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'Eyelids' printed from https://nrich.maths.org/
Let $O$ be the centre of the circle and let the points where the
arcs meet be $C$ and $D$ respectively. $ABCD$ is a square since its
sides are all equal to the radius of the arc $CD$ and $\angle
ACB=90^{\circ}$ (angle in a semicircle).
In triangle $OCB$, $CB^2 = OC^2 + OB^2$; hence $CB=\sqrt{2}$ cm.
The area of the segment bounded by arc $CD$ and diameter $CD$ is
equal to the area of section $BCD - $ the area of the triangle
$BCD$, i.e.
$$\left(\frac{1}{4}\pi\left(\sqrt{2}\right)^2-\frac{1}{2}\times\sqrt{2}\times
\sqrt{2}\right)\textrm{cm}^2$$,i.e. $(\frac{1}{2}\pi - 1)$
cm$^2$.
The unshaded area in the original figure is, therefore, $(\pi-2)$
cm$^2$. Now the area of the circle is $\pi$ cm$^2$, and hence the
shaded area is 2 cm$^2$.