Folding Paper

Let the rectangular paper be $ABCD$ , such that $\overline{AB}=\overline{CD} = w$ and $\overline{AD}=\overline{BC} = l$ . By Pythagorean theorem the length of the diagonal fold $\overline{AC} = \sqrt{l^2+w^2}$ . Let $F$ be the midpoint of the fold $\overline{AC}$ and $\overline{EF}$ be perpendicular to $\overline{AC}$ . By symmetry $E$ is the intersecting point of $\overline{BC}$ and $\overline{AD}$ .
We note that $\triangle AEF$ and $\triangle CEF$ have the same area and they are similar to $\triangle ADC$ . Therefore,

$\begin{aligned} \dfrac {\overline{EF}}{\overline{AF}} & = \dfrac {\overline{CD}}{\overline{AD}} = \frac wl \\ \implies \overline{EF} & = \frac wl \color{#3D99F6} \overline{AF} & \small \color{#3D99F6} \text{Note that }\overline{EF} = \frac {\overline{AC}}2 \\ & = \frac wl \cdot \frac {\sqrt{l^2+w^2}}2 \end{aligned}$

The reduced visible area is $\triangle AEC$ ; then we have:

$\begin{aligned} [AEC] & = 30\%[ABCD] \\ \frac 12 \cdot \overline{AC} \cdot \overline{EF} & = \frac 3{10}wl \\ \frac 12 \cdot \sqrt{l^2+w^2} \cdot \frac {w\sqrt{l^2+w^2}}{2l} & = \frac 3{10}wl \\ \frac {w(l^2+w^2)}{4l} & = \frac 3{10}wl \\ 5l^2+5w^2 & = 6l^2 \\ l^2 & = 5w^2 \\ \frac {l^2}{w^2} & = 5 \\ \implies \frac lw & = \boxed{\sqrt 5} \end{aligned}$