Question 1142743
<font face="Times New Roman" size="+2">
*[illustration Figure1]


If the side length of ABCD is *[tex \Large x\ -\ 1], then the area of ABCD is *[tex \Large x^2\ -\ 2x\ +\ 1].


The difference in the area of rectangle ABFE and square ABCD is *[tex \Large x^2\ +\ x\ -\ 2\ -\ (x^2\ -\ 2x\ +\ 1)\ =\ 3x\ -\ 3], and this is the area of rectangle DCFE.


One side of rectangle DCFE, namely DC, must have length *[tex \Large x\ -\ 1],
so either side orthogonal to side DC, one of which is side CF, must have a measure of:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3x\ -\ 3}{x\ -\ 1}\ =\ 3\ \text{cm}]
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  
								
{{n}\choose{r}}
</font>