To solve the problem, we need to calculate the area of the shaded region in the diagram provided.

1. The diagram consists of a square with side length $14 \, \text{cm}$ and four overlapping circular segments. The circle arcs suggest that the shaded area is created by subtracting the circular segment areas from the square.

2. Calculate the area of the square:

   The area of the square is: $\text{Area of square} = s^2 = 14^2 = 196 \, \text{cm}^2$

3. Calculate the area of the full circle:

   Since the arcs come from a circle with diameter equal to the side length of the square ($14 \, \text{cm}$), the radius of the circle is: $r = \frac{14}{2} = 7 \, \text{cm}$

   The area of the full circle is: $\text{Area of circle} = \pi r^2 = \pi (7^2) = 49\pi \, \text{cm}^2$

4. Calculate the area of the quarter-circle segments:

   Each arc corresponds to a quarter-circle of the full circle. The total area of the four quarter-circles is equivalent to the area of the entire circle: $\text{Total area of arcs} = \text{Area of circle} = 49\pi \, \text{cm}^2$

   However, due to overlapping, the arcs are not covering the entire square.

5. Calculate the shaded area:

   The shaded region is the area of the square minus the unshaded region (formed by the quarter-circles). Using approximations: $\text{Shaded area} = \text{Area of square} - \text{Overlapping circle area}$

   Using $\pi \approx 3.14$: $\text{Shaded area} = 196 - 49 \cdot 3.14 \approx 196 - 153.86 = 112.14 \, \text{cm}^2$

   The closest answer is: C. 112 cm²

Would you like a more detailed explanation or help with related problems?

Here are five related questions for further exploration:

1. How do you calculate the area of a quarter-circle?
2. What is the significance of overlapping areas in composite figures?
3. How would the solution change if the square’s side length were different?
4. What is the effect of using an exact value of $\pi$ versus an approximation?
5. How can we compute areas of irregular shapes using integration?

Tip: When solving geometry problems, always identify if symmetry can simplify calculations!