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Diameters $PQ$ and $RS$ cross at the center of the circle, which we call $O$.

The area of the shaded region is the sum of the areas of $△POS$ and $△ROQ$ plus the sum of the areas of sectors $POR$ and $SOQ$.

Each of $△POS$ and $△ROQ$ is right-angled and has its two perpendicular sides of length 4 (the radius of the circle).

Therefore, the area of each of these triangles is $\frac{1}{2}(4)(4)=8$.

Each of sector $POR$ and sector $SOQ$ has area $\frac{1}{4}$ of the total area of the circle, as each has central angle $90^{\circ }$ (that is, $\angle POR=\angle SOQ=90^{\circ }$) and $90^{\circ }$ is one-quarter of the total central angle.

Therefore, each sector has area $\frac{1}{4}(\pi (4^2))=\frac{1}{4}(16\pi )=4\pi$.

Thus, the total shaded area is $2(8)+2(4\pi )=\boxed{16+8\pi }$.