In the figure (i) given below, DE || BC. If DE = 6 cm, BC = 9 cm and area of △ADE = 28 sq. cm, find the area of △ABC.

Considering △ADE and △ABC,

∠ 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.

Let the area of △ABC be x cm².

$\therefore \dfrac{\text{Area of △ADE}}{\text{Area of △ABC}} = \dfrac{DE^2}{BC^2} \\[1em] \Rightarrow \dfrac{28}{x} = \dfrac{6^2}{9^2} \\[1em] \Rightarrow \dfrac{28}{x} = \dfrac{36}{81} \\[1em] \Rightarrow x = \dfrac{28 \times 81}{36} \\[1em] \Rightarrow x = \dfrac{2268}{36} \\[1em] \Rightarrow x = 63.$

Hence, area of △ABC is 63 cm².