Find the radius of a circle touching two arcs within a square

Question

$ABCD$ is a square having each side = $6$ cm. Two arcs are drawn within the square as shown in the figure. Find the radius of the circle (centre $P$ in the figure) which touches both the arcs and the side of the square.

Answer 1

It is only a matter of pythagorean theorem:

Answer 2

let the radius of the little circle be $r$. Let X be the point on BC where this little circle is tangential to BC. Let the angle PBX be $\phi$. Consider the triangle PBX \begin{eqnarray*} sin \phi = \frac{r}{6-r} \end{eqnarray*} Now condiser the triangle ABP ... Its sides are of length $6,6+r , 6-r$ and the angle at $B$ satisfies $cos \theta = \frac{r}{6-r}$. Now apply the cosine rule to this triangle. \begin{eqnarray*} (6+r)^2=6^2+(6-r)^2-2 \times 6 (6-r) \frac{r}{6-r} \end{eqnarray*} Solving this we have $\color{red}{r=1}$.