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If the semicircle with diameter $PQ$ is rotated through $180^\circ$ about $Q$, the new shape formed has the same area as the original shape. It consists of a semicircle of diameter $6\;\mathrm{cm}$ and a semicircle of diameter $2\;\mathrm{cm}$.

So the area of the original shape is $$(\frac{1}{2}\times \pi \times 3^2+\frac{1}{2}\times \pi \times 1^2){\mathrm{cm}}^2=5\pi {\mathrm{cm}}^2.$$