The Golden Area.

Label the diagram as follows and draw $AD$ and $DE$ :

Since $\angle B = \angle E = 90°$ , $BD = AE = 2$ , and $AD = AD$ , $\triangle AED \cong \triangle DBA$ by the hypotenuse-leg congruence theorem.

Since $AB = DE = 1$ by congruent parts, $\angle DCE = \angle ACB$ by vertical angles, and $\angle B = \angle E = 90°$ , $\triangle CED \cong \triangle CBA$ by the AAS congruence theorem.

Letting $BC = CE = x$ , then $AC = CD = \sqrt{x^2 + 1}$ by Pythagorean's Theorem, and $AE = AC + CE = \sqrt{x^2 + 1} + x = 2$ which solves to $x = \frac{3}{4}$ .

Therefore the golden area is $A_G = 2 \cdot \frac{1}{2} \cdot 1 \cdot \frac{3}{4} = \frac{3}{4}$ , the white area is $A_W = 3 \cdot 1 \cdot 2 - A_G = 6 - \frac{3}{4} = \frac{21}{4}$ , and the fraction is $\frac{A_G}{A_W} = \frac{\frac{3}{4}}{\frac{21}{4}} = \boxed{\frac{1}{7}}$ .

Due to symmetry, the two golden right triangle are congruent. Let the smallest vertex angle of the golden triangle be $\theta$ as shown. We note that:

$\begin{aligned} \cos \theta + 2 \sin \theta & = 2 & \small \blue{\text{Using half-angle tangent substitution}} \\ \frac {1-t^2}{1+t^2} + \frac {4t}{1+t^2} & = 2 & \small \blue{\text{and let }t = \tan \frac \theta 2} \\ 3t^2 - 4t + 1 & = 0 \\ (3t-1)(t-1) & = 0 \\ \implies t & = \frac 13 & \small \blue{t = 1 \text{ is rejected because } \theta \ne \frac \pi 2} \end{aligned}$

Note that the area of a golden triangle is $\dfrac 12 \tan \theta$ . Therefore, area of the golden region $A_g = \tan \theta = \dfrac {2t}{1-t^2} = \dfrac 34$ . The area of the figure is $A_F = 6 - A_g = 6 - \dfrac 34 = \dfrac {21}4$ and $\dfrac {A_g}{A_F} = \dfrac 34 \times \dfrac 4{21} = \boxed{\frac 17}$ .