1965 AMC 12 Problem 35

The crease is the perpendicular bisector of a rectangle diagonal

If the width is $w\lt4,$ its intersections with the long sides differ horizontally by $\frac{w^2}{5}$

Place the rectangle at $(0,0),$ $(5,0),$ $(5,w),$ and $(0,w).$ The crease sending $(0,0)$ to $(5,w)$ is $$5x+wy=\frac{25+w^2}{2}.$$ Because $w\lt4,$ it meets the two horizontal sides. Between those intersections the vertical change is $w$ and the horizontal change is $\frac{w^2}{5}.$ Thus $$6=w^2+\frac{w^4}{25}.$$ Setting $z=w^2$ gives $z^2+25z-150=0,$ so $z=5$ and $w=\sqrt5.$

Therefore, the correct answer is D.