Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

enter image description here

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?

share|improve this question
add comment

3 Answers 3

up vote 1 down vote accepted

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:

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

share|improve this answer
    
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
show 10 more comments

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 $a$?
share|improve this answer
    
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
add comment

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, "$\sqrt{x}$":

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

share|improve this answer
    
thanks very much for your help –  dato datuashvili Aug 9 '13 at 7:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.