The problem is to solve for $x$ given the area of a trapezium is 15 cm². The parallel sides of the trapezium are $(x-4)$ cm and $(x+2)$ cm, and the height is $x$ cm.

1. Problem Description

The problem is to solve for $x$ given the area of a trapezium is 15 cm². The parallel sides of the trapezium are $(x-4)$ cm and $(x+2)$ cm, and the height is $x$ cm.

2. Solution Steps

The area of a trapezium is given by the formula:

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

In this case, the area is 15 cm², the parallel sides are $(x-4)$ cm and $(x+2)$ cm, and the height is $x$ cm.

So, we can write the equation:

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

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

$15 = (x - 1) \times x$

$15 = x^2 - x$

$x^2 - x - 15 = 0$

Solving the quadratic equation $x^2 - x - 15 = 0$

We can use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=1$, $b=-1$, and $c=-15$.

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-15)}}{2(1)}$

$x = \frac{1 \pm \sqrt{1 + 60}}{2}$

$x = \frac{1 \pm \sqrt{61}}{2}$

$x = \frac{1 \pm 7.81}{2}$ (approximately)

So we have two possible values for $x$:

$x_1 = \frac{1 + 7.81}{2} = \frac{8.81}{2} = 4.405$

$x_2 = \frac{1 - 7.81}{2} = \frac{-6.81}{2} = -3.405$

Since $x$ represents the height, it must be positive. Also, $x-4$ must be positive since it is a side, so $x>4$. Therefore, $x=4.405$ is the plausible solution.

The sides of trapezium should be $x-4 > 0$ and $x+2>0$. So $x>4$.

$x = \frac{1 + \sqrt{61}}{2}$

3. Final Answer

$x = \frac{1 + \sqrt{61}}{2}$