A picture frame measures 24 cm. What is the area of the photograph fixed inside the frame?

From the diagram:
The diameter of the inner circle is $28 \text{ cm}$.
The radius of the inner circle, $r$, is half of its diameter:

$$r = \frac{28 \text{ cm}}{2} = 14 \text{ cm}$$

The total diameter of the outer circle is the sum of the inner diameter and the two widths of the ring:

$$D = 7 \text{ cm} + 28 \text{ cm} + 7 \text{ cm} = 42 \text{ cm}$$

The radius of the outer circle, $R$, is half of its diameter:

$$R = \frac{42 \text{ cm}}{2} = 21 \text{ cm}$$

The area of an annulus is the area of the outer circle minus the area of the inner circle.

$$A = \pi R^2 - \pi r^2 = \pi (R^2 - r^2)$$

We will use the approximation $\pi \approx \frac{22}{7}$.

$$A = \frac{22}{7} ((21 \text{ cm})^2 - (14 \text{ cm})^2)$$ $$A = \frac{22}{7} (441 \text{ cm}^2 - 196 \text{ cm}^2)$$ $$A = \frac{22}{7} (245 \text{ cm}^2)$$ $$A = 22 \times \frac{245}{7} \text{ cm}^2$$ $$A = 22 \times 35 \text{ cm}^2$$ $$A = 770 \text{ cm}^2$$

The area of the shaded region is $\text{770 cm}^2$.