# easy ratio question, need help please.

• Oct 4th 2010, 05:13 AM
Nguyen
easy ratio question, need help please.
Hey all, I have a easy question that I have found two different answers for (stupid me). Can anyone check my working out and tell me which one is correct. Thanks

The question is to find the value of side x in the rectangle below, the other side being 1. The rectangle is made up of two unit squares and a remaining rectangle that is a scaled down version of the original rectangle.

|-side x--------|

|-----|-----|---|
|-----|-----|---| 1
|-----|-----|---|

1st way: Combining the two unit squares we know that the side lengths of the small rectangle are (x-2) and 1.

|-side x-------|

|----------|---|
|----------|---| 1
|----------|---|
x-2 (meant to be side of small rectangle)

ratio: x/1 = 1/(x-2) we get x= 1+ sqrt(2)

but:
2nd way: Take one unit square away and x becomes (x-1), using the above diagram we get the ratio:

(x-1)/1 = 1/(x-2) and so x= (3 +sqrt(5))/2

I don't know what my error in thinking here is. Can anyone help?

Thanks heaps everyone, this forum is great!
• Oct 4th 2010, 07:45 AM
HallsofIvy
If I understand this correctly, you are saying that the rectangle with lenght x and width 1 can be partitioned into two unit squares and a "scaled down" rectangle similar to the orginal one. That is, the "scaled down" rectangle has sides of length $\displaystyle \alpha x$ and $\displaystyle \alpha$ for some $\displaystyle 0< \alpha< 1$. The we must have $\displaystyle \alpha x= 1$, the width of the original rectangle and [tex]\alpha= x- 2[tex]. Putting that value of $\displaystyle \alpha$ into the first equation, $\displaystyle \alpha x= (x- 2)x= x^2- 2x= 1$. That also gives your ratio, $\displaystyle \frac{x}{1}= \frac{1}{x- 2}$ Completing the square by adding 1 to both sides, $\displaystyle x^2- 2x+ 1= (x- 1)^2= 2$, $\displaystyle x- 1= \pm\sqrt{2}$ so that $\displaystyle x= 1\pm\sqrt{1}$. Since x, as the lenght of a side of a rectangle, must be positive, $\displaystyle x= 1+ \sqrt{2}$, your first solution, is the only solution.

"Taking one square away" we still have $\displaystyle \alpha x= 1$ but now $\displaystyle x= 2+ \alpha$ or $\displaystyle \alpha= x- 2$ which reduces to the same thing.

Your "taking one square away" seems to be starting with a completly different problem in which the original rectangle is made up of only one square together with the "scaled down" rectangle.