In the adjoining figure, ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are mid-points of the non-parallel sides. The ratio of area of ABFE and area of EFCD is

1. a : b

2. (3a + b) : (a + 3b)

3. (a + 3b) : (3a + b)

4. (2a + b) : (3a + b)

It is given that

AB = a cm

DC = b cm

AB || DC

E and F are the mid-points of AD and BC

Consider h as the distance between AB, CD and EF

Now 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

and

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)

Adding equations (1) and (2)

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

EF = $\dfrac{1}{2}$ (AB + CD)

EF = $\dfrac{1}{2}$ (a + b)

Here,

Area of trapezium ABFE = $\dfrac{1}{2}$ [sum of parallel sides] × [distance between parallel sides]

Substituting the values,

$\text{Area of trap. ABFE} = \dfrac{1}{2} \Big[a + \dfrac{1}{2} (a + b)\Big] × h \\[1em] = \dfrac{1}{2}\Big[\dfrac{2a + a + b}{2}\Big]h \\[1em] = \dfrac{h}{4}[3a + b].$

Similarly,

$\text{Area of trap. EFCD } = \dfrac{1}{2}\Big[b + \dfrac{1}{2}(a + b)\Big] × h \\[1em] = \dfrac{1}{2}\Big[\dfrac{2b + a + b}{2}\Big]h \\[1em] = \dfrac{h}{4}[3b + a].$

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

By substituting the values,

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

Hence, Option 2 is the correct option.