In the adjoining figure, $$ABC$$ is a right-angled triangle, $$\angle ABC = 90^\circ$$, $$BC = 12\,\text{cm}$$, area of $$\triangle ABC = 30\,\text{cm}^2$$, area of square $$ADEF = 9\,\text{cm}^2$$. Then area of $$ADEC$$ is ______ $$\text{cm}^2$$.

From $$\frac{1}{2} \times AB \times 12 = 30$$ we get $$AB = 5$$, so $$AC = \sqrt{5^2 + 12^2} = 13$$. The square has side 3, so $$AD = 3$$ along $$AC$$, leaving $$DC = 13 - 3 = 10$$, and $$E$$ lies at a distance 3 from the line $$AC$$ because $$DE = 3$$ and $$DE$$ is perpendicular to $$AD$$. As $$A$$, $$D$$, $$C$$ are collinear, the quadrilateral $$ADEC$$ has the same area as $$\triangle DEC$$, which is $$\frac{1}{2} \times 10 \times 3 = 15$$.