In the given figure, BC is parallel to DE. Area of triangle ABC = 25 cm², Area of trapezium BCED = 24 cm² and DE = 14 cm. Calculate the length of BC.

Also, find the area of triangle BCD.

Area of △ADE = Area of △ABC + Area of trapezium BCED = 25 + 24 = 49 cm².

Given,

BC || DE.

∠ABC = ∠ADE [Corresponding angles are equal]

∠ACB = ∠AED [Corresponding angles are equal]

∴ △ABC ~ △ADE [By AA]

We know that,

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

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

Let height of trapezium BCED be h cm.

Area = $\dfrac{1}{2}$ × (Sum of || sides) × h

⇒ 24 = $\dfrac{1}{2}$ × (BC + DE) × h

⇒ 24 × 2 = (BC + DE) × h

⇒ 48 = (10 + 14) × h

⇒ 24h = 48

⇒ h = 2 cm.

Area of △BCD = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × BC × h

= $\dfrac{1}{2}$ × 10 × 2

= 10 cm².

Hence, BC = 10 cm and area of △BCD = 10 cm².