A square in a quarter circle

Say the side of the square is $s$ . In triangle $OPQ$ , $OP=1,\;\;\;PQ=s,\;\;\;OQ=\frac{s}{\sqrt2},\;\;\;\angle OQP=135^\circ$ so by the cosine rule, $\begin{aligned} OP^2 &=PQ^2+OQ^2-2PQ\cdot OQ\cos \angle OQP \\ 1 &=s^2+\frac12 s^2+2\cdot s \cdot \frac{s}{\sqrt2} \cdot \frac{1}{\sqrt2} \\ 1 &=\frac52 s^2 \end{aligned}$

and $\boxed{s=\sqrt{\frac25}}$ .

Let's note the square's edge with length of s, and taking the rightmost vertex into consideration, we have
1² = (x-coordinate)² + (y-coordinate)²
= (s × √2)² + (s ÷ √2)²
= 2s² + s² / 2
= 5s² / 2
s² = 2 / 5
s = √(2/5)