To solve the problem, we need to find the length of $DC$ in the trapezium $ABCD$, given that $AB = (x + 3)$ cm, $DC = (2x - 3)$ cm, and $BE = EC$. We also know the area of the trapezium is 15 cm².

The formula for the area of a trapezium is: $\text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height}$

In this case, the parallel sides are $AB$ and $DC$, and the height is $AD$. First, we need to find the height.

Given: $AB = x + 3 \text{ cm}$ $DC = 2x - 3 \text{ cm}$

The height $AD$ can be found using the right triangle $BEC$: $BE = EC = \frac{DC - AB}{2} = \frac{(2x - 3) - (x + 3)}{2} = \frac{2x - 3 - x - 3}{2} = \frac{x - 6}{2}$

Using the Pythagorean theorem in triangle $BEC$: $AD^2 + BE^2 = AB^2$

$AD^2 + \left( \frac{x - 6}{2} \right)^2 = (x + 3)^2$

$AD^2 + \frac{(x - 6)^2}{4} = (x + 3)^2$

Simplify and solve for $AD$: $AD^2 + \frac{x^2 - 12x + 36}{4} = x^2 + 6x + 9$

Multiply through by 4 to clear the fraction: $4AD^2 + x^2 - 12x + 36 = 4x^2 + 24x + 36$

Simplify: $4AD^2 + x^2 - 12x + 36 - 4x^2 - 24x - 36 = 0$

$4AD^2 - 3x^2 - 36x = 0$

[ 4AD^2 = 3