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 $$\left(\frac{1}{2}\times\pi\times 3^2+\frac{1}{2}\times\pi\times 1^2\right)\;\mathrm{cm}^2 = 5\pi\;\mathrm{cm}^2\;.$$