a rectangle $ABCD$, which measure $9 ft$ by $12 ft$, is folded once perpendicular to diagonal AC 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. 
 A: 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
$$\lvert FG\rvert=\sqrt{9^2+\left(\frac{27}4\right)^2}=\sqrt{\frac{1296+729}{16}}=\sqrt{\frac{2025}{16}}=\frac{45}4$$
A: Your folding line is perpendicular to the rectangle's diagonal, which is a hypotenuse of a right triangle with legs 9 and 12 feet, so the folding line itself is a hypotenuse of a right trianlge with legs $9\times\frac 9{12}$ and  $12\times\frac 9{12}$ — so its length is $$\sqrt{\left(9\times\frac 9{12}\right)^2 + \left(12\times\frac 9{12}\right)^2} = \frac9{12}\sqrt{81 + 144} = \frac9{12}\times 15= \frac{45}4$$
