From the question it is given that,
AB∥DC
AB=2DC,AD=3cm,BC=4cm
Now consider △EAB,
EA/DA=EB/CB=AB/DC=2DC/DC=2/1
(i) EA=2,DA=2×3=6cm
Then, ED=EA−DA
=6−3
=3cm
(ii) EB/CB=2/1
EB=2CB
EB=2×4
EB=8cm
(iii) Now, consider the △EAB,DC∥AB
So, △EDC∼△EAB
Therefore, area of △EDC/area of △ABE=DC2/AB2
area of △EDC/area of △ABE=DC2/(2DC)2
area of △EDC/area of △ABE=DC2/4DC2
area of △EDC/area of $$\triagle ABE = \dfrac {1}{4}$$
Therefore, area of ABE=4 area of △EDC
Then, area of △EDC+ area of trapezium ABCD=4 area of △EDC
Area of trapezium ABCD=3 area of △EDC
So, area of △EDC/area of trapezium ABCD=1/3
Therefore, area of △EDC: area of trapezium ABCD=1:3