In the adjoining figure, DE || BC and all measurements are in centimeters. The length of AE is

1. 2 cm

2. 2.25 cm

3. 3.5 cm

4. 4 cm

Given DE || BC.

Considering △ABC and △ADE,

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

Hence, by AA axiom △ABC ~ △ADE.

Let the length of AE be x cm.

Since, both the triangles are similar,

$\therefore \dfrac{AD}{AB} = \dfrac{AE}{AC} \\[1em] \Rightarrow \dfrac{AD}{AD + DB} = \dfrac{AE}{AE + EC} \\[1em] \Rightarrow \dfrac{3}{3 + 4} = \dfrac{x}{x + 3} \\[1em] \Rightarrow \dfrac{3}{7} = \dfrac{x}{x + 3} \\[1em] \Rightarrow 3(x + 3) = 7x \\[1em] \Rightarrow 3x + 9 = 7x \\[1em] \Rightarrow 4x = 9 \\[1em] \Rightarrow x = 2.25$

∴ Length of AE = 2.25 cm.

Hence, Option 2 is the correct option.