Hello, gearshifter!

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

A piece of wire $\displaystyle x$ 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 $\displaystyle A$ enclosed in the two shapes in terms of $\displaystyle x$. Code:

: 14 P x-14 :
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A x B

The wire is: .$\displaystyle AB \:=\:x$

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

Length $\displaystyle AP = 14$ is bent into a circle.

The circumference of a circle is: .$\displaystyle C \:=\:2\pi r$

Since the circumference is 14, we have: .$\displaystyle 2\pi r \:=\:14 \quad\Rightarrow\quad r \:=\:\tfrac{7}{\pi}$

The area of a circle is: .$\displaystyle A \:=\:\pi r^2$

. . So we have: .$\displaystyle A_c \:=\:\pi\left(\tfrac{7}{\pi}\right)^2 \quad\Rightarrow\quad\boxed{ A_c\:=\:\frac{49}{\pi}}$

The other piece, $\displaystyle PB = x-14$, is bent into a rectangle.

. . Its length is twice its width. Code:

2W
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| |
W | | W
| |
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2W

Its perimeter is $\displaystyle x-14.$

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

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

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

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