# Thread: Express in terms of X

1. ## Express in terms of X

A piece of wire x m long is cut into two pieces, the length of the first piece being 14 m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.

Thank you!

2. ## Word Problem

Hello,

Have a look here for a similar starting : http://www.mathhelpforum.com/math-he...entiation.html, which should give you ideas as to how you can express this in terms of x.

3. Hello, gearshifter!

A tricky one . . . it requires baby-talking our way through it.

A piece of wire $x$ m long is cut into two pieces.
The length of the first piece is 14 m and is bent into a circle.
The other is bent into a rectangle with length twice the width.
Give an expression for the total area $A$ enclosed in the two shapes in terms of $x$.
Code:
      :    14     P   x-14    :
* - - - - - * - - - - - *
A         x             B
The wire is: . $AB \:=\:x$

It is cut at point $P\!:\;\;AP = 14,\;\;PB \:=\:x-14$

Length $AP = 14$ is bent into a circle.
The circumference of a circle is: . $C \:=\:2\pi r$
Since the circumference is 14, we have: . $2\pi r \:=\:14 \quad\Rightarrow\quad r \:=\:\tfrac{7}{\pi}$
The area of a circle is: . $A \:=\:\pi r^2$
. . So we have: . $A_c \:=\:\pi\left(\tfrac{7}{\pi}\right)^2 \quad\Rightarrow\quad\boxed{ A_c\:=\:\frac{49}{\pi}}$

The other piece, $PB = x-14$, is bent into a rectangle.
. . Its length is twice its width.
Code:
            2W
* - - - - - - *
|             |
W |             | W
|             |
* - - - - - - *
2W
Its perimeter is $x-14.$

We have: . $6W \:=\:x-14 \quad\Rightarrow\quad W \:=\:\frac{x-14}{6}$

And the length is: . $L \:=\:2W \:=\:\frac{x-14}{3}$

The area of the rectangle is: . $A \:=\:L\cdot W\quad\Rightarrow\quad\boxed{ A_r \:=\:\frac{(x-14)^2}{18}}$

Therefore, the total area is: . $\boxed{{\color{blue}A \;=\;\frac{49}{\pi} + \frac{(x-14)^2}{18}}}$ square meters.