# find width of path

suppose we have following question

A square garden is surrounded by a path of uniform width. If the path and the garden both have an area of $x$, then what is the width of the path in terms of $x$?

so we have following picture right

because we dont know if path of uniform width is square or not how can i find width?if area of square is $x$,then length is $\sqrt{x}$,but what about second figure?suppose it's length are $a$ and $b$,then $a*b=x$,then how can i continue?

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You would have

$$4 w \sqrt{x} + 4 w^2 = x$$

(i.e., four rectangles + four squares that make up path, width of path = $w$)

Solve for $w$:

$$2 w = \frac{-\sqrt{x} \pm \sqrt{2 x}}{2}$$

Choose the positive solution:

$$w=\frac{\sqrt{2}-1}{4} \sqrt{x}$$

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a little detailed please –  dato datuashvili Aug 8 '13 at 21:52
@dato: where are you having trouble understanding this solution? The setup or the algebra? –  Ron Gordon Aug 8 '13 at 21:53
setup of course,what is denoting by what? –  dato datuashvili Aug 8 '13 at 21:54
$w$ is the unknown path width. $x$ is, as you denoted, the area of both the garden and the path. –  Ron Gordon Aug 8 '13 at 21:55
but i saw,why it is equal to $x$,we already eliminated it,it should equal to zero,is not it –  dato datuashvili Aug 8 '13 at 21:55

Call the width of the path $a$.

Like you say, the width of the inner square is $\sqrt{x}$. The width (and height) of the outer square is then $2a+\sqrt{x}$.

1. What is the area of the outer square?
2. What is the area of the path?
3. What can you conclude about $a$?
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sorry we are denoting with as a full length of half part? –  dato datuashvili Aug 8 '13 at 21:49
I'm denoting by $a$ the width of one part of the path (so that if you draw a horizontal line through the middle of your figure, the first segment has length $a$, then $x$, then $a$ again.) –  user7530 Aug 8 '13 at 21:57

Define \begin{align} g &:= \text{width of garden (inner square)} \\ w &:= \text{width of path} \\ s &:= \text{width of outer square} = g + 2 w \end{align}

Then \begin{align} \text{area of garden (inner square)} &= g^2 \\ \text{area of path (outer sq., minus inner sq.)} &= s^2 - g^2 \\ &= ( g + 2 w )^2 - g^2 \\ &= g^2 + 4 g w + 4 w^2 - g^2 \\ &= 4 w^2 + 4 g w \end{align}

We know that each of the two areas equals $x$. For notational simplicity, temporarily write "$y$" for "$\sqrt{x}$", so that $x = y^2$. \begin{align} g^2 &= x = y^2 &(1) \\ 4 w^2 + 4 g w &= x = y^2 &(2) \end{align}

From equation $(1)$, we get that $g = y$. Substituting that into $(2)$ gives $$4 w^2 + 4 w y = y^2 \qquad \to \qquad 4 w^2 + 4 w y - y^2 = 0$$

Now, simply solve the quadratic equation for $w$:

$$w = \frac{- 4 y \pm \sqrt{(4y)^2-4\cdot 4\cdot(-y^2)}}{2\cdot 4} = \frac{-4y \pm \sqrt{16y^2+16y^2}}{8} = \frac{-4y\pm 4y\sqrt{2}}{8} = \frac{y}{2}\left(-1\pm\sqrt{2}\right)$$

We (presumably) want the positive root, so (since $\sqrt{2} > 1$) take "$\pm$" to be "$+$"; also, replace "$y$" with its defined value, "$\sqrt{x}$":

$$w = \frac{\sqrt{x}}{2}\left(\sqrt{2} - 1\right)$$

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thanks very much for your help –  dato datuashvili Aug 9 '13 at 7:57