Problem 458

Since point $R$ is on $PQ$, then $PQ = PR + QR = 6+4=10$.

Semi-circles with diameters of 10, 6 and 4 have radii of 5, 3 and 2, respectively.

A semi-circle with diameter $PQ$ has area $\frac{1}{2}\times\pi \times 5^2 = \frac{25}{2}\pi$.

A semi-circle with diameter $PR$ has area $\frac{1}{2}\times\pi \times 3^2 = \frac{9}{2}\pi$.

A semi-circle with diameter $QR$ has area $\frac{1}{2}\times\pi \times 2^2 = 2\pi$.

The shaded region consists of a section to the left of $PQ$ and a section to the right of $PQ$.

The area of the section to the left of $PQ$ equals the area of the semi-circle with diameter $PQ$ minus the area of the semi-circle with diameter $PR$.

This area is thus $\frac{25}{2}\pi - \frac{9}{2}\pi = 8\pi$.

The area of the section to the right of $PQ$ equals the area of the semi-circle with diameter $QR$.

This area is thus $2\pi$.

Therefore, the area of the entire shaded region is $8\pi + 2\pi = 10\pi$.

Since the large circle has radius 5, its area is $\pi \times 5^2 = 25\pi$.

Since the area of the shaded region is $10\pi$, then the area of the unshaded region is equal to $25\pi - 10\pi = 15\pi$.

The ratio of the area of the shaded region to the area of the unshaded region is $10\pi : 15\pi$ which is equivalent to $2:3$.