To find the area of the shaded region, we can follow these steps:

1. Identify the Geometry: The shaded area is the region between a circle and a square inscribed within it.
2. Determine Dimensions: The square has a side length of 5 cm, as shown in the image.

Step-by-Step Solution

1. Calculate the Area of the Square: $\text{Area of square} = \text{side}^2 = 5 \, \text{cm} \times 5 \, \text{cm} = 25 \, \text{cm}^2$

2. Calculate the Diameter of the Circle: Since the square is inscribed in the circle, the diagonal of the square is equal to the diameter of the circle.

   • For a square with side $s$, the diagonal $d$ can be calculated using the Pythagorean theorem: $d = s\sqrt{2} = 5\sqrt{2} \approx 7.07 \, \text{cm}$

3. Calculate the Radius of the Circle: The radius $r$ of the circle is half of the diameter: $r = \frac{d}{2} = \frac{5\sqrt{2}}{2} \approx 3.54 \, \text{cm}$

4. Calculate the Area of the Circle: $\text{Area of circle} = \pi r^2 = \pi \left( \frac{5\sqrt{2}}{2} \right)^2 = \pi \times \frac{25}{2} \approx 39.27 \, \text{cm}^2$

5. Calculate the Shaded Area: The shaded area is the area of the circle minus the area of the square: $\text{Shaded area} = \text{Area of circle} - \text{Area of square} \approx 39.27 \, \text{cm}^2 - 25 \, \text{cm}^2 = 14.27 \, \text{cm}^2$

Final Answer

The area of the shaded region is approximately $14.27 \, \text{cm}^2$.

Would you like further details or have any questions about this solution?

Here are some related questions you might find interesting:

1. How would the shaded area change if the square's side length were increased?
2. What if the circle's radius was given instead of the square's side?
3. How do we derive the formula for the diagonal of a square?
4. What would be the area of the shaded region if the circle was circumscribed around a different polygon?
5. Can we find the shaded area if only the perimeter of the square is given?

Tip: Always check if a shape is inscribed or circumscribed, as it affects how we relate dimensions between the shapes involved.