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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

Learn how to find the height of trapezium ABCD using the area and given side lengths. This problem involves applying the Pythagorean theorem and the formula for the area of a trapezium. Suitable for high school students studying geometry.

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