Question

there is a trap ABCD and E and F are mid points.AB//BC.AB=50,CD=30.prove ar(DEFC) = 7/9 ar (AEFB)

Answer 1

Let E be the mid point of AD and F be the mid point of CB.
Const : Join EF, AC  and construct a perpendicular from A on CD and mark the point on CD as G and the point on AG intersecting EF as H.

Now, In triangle ADC, let O be the mid point of AC
Thus, EO // CD and EO = 1/2 CD = 25
Similarly, OF = 1/2 AB =15
So, EF = OE + OF = 25 +15 = 40

In triangle ADG,
EH // CD and E 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(ABFE) = (1/2(40+30) x AH : ar(DEFC) = 1/2(40+50) x GH
as AH = GH,
= 1/2 x 70 = 1/2 x 90
= 70 : 90
= 7 : 9

Hope it helps !!

Answer 2

Hope it helps . Make sure to do the construction and prove AB half of the side