EF : FD = 2 : 3

ED = EF + FD

So EF : ED = 2 : (2 + 3) = 2 : 5

Area of triangle EFG: Area of triangle EDC
= 2² : 5²
= 4 : 25


Area of triangle EFG : Area of trapezium CDFG

= Area of triangle EFG : (Area of triangle EDC - area of triangle EFG)

= 4 : (25 - 4)
= 4 : 21


So area of triangle EFG/area of trapezium CDFG = 4/21




ii)

Since EF:ED = 2 : 5,

Then FG : DC also = 2 : 5

So FG/DC = 2/5
FG = 2/5 DC


let height of trapezium be h. Notice the triangle AFD and trapezium ABCD have the same height.



Area of AFD / Area of trapezium ABCD

= (½ x h x AF)/(½ x h x (AB + DC) )

= (½DC)/(AF + FG + BG + DC)

[ The ½h cancels out. And AB = AF + FG + BG]

= ½DC / (½DC + 2/5DC + ½DC + DC)
[AF = BG = ½DC]

= ½DC/ (12/5 DC)

= ½ / (12/5)
= ½ x 5/12
= 5/24