Let $O$ be the centre of the circle and let the points where the arcs meet be C and D respectively. $\mathrm{ABCD}$ is a square since its sides are all equal to the radius of the arc $\mathrm{CD}$ and $\angle \mathrm{ACB}={90}^{\circ }$ (angle in a semicircle).

In triangle $\mathrm{OCB}$, ${\mathrm{CB}}^2={\mathrm{OC}}^2+{\mathrm{OB}}^2$; hence $\mathrm{CB}=\sqrt{2}\mathrm{cm}$. The area of the segment bounded by arc $\mathrm{CD}$ and diameter $\mathrm{CD}$ is equal to the area of sector $\mathrm{BCD}$ - the area of triangle $\mathrm{BCD}$, i.e.

${\displaystyle (\frac{1}{4}\pi {\left(\sqrt{2}\right)}^2-\frac{1}{2}\times \sqrt{2}\times \sqrt{2}){\mathrm{cm}}^2}$

The unshaded area in the original figure is, therefore, $(\pi -2){\mathrm{cm}}^2$. Now the area of the circle is $\pi {\mathrm{cm}}^2$ and hence the shaded area is $2{\mathrm{cm}}^2$.