Smallest square

First, notice that the three rotations are add up to 90 degrees counterclockwise so each one is of 30 degrees.

Then, we calculate the length $x$ of the second square side:

From the upper left triangle we notice that $\displaystyle\frac{a}{1-a}=\frac{1}{\sqrt{3}}$ , hence solving for $a$ we find the side of the triangle

$a=\frac{1}{2}\left(\sqrt{3} - 1\right)$

Via the Pythagorean theorem we find the length we are looking for

$x=\sqrt{a^2+(1-a)^2}=\sqrt{2a^2-2a+1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3} - 1\right)^2}=\sqrt{3} - 1$

We notice the proportionality of the subsequent squares sides lengths $\displaystyle\frac{1}{x}=\frac{x}{y}=\frac{y}{z}$ , from where we obtain $y=x^2=4-2\sqrt{3}$ , and later $z=\frac{y^2}{x}=\frac{28-16\sqrt{3}}{\sqrt{3} - 1}=6\sqrt{3}-10$

Finally the area of the smallest square is $z^2=208-120\sqrt{3}\approx 0.1539$

We can solve this problem by similarity. We have similar squares with area ratio = 1/(a^2+(1 – a)^2) = 1/(1-2a+a^2). The value of a can be found by trigonometry. a/(1- a) =tan 30 = 1/√3, a=1/(1+√3), so area ratio = 0.535898 The area of the smallest is the forth term (a,ar,ar^2,ar^3) in the sequence = ar^3 = 1(0.535898)^3 = 0.1539