The problem describes a trapezium with an area of $15 \text{ cm}^2$. The lengths of the parallel sides are $(x-4)$ cm and $(x+2)$ cm, and the height is $x$ cm. The task is to form a quadratic equation based on the situation.

1. Problem Description

The problem describes a trapezium with an area of $15 \text{ cm}^2$. The lengths of the parallel sides are $(x-4)$ cm and $(x+2)$ cm, and the height is $x$ cm. The task is to form a quadratic equation based on the situation.

2. Solution Steps

The area of a trapezium is given by the formula:

$Area = \frac{1}{2} \times (sum \ of \ parallel \ sides) \times height$

In this case, the area is $15 \text{ cm}^2$, the parallel sides are $(x-4)$ cm and $(x+2)$ cm, and the height is $x$ cm. Substituting these values into the formula, we get:

$15 = \frac{1}{2} \times ((x-4) + (x+2)) \times x$

Simplifying the expression inside the parentheses:

$15 = \frac{1}{2} \times (2x - 2) \times x$

Multiplying both sides by 2:

$30 = (2x - 2) \times x$

Expanding the right side:

$30 = 2x^2 - 2x$

Rearranging the equation to form a quadratic equation in the standard form $ax^2 + bx + c = 0$:

$2x^2 - 2x - 30 = 0$

We can divide the equation by 2 to simplify it:

$x^2 - x - 15 = 0$

3. Final Answer

The quadratic equation that represents the situation is $x^2 - x - 15 = 0$.