In the figure, ABCD is a trapezium with parallel sides AB = x and CD = y. E and F are mid-points of the non-parallel sides AD and BC respectively. The ratio of ar (ABFE) and ar (EFCD) is :

1. x : y

2. (3x + y) : (x + 3y)

3. (x + 3y) : (3x + y)

4. (2x + y) : (3x + y)

Join BD which intersects EF at M.

In ∆ABD,

E is the midpoint of AD and EM || AB

By midpoint theorem,

M is the midpoint of BD

EM = $\dfrac{1}{2} AB$ ….(1)

In ∆CBD,

F is mid-point of BC and M is mid-point of BD so by mid-point theorem,

MF = $\dfrac{1}{2} CD$ ….(2)

So EF ∥ AB and EF ∥ CD

That means:

AB ∥ EF ∥ CD

Adding equations (1) and (2), we get:

EM + MF = $\dfrac{1}{2} AB + \dfrac{1}{2} CD = \dfrac{x + y}{2}$

Since:

AB ∥ EF ∥ CD

Let total height between AB and CD = H

EF lies exactly halfway between them,

∴ Height of trapezium EFCD = Height of trapezium ABEF = $\dfrac{H}{2}$ = h (let)

Area of trapezium = $\dfrac{1}{2} \times (\text{sum of parallel sides}) \times h$

Area of trapezium ABFE

$= \dfrac{1}{2} \times \Big(x + \dfrac{x + y}{2}\Big) \times h \\[1em] = \dfrac{1}{2} \times \Big(\dfrac{3x + y}{2}\Big) h \\[1em] = \dfrac{h}{4} \times (3x + y).$

Area of trapezium EFCD

$= \dfrac{1}{2} \times \Big(y + \dfrac{x + y}{2}\Big) \times h \\[1em] = \dfrac{1}{2} \times \Big(\dfrac{x + 3y}{2}\Big) h \\[1em] = \dfrac{h}{4} \times (x + 3y).$

Required ratio = Area of trapezium ABFE / Area of trapezium EFCD

By substituting the values,

$\text{Ratio} = \dfrac{\dfrac{h}{4}\times (3x + y)}{\dfrac{h}{4} \times (x + 3y)} \\[1em] \text{Ratio} = \dfrac{(3x + y)}{(x + 3y)} \\[1em] \text{Ratio} = (3x + y) : (x + 3y).$

The ratio of ar (ABFE) and ar (EFCD) is (3x + y) : (x + 3y).

Hence, option 2 is the correct option.