To solve the problem, let's carefully analyze the given information and break it down step by step.

Given:

1. Circle Information:

   • The radius of the circle is $OA = 2$, so the area of the circle is $\pi r^2 = \pi (2)^2 = 4\pi$.

2. Square Information:

   • The square $OBDF$ has side length $OD = 8$. Thus, the area of the square is $8 \times 8 = 64$.

3. Shadowed Region:

   • The shadowed region is part of the circle outside triangle $\triangle OAB$, where $\triangle OAB$ lies within the square.

We aim to find the area of the shadowed region, express it as $a - \frac{\pi}{b}$, and determine $a + b$.

Step 1: Area of Sector $OAB$

The circle's sector $OAB$ subtends a $90^\circ$ angle at the center $O$. This sector covers $\frac{90^\circ}{360^\circ} = \frac{1}{4}$ of the total circle area.

$\text{Area of Sector } OAB = \frac{1}{4} \times 4\pi = \pi$

Step 2: Area of Triangle $\triangle OAB$

Triangle $\triangle OAB$ is a right triangle with both legs equal to the radius of the circle, $OA = OB = 2$. The area of $\triangle OAB$ is:

$\text{Area of } \triangle OAB = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2$

Step 3: Area of Shadowed Region

The shadowed region is the area of the sector $OAB$ minus the area of $\triangle OAB$:

$\text{Area of Shadowed Region} = \text{Area of Sector } OAB - \text{Area of } \triangle OAB$

$\text{Area of Shadowed Region} = \pi - 2$

Step 4: Expressing the Shadowed Area as $a - \frac{\pi}{b}$

We compare the shadowed area expression $\pi - 2$ with $a - \frac{\pi}{b}$:

• Here, $a = 2$ and $\frac{\pi}{b} = \pi$. Thus, $b = 1$.

Finally, $a + b = 2 + 1 = 3$.

Final Answer:

$\boxed{3}$

Let me know if you would like further clarification or a more detailed breakdown!

Related Questions:

1. How do you calculate the area of a circle sector with any angle?
2. Why does the shadowed region depend on the subtraction of triangle and sector areas?
3. What is the significance of the circle's radius in determining the solution?
4. How do the properties of a square relate to solving geometry problems like this?
5. How would the solution change if $OD$ (the side of the square) were different?

Tip:

When solving geometry problems, always break them into simpler parts (e.g., sectors, triangles) and use known formulas for each!