Let X be the mid point of AD and Y be the mid point of CB.
Const : Join XY, AC and construct a perpendicular from A on CD and mark the point on CD as G and the point on AG intersecting XY as H.
Now, In triangle ADC, let O, the intersection point of AC and XY, be the mid point of AC
Thus, XO // CD and XO = 1/2 CD = 15 cm
Similarly, OY = 1/2 AB = 15 cm
So, EF = OE + OF = 25 +15 = 40 cm
In triangle ADG,
XH // CD and X is the mid point.
Therefore, H is also the mid point of AG (converse MPT)
AH = GH ................ (i)
Now we compare the area of the trapeziums
ar(DCYX) = (1/2(40+30) x AH : ar(XYBA) = 1/2(40+50) x GH
as AH = GH,
= 1/2 x 70 = 1/2 x 90
= 70 : 90
= 7 : 9
Hope it helps !!