In the figure (i) given below, DE || BC and the ratio of the areas of △ADE and trapezium DBCE is 4 : 5. Find the ratio of DE : BC.

Given, ratio of the areas of △ADE and trapezium DBCE = 4 : 5.

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of trapezium DBCE}} = \dfrac{4}{5} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ABC - Area of △ADE}} = \dfrac{4}{5} \\[1em] \Rightarrow 5 \text{ Area of △ADE} = 4 (\text{Area of △ABC - Area of △ADE}) \\[1em] \Rightarrow 9 \text{ Area of △ADE} = 4 \text{Area of △ABC} \\[1em] \Rightarrow \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{4}{9}.$

Considering △ABC and △ADE,

∠ A = ∠ A (Common angles)
∠ ADE = ∠ ABC (Corresponding angles are equal)

Hence, by AA axiom △ADE ~ △ABC.

We know that, the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides.

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{DE^2}{BC^2} \\[1em] \Rightarrow \dfrac{4}{9} = \Big(\dfrac{DE}{BC}\Big)^2 \\[1em] \Rightarrow \dfrac{DE}{BC} = \sqrt{\dfrac{4}{9}} \\[1em] \Rightarrow \dfrac{DE}{BC} = \dfrac{2}{3}.$

Hence, the ratio of DE : BC is 2 : 3.