This was the question I attempted to answer. I am CalculatorUser and I tried my best to solve it, however, I couldn't and I need help.
Three 2 x 1 rectangles are arranged as shown. What fraction of the figure is golden?
Go to the link if you want to see my attempt, I really tried hard, but there is no one to help..
Let the two segment lengths along the long side of the middle rectangle $a$ and $b$. Then,
$$a+b=2,\>\>\>\>\>1+a^2=b^2$$
where the hypotenuse of the golden triangle is $b$, due to the SAA-congruency of golden and grey triangles. Solve to obtain $a=\frac34$ and $b=\frac54$. Thus, the two golden areas sum up to $\frac{3}{4}$, which is $\frac{3}{8}$ of one rectangle.