In triangle ABC, the point D lies on AB and E lies on AC, with DE ∥ BC…

Step 1. Similar triangles. $\angle A$ is common, and $\angle ADE = \angle B$ because $DE \parallel BC$. By AA, $\triangle ADE \sim \triangle ABC$.

Step 2. The ratio. $AB = 4 + 6 = 10$ cm, so $\dfrac{AD}{AB} = \dfrac{4}{10} = \dfrac{2}{5}$.

Step 3. The area of ADE. $\dfrac{\text{ar}(ADE)}{\text{ar}(ABC)} = \dfrac{4}{25}$, so $\text{ar}(ADE) = 150 \times \dfrac{4}{25} = 24$.

Step 4. The trapezium. $\text{ar}(DBCE) = 150 - 24 = 126$.