congruence criterion)

DY = YP and DC = BP
Y is the id point of AD

XY || AP and XY = 1/2 AP (mid point theorem)
XY = 1/2AP = 1/2(AB + BP) = 1/2 (AB + DC) = 1/2(50 + 30) = 1/2 x 80 = 40 cm

Since X and Y are mid points of AD and BC respectively.
trapezium DCYX and ABYX are of same height say h cm.

Now area of DCYX / Area of ABYX = 1/2 (DC + XY)x h / 1/2 ((AB x XY)H = 30+ 40 / 50 + 40 = 70 / 90 = 7/ 9
9 ar (DCXY) = 7 ar (XYBA)


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