1. As shown in Figure 2, two diameters $A C$ and $B D$ of the large circle intersect perpendicularly at point $O$. Four semicircles are drawn outside the large circle with $A B$, $B C$, $C D$, and $D A$ as diameters, respectively. The total area of the four "crescent" shaded regions in the figure is $2 \mathrm{~cm}^{2}$. Then the radius of the large circle is $\qquad$ $\mathrm{cm}$.
$=.1 .1$.
By the Pythagorean theorem, we know $A D^{2}+C D^{2}=A C^{2}$, so the area of the upper half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A D$ and $C D$. Similarly, the area of the lower half of the large circle is equal to the sum of the areas of the two semicircles with diameters $A B$ and $B C$. Therefore, the area of the square $A B C D$ is equal to the total area of the four "lunar" shapes. It is easy to calculate that the radius of the large circle $O D$ is $1 \mathrm{~cm}$.