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To solve the problem, we need to find the value of \(x\) for which the area of the trapezium \(ABCD\) is 15 cm\(^2\). The given dimensions are \(AB = (x + 3)\) cm, \(DC = (2x - 3)\) cm, and the height of the trapezium is the same as the height of the right triangle \(BEC\). 

First, let's analyze the right triangle \(BEC\):
- \(BE = EC\)
- \(BC = x + 3\) cm
- \(DC = (2x - 3)\) cm

Using the Pythagorean theorem in \(\triangle BEC\):
\[
BE^2 + EC^2 = BC^2
\]
Since \(BE = EC\), let's denote them both as \(y\):
\[
y^2 + y^2 = (x + 3)^2
\]
\[
2y^2 = (x + 3)^2
\]
\[
y^2 = \frac{(x + 3)^2}{2}
\]
\[
y = \sqrt{\frac{(x + 3)^2}{2}}
\]

Now, we can calculate the height \(h\) of the trapezium \(ABCD\) using \(y\), which is the same as \(BE\):
\[
h = \sqrt{\frac{(x + 3)^2}{2}}
\]

The area \(A\) of the trapezium \(ABCD\) is given by:
\[
A = \frac{1}{2} \times (AB + DC) \times h
\]
Given \(A = 15\) cm\(^2\), we have:
\[
15 = \frac{1}{2} \times ((x + 3) + (2x - 3)) \times \sqrt{\frac{(x + 3)^2}{2}}
\]
\[
15 = \frac{1}{2} \times (3x) \times \sqrt{\frac{(x + 3)^2}{2}}
\]
\[
15 = \frac{3x}{2} \times \sqrt{\frac{(x + 3)^2}{2}}
\]
\[
15 = \frac{3x}{2} \times \frac{x + 3}{\sqrt{2}}
\]
\[
15 = \frac{3x (x + 3)}{2\sqrt{2}}
\]
\[
30\sqrt{2} = 3x (x + 3)
\]
\[
30\sqrt{2} = 3x^2 + 9x
\]
\[
3x^2 + 9x - 30\sqrt{2} = 0
\]

This is a quadratic equation in \(x\):
\[
3x^2 + 9x - 30\sqrt{2} = 0
\]

Now, we solve the quadratic equation using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 3\), \(b = 9\), and \(c = -30\sqrt{2}\):
\[
x = \frac{-9 \pm \sqrt{9^2 - 4 \cdot 3 \cdot (-30\sqrt{2})}}{2 \cdot 3}
\]
\[
x = \frac{-9 \pm \sqrt{81 + 360\sqrt{2}}}{6}
\]
\[
x = \frac{-9 \pm \sqrt{81 + 360\sqrt{2}}}{6}
\]
Using \(\sqrt{2} \approx 1.41\), we get:
\[
x = \frac{-9 \pm \sqrt{81 + 360 \cdot 1.41}}{6}
\]
\[
x = \frac{-9 \pm \sqrt{81 + 508.8}}{6}
\]
\[
x = \frac{-9 \pm \sqrt{589.8}}{6}
\]
Using \(\sqrt{589.8} \approx 24.28\), we get:
\[
x = \frac{-9 \pm 24.28}{6}
\]

Thus, we have two solutions:
\[
x = \frac{24.28 - 9}{6} \approx \frac{15.28}{6} \approx 2.5
\]
\[
x = \frac{-9 - 24.28}{6} \approx \frac{-33.28}{6} \approx -5.5 \, (\text{not valid since } x > 0)
\]

Thus, the valid solution is \(x \approx 2.5\).

Finally, using \(x = 2.5\):
\[
DC = 2x - 3 = 2(2.5) - 3 = 5 - 3 = 2 \text{ cm}
\]

So, the length of \(DC\) is 2.0 cm.

If you have any questions or need further details, please let me know!

Here are some related questions:
1. How do you derive the formula for the area of a trapezium?
2. What is the Pythagorean theorem and how is it applied here?
3. How do you solve quadratic equations using the quadratic formula?
4. Why is it important to consider the validity of solutions in the context of the problem?
5. How can you approximate square roots without a calculator?
6. What are some common mistakes to avoid when solving quadratic equations?
7. How do you determine the height of a trapezium in general?
8. What is the significance of significant figures and decimal places in mathematical problems?

**Tip:** Always double-check your units and ensure consistency throughout your calculations to avoid errors.

Learn how to solve a problem involving the area of a trapezium using quadratic formulas and the Pythagorean theorem. Detailed step-by-step solution provided. Suitable for advanced high school students.

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