A rectangle $ABCD$, which measure $9 ft$ by $12 ft$, is folded once perpendicular to diagonal AC so that the opposite vertices A and C coincide. Find the length of the fold. So I tried to fold a rectangular paper but there are spare edges. So the gray is my fold and I'm not sure if its in the middle of my diagonal AC? If it is then now I need to solve the A to the center and center to A. Where I'm so confused. 
[Math] a rectangle $ABCD$, which measure $9 ft$ by $12 ft$, is folded once perpendicular to diagonal AC
geometry
Related Solutions
$\hskip 2.2in$ 
The above figure was done using grapher on mac osx.
Let $l$ be the length (i.e. the sides $AD$ and $BC$) and $b$ be the breadth (i.e. the sides $AB$ and $CD$). Once you get the diagonal vertices together, i.e. when $D$ coincides with $B$, the length $EB$ is the same as the length $ED = l-a$.
Hence, for the right triangle, we have that $$a^2 + b^2 = (l-a)^2\\ b^2 = l^2 - 2al\\ a = \frac{l^2 - b^2}{2l}$$ The length of $BF$ is $l-a$ and is given by $$l-a = l - \frac{l^2 - b^2}{2l} = \frac{2l^2 - l^2 + b^2}{2l} = \frac{l^2 + b^2}{2l}$$ The distance between the two points is $$EF^2 = r^2 = b^2 + (l-2a)^2 = b^2 + \left(l - \frac{l^2 - b^2}{l} \right)^2 = b^2 + \frac{b^4}{l^2}$$ This is so since the vertical distance between $E$ and $F$ is $l-2a$.
You are given that $r = l$ and hence $$l^2 = b^2 + \frac{b^4}{l^2}$$If we let $$\frac{l}{b} = x,$$ then we get that $$x^2 = 1 + \frac1{x^2}\\ x^4 = x^2 + 1$$ which gives us that $$x = \sqrt{\frac{1}{2} (1+\sqrt{5})} = \sqrt{\phi}$$ where $\phi$ is the golden ratio.
The $1:\sqrt{2}$ ratio ensures exactly that. That is the idea behind the ISO 216 standard for paper sizes, which was adopted from the German DIN 476 standard.
Its most common usage is the A series which especially in Europe is a collection of very common paper sizes. The base size, A0, has an area of a square meter, and every next smaller paper size is constructed by folding it in half.
Best Answer
The length is
$$\frac{45}4\,\text{ft}=11.25\,\text{ft}$$
One way to obtain this result is using coordinates. Use $E$ as the center of the coordinate system, then the corners are $A,B,C,D=\left(\pm\frac92,\pm\frac{12}2\right)$. The diagonal $AC$ has the equation $y=\frac{12}{9}x=\frac43x$ so the perpendicular line has $y=-\frac34x$. For $x=-\frac92$ you get $y=\frac{27}{8}$, so you have $F=\left(-\frac92,\frac{27}8\right)$ and likewise $G=\left(\frac92,-\frac{27}8\right)$. The line between these two has a length of
$$\lvert FG\rvert=\sqrt{9^2+\left(\frac{27}4\right)^2}=\sqrt{\frac{1296+729}{16}}=\sqrt{\frac{2025}{16}}=\frac{45}4$$