1. ## golden ratio question

A rectangle has width 1 unit and length x units.
A square is cut off it.
the rectangle left behind is similar to the original rectangle.
Find the value of x in surd form.
The value of x is called Golden ratio

MY attempt:
area of the original rectangle = x sq units

area of square = 1 sq units
are of b ( small rectangle) = x-1 sq units

i found the golden ratio by doing this:
$x = \frac{1}{x-1}$
$x(x-1)=1$
$x^2 -x=1$
$(x-\frac{1}{2})^2-\frac{1}{4}=1$
$(x-\frac{1}{2})^2 = \frac{5}{4}$
$x = \frac{1}{2}\pm\frac{\sqrt{5}}{2}$

since width and length are positive units

thus $x = \frac{1}{2} + \frac{\sqrt{5}}{2}$

I found the answer but I read off a forum that i needed to set up the equations relating ratios involving x

but when i tihnk about it why is $x = \frac{1}{x-1}$ this is the part im confused about
in my head 3 = (6/2) thus i thought for a long time that the ratio should be $1 = \frac{x}{x-1}$ not $x = \frac{1}{x-1}$

Pleqase can some explain more in order to deepen my understanding thank you

2. ## Re: golden ratio question

rectangle 1 is length $x$, and width $1$

a little thought shows that $x < 2$ and you must swap the orientation of the 2nd rectangle with respect to the first

rectangle 2 is thus length $1$ and width $x-1$

the two rectangles are similar so

$\dfrac{x}{1} = \dfrac{1}{x-1}$

$x = \dfrac {1}{x-1}$

3. ## Re: golden ratio question

Originally Posted by romsek
rectangle 1 is length $x$, and width $1$

a little
thank you very much i cant add to your reputation. It does not allow me to.

4. ## Re: golden ratio question

Originally Posted by bigmansouf
thank you very much i cant add to your reputation. It does not allow me to.
you can always "thank" a responder by clicking the "thanks" link in the bottom right hand corner of a post.