In the given figure, DE ∥ BC and DE : BC = 3 : 5. alculate ar(ΔADE) : ar(trap. BCED).

It is given that DE ∥ BC.

∠ADE = ∠ABC [Corresponding angles are equal]

∠AED = ∠ACB [Corresponding angles are equal]

∴ ΔADE ∼ ΔАВС (By A.A. axiom)

We know that,

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\dfrac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} = \dfrac{(BC)^2}{(DE)^2}$

Subtracting 1 from both sides, we get:

$\Rightarrow \dfrac{\text{ar(ΔABC)}}{\text{ar(ΔADE)}} - 1 = \dfrac{(5)^2}{(3)^2} - 1 \\[1em] \Rightarrow \dfrac{\text{ar(ΔABC)} - \text{ar(ΔADE)}}{\text{ar(ΔADE)}} = \dfrac{25}{9} - 1 \\[1em] \Rightarrow \dfrac{\text{ar(BCED)}}{\text{ar(ΔADE)}} = \dfrac{25 - 9}{9} \\[1em] \Rightarrow \dfrac{\text{ar(BCED)}}{\text{ar(ΔADE)}} = \dfrac{16}{9} \\[1em] \Rightarrow \dfrac{\text{ar(ΔADE)}}{\text{ar(trap. BCED)}} = \dfrac{9}{16}.$

Hence, ar(ΔADE) : ar(trap. BCED) = 9 : 16.