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

∠ 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.

$(\frac{1}{4} π {(\sqrt{2})}^{2} - \frac{1}{2} \times \sqrt{2} \times \sqrt{2}) {cm}^{2},$

i.e. $(\frac{1}{2} π - 1)$ cm $^{2}$ .

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