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**pflo** You are looking for the difference in perimeters, or $\displaystyle P_{r} - P_{s}$.

$\displaystyle P_{r}=2l + 2w$ or $\displaystyle P_{r}=2(xw) + 2w$. Factoring yields $\displaystyle P_{r}=2w(x+1)$.

To get the $\displaystyle w$ out of the equation, you can use the fact that areas are the same. The perimeter of the square is $\displaystyle L=4s$, which means the length of a side is $\displaystyle \frac{L}{4}$, making the area $\displaystyle \frac{L^2}{16}$. The area of the rectangle is $\displaystyle (wx)w$ or $\displaystyle w^2x$. The relationship is then $\displaystyle w^2x=\frac{L^2}{16}$ and $\displaystyle w=\sqrt{\frac{L^2}{16x}}=\frac{L}{4\sqrt{x}}$.

Plugging this into the formula for the rectangle perimeter, you can see that $\displaystyle P_{r}=2\frac{L}{4\sqrt{x}}(x+1)=\frac{L}{2\sqrt{x} }(x+1)$.

The differences in perimeters (the answer to the original part b question) is then $\displaystyle \frac{L}{2\sqrt{x}}(x+1)-L$. You can rationalize this, factor it, distribute, or whatever other simplification you want. The form I kinda like is $\displaystyle L(2\sqrt{x}+\frac{\sqrt{x}}{2x}-1)$.

Using this final version of the formula to solve part a (plugging in 4 for x and 160 for L), I get 520 mm. If that's the right answer to part a then I'm pretty comfortable with it.