Folded Parallelgram Area Problem

Question

we are trying to solve this problem shown in the attached capture:

Having very little background on this, I am not sure how to proceed in solving this problem and any hints or solutions would be very much appreciated. Thanks in advance

Answer 1

Let longer side be $a$ and shorter side be $b$ then area of the original parallelogram is $a\cdot h$, area of the folded parallelogram is $b \cdot h$ where h is height. Area of the shaded area after the first fold is $a\cdot h- b \cdot h$. Thus, $\Large{\frac{(a-b)h}{ah}=\frac{3}{8}}$ or $b=\frac{5}{8}a$. So the first shaded parallelogram has sides $\frac{5}{8}a$ and $\frac{3}{8}a$.
We can set $a=8$ without a loss of generality. Then the original parallelogram has sides $8$ and $5$, the first shaded parallelogram has sides $5$ and $8-5=3$ and the second shaded parallelogram has sides $3$ and $5-3=2$. The ratio of areas of the second shaded parallelogram and the first shaded parallelogram is $\frac{2}{5}$ (both have the same height). The answer to the puzzle is $\Large{\frac{3}{8}\cdot \frac{2}{5}=\frac{3}{20}}$