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$^2$.

*This problem is taken from the UKMT Mathematical Challenges.**View the archive of all weekly problems grouped by curriculum topic*

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