In the given figure, AB and DE are perpendiculars to BC.

(i) Prove that △ABC ~ △DEC.

(ii) If AB = 6 cm, DE = 4 cm and AC = 15 cm, calculate CD.

(iii) Find the ratio of the area of △ABC : area of △DEC.

(i) Considering △DEC and △ABC,

∠ C = ∠ C (Common angles)
∠ ABC = ∠ DEC (Both angles are equal to 90°)

Hence, by AA axiom △DEC ~ △ABC.

(ii) Since △DEC ~ △ABC, so, ratio of their corresponding sides will be equal

$\therefore \dfrac{AB}{DE} = \dfrac{AC}{CD} \\[1em] \Rightarrow \dfrac{6}{4} = \dfrac{15}{CD} \\[1em] \Rightarrow CD = \dfrac{15 \times 4}{6} \\[1em] \Rightarrow CD = 10 \text{ cm}$

Hence, the length of CD = 10 cm.

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

$\therefore \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{AB^2}{DE^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{6^2}{4^2} \\[1em] \Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{36}{16} \\[1em] \Rightarrow \dfrac{\text{Area of △ABC}}{\text{Area of △DEC}} = \dfrac{9}{4} \\[1em].$

Hence, the ratio of the area of △ABC : area of △DEC = 9 : 4.