1. **Stating the problem:** We have two similar triangles, ABC and ADE, with given side lengths: $AB=15$ cm, $BD=7.5$ cm, $BC=5$ cm, and $CF=4$ cm. We need to find the area of triangle ADE. 2. **Similarity and scale factor:** Since triangles ABC and ADE are similar, their corresponding sides are proportional. The scale factor from ABC to ADE is given by the ratio of corresponding sides. Given $AB=15$ cm and $AD=AB+BD=15+7.5=22.5$ cm, the scale factor is: $$\text{scale factor} = \frac{AD}{AB} = \frac{22.5}{15} = 1.5$$ 3. **Area ratio:** The ratio of areas of similar triangles is the square of the scale factor: $$\frac{\text{Area}_{ADE}}{\text{Area}_{ABC}} = (1.5)^2 = 2.25$$ 4. **Area of triangle ABC:** Using the formula for the area of a triangle with two sides and the included angle: $$\text{Area} = \frac{1}{2} ab \sin C$$ Here, $a=AB=15$ cm, $b=BC=5$ cm, and angle $C$ is $90^\circ$ (since $CF$ is perpendicular to $AE$), so $\sin 90^\circ = 1$. Calculate area of ABC: $$\text{Area}_{ABC} = \frac{1}{2} \times 15 \times 5 \times 1 = \frac{1}{2} \times 75 = 37.5 \text{ cm}^2$$ 5. **Calculate area of ADE:** Using the area ratio: $$\text{Area}_{ADE} = 2.25 \times 37.5 = 84.375 \text{ cm}^2$$ **Final answer:** The area of triangle ADE is $84.375$ cm$^2$.