In the given figure, BC ∥ DE, area (ΔABC) = 25 cm², area (trap. BCED) = 24 cm² and DE = 14 cm. Calculate the length of BC.

Area of ΔADE = Area of ΔABC + Area of trapezium BCED = 25 + 24 = 49 cm².

Given,

BC ∥ DE.

In ΔABC and ΔADE,

∠ABC = ∠ADE [Corresponding angles are equal]

∠ACB = ∠AED [Corresponding angles are equal]

∴ ΔABC ∼ ΔADE by (By A.A. axiom)

We know that,

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

$\therefore \dfrac{\text{Area of ΔABC}}{\text{Area of ΔADE}} = \dfrac{BC^2}{DE^2} \\[1em] \Rightarrow \dfrac{25}{49} = \dfrac{BC^2}{14^2} \\[1em] \Rightarrow BC^2 = \dfrac{25}{49} \times 196 \\[1em] \Rightarrow BC^2 = 100 \\[1em] \Rightarrow BC = 10 \text{ cm.}$

Hence, BC = 10 cm.