# Algebra help. (word problem)

• September 13th 2008, 02:51 PM
hana_102
Algebra help. (word problem)
I am really having difficulty with this question, I just dont seem to get 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. If the sum of the areas of the square and rectangle is A square meters show that;

$A = 16-3x + \frac{x^2}{4}$

I don't really know how to get to this formula, it does not say that the the wire is cut into 2 equal parts, so i can't assume that can i ?

I understand were the number 16 comes from and that's about it.
• September 13th 2008, 03:05 PM
Shyam
Quote:

Originally Posted by hana_102
I am really having difficulty with this question, I just dont seem to get 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. If the sum of the areas of the square and rectangle is A square meters show that;

$A = 16-3x + \frac{x^2}{4}$

I don't really know how to get to this formula, it does not say that the the wire is cut into 2 equal parts, so i can't assume that can i ?

I understand were the number 16 comes from and that's about it.

${\text{Let side of square be }} = a \hfill \\$

${\text{Perimeter of square + perimeter of rectangle = 18}} \hfill \\$

$4a + 2\left( {1 + x} \right) = 18 \hfill \\$

$4a + 2 + 2x = 18 \hfill \\$

$4a = 18 - 2 - 2x \hfill \\$

$a = \frac{{16 - 2x}}
{4} \hfill \\$

$a = \frac{{8 - x}}
{2}...............................{\text{eqn(1)}} \hfill \\$

${\text{A = area of square + area of rectangle}} \hfill \\$

$A = \left( {a^2 } \right) + \left( {1 \times x} \right) \hfill \\$

$A = a^2 + x \hfill \\$

${\text{Put the value of }}a{\text{ from eqn (1)}} \hfill \\$

$A = \left( {\frac{{8 - x}}
{2}} \right)^2 + x \hfill \\$

$A = \frac{{64 - 16x + x^2 }}
{4} + x \hfill \\$

$A = 16 - 4x + \frac{{x^2 }}
{4} + x \hfill \\$

$A = 16 - 3x + \frac{{x^2 }}
{4} \hfill \\
$

• September 13th 2008, 03:19 PM
hana_102
Thank you I understand it now, that's a very good explanation, but I would have never thought of using the perimeters to work it out in the first place.
(Worried)
• September 13th 2008, 03:35 PM
Soroban
Hello, hana_102!

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

Quote:

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

Code:

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

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

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

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

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

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

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