Answers

2015-03-21T16:34:40+05:30
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 !!
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2015-03-21T16:39:07+05:30
join db
let it intersect xy in e
by M.P.T,xe=1/2*50=25    and  ey=15
hence,xy=25+15=40
draw a perpendicular from the top vertex
height of abxy=that of dcxy
now,
ar.of a trapezium=1/2*h*(a+b)
thus,
ar.abyx=1/2*h*(50+40)
ar.dcxy=1/2*h*(30+40)
area of dcyx/ar.abyx= [1/2*h*70]/[1/2*h*90]=70/90=7/9

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