Hello, hana_102!

We just have to baby-talk our way through it . . .

A piece of wire which is 18 meters long is cut into two parts.

The first part is bent to form the four sides of a square.

The second part is bent to form the fours sides of a rectangle.

The breadth of the rectangle is one meter. And its length is x meters.

Show that the total area of the square and rectangle is:

. . $\displaystyle A \;= \;16-3x + \frac{x^2}{4} $ Code:

x
* - - - - - *
| |
1 | | 1
| |
* - - - - - *
x

The rectangle has an area of: .$\displaystyle x\cdot1 \:=\:\boxed{x\text{ m}^2}$

The rectangle has a perimeter of $\displaystyle 2x + 2$ meters.

This leaves: .$\displaystyle 18 - (2x+2) \:=\:16-2x$ meters for the square.

The side of the square is: .$\displaystyle \frac{16-2x}{4} \:=\:\frac{8-x}{2}$ meters.

The area of the square is: .$\displaystyle \left(\frac{8-x}{2}\right)^2 \;=\;\frac{64-16x + x^2}{4} \;=\;\boxed{16 - 4x + \frac{x^2}{4}\text{ m}^2}$

Therefore, the total area is: .$\displaystyle A \;=\;x + \left(16 - 4x + \frac{x^2}{4}\right) \;=\;{\color{blue}16 - 3x + \frac{x^2}{4}}$