In the given figure, DE || BC. If DE = 4 cm, BC = 6 cm and ar(ΔADE) = 20 cm², find the area of ΔABC.

Considering ΔADE and ΔABC,

∠A = ∠A [Common angles]

∠ADE = ∠ABC [Corresponding angles are equal]

∴ ΔADE ∼ ΔABC (By A.A. axiom)

We know that,

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

Let the area of ΔABC be x cm².

$\therefore \dfrac{\text{Area of ΔADE}}{\text{Area of ΔABC}} = \Big(\dfrac{{DE}}{{BC}}\Big)^2 \\[1em] \Rightarrow \dfrac{\text{Area of ΔADE}}{\text{Area of ΔABC}} = \dfrac{{DE}^2}{{BC}^2} \\[1em] \Rightarrow \dfrac{20}{x} = \dfrac{4^2}{6^2} \\[1em] \Rightarrow \dfrac{20}{x} = \dfrac{16}{36} \\[1em] \Rightarrow x = \dfrac{20 \times 36}{16} \\[1em] \Rightarrow x = \dfrac{720}{16} \\[1em] \Rightarrow x = 45 \text{ cm}^2$

Hence, area of ΔABC = 45 cm².