ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are the mid-points of the non-parallel sides. The ratio of ar.(ABFE) and ar.(EFCD) is:

1. a : b

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

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

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

We know that,

The line segment connecting the midpoints of the non-parallel sides of a trapezium is parallel to the parallel sides and its length is half the sum of the lengths of the parallel sides.

AB || EF || DC

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

By formula,

Area of trapezium = $\dfrac{1}{2}$ × Sum of parallel sides × Distance between them

From figure,

E is the mid-point of AD, so AE = ED = x (let)

Area of trapezium ABFE = $\dfrac{1}{2}$ × (AB + EF) × AE …….(1)

Area of trapezium EFCD = $\dfrac{1}{2}$ × (EF + CD) × DE …….(2)

Dividing equation (1) from (2), we get :

$\Rightarrow \dfrac{\text{Area of trapezium ABFE}}{\text{Area of trapezium EFCD}} = \dfrac{\dfrac{1}{2} \times (AB + FE) \times AE}{\dfrac{1}{2} \times (EF + CD) \times DE} \\[1em] = \dfrac{\Big[a + \dfrac{1}{2}(a + b)\Big] \times x}{\Big[b + \dfrac{1}{2}(a + b)\Big] \times x} \\[1em] = \dfrac{\dfrac{2a + a + b}{2}}{\dfrac{2b + a + b}{2}} \\[1em] = \dfrac{3a + b}{3b + a}.$

Area of trapezium ABFE : Area of trapezium EFCD = (3a + b) : (3b + a).

Hence, option 2 is the correct option.