This divides the diagram into 4 equal squares where the shaded region in each of these smaller squares is: $r^2-2 A$.
The area of region $A$ can then be found with the equation: region $A=r^2-\frac{1}{4} \pi r^2$.
Substitute the expression for the area of region $A$ back into the shaded region formula:

$$
\(\begin{array}{ll}
r^2-2 A & \text { Shaded reg } \\
r^2-2\left(r^2-\frac{1}{4} \pi r^2\right) & \text { Substitute } \\
r^2-2 r^2\left(1-\frac{1}{4} \pi\right) & \text { Factor } \\
r^2\left(1-2\left(1-\frac{1}{4} \pi\right)\right) & \text { Factor } \\
r^2\left(1-2+\frac{1}{2} \pi\right) & \text { Simplify } \\
r^2\left(-1+\frac{1}{2} \pi\right) & \text { Simplify }
\end{array}\)
$$


Shaded region formula

The area of the un-shaded region is then expressed:

$$
\(\begin{array}{rlrl}
\text { un-shaded area } & =\text { square }- \text { shaded area } \\
\begin{aligned}
\text { un-shaded area } & =r^2-\left(r^2\left(-1+\frac{1}{2} \pi\right)\right) & & \text { Substitution } \\
& =r^2\left(1-1\left(-1+\frac{1}{2} \pi\right)\right) & & \text { Distributive } \\
& =r^2\left(1+1-\frac{1}{2} \pi\right) & & \text { Simplify the signs } \\
& =r^2\left(2-\frac{1}{2} \pi\right) & & \text { Simplify }
\end{aligned} .
\end{array}\)
$$


Now compare the two areas:
shaded region VS unshaded region


$$
\(\begin{aligned}
r^2\left(-1+\frac{1}{2} \pi\right) & ? r^2\left(2-\frac{1}{2} \pi\right) & & \text { Substitution } \\
\left(-1+\frac{1}{2} \pi\right) & ?\left(2-\frac{1}{2} \pi\right) & & \text { Divide by } r^2 \\
\pi & >3 & & \text { Add } \frac{\pi}{2} \text { to both sides and simplify }
\end{aligned}\)
$$


From this you see that the shaded area of each small square is greater than the un-shaded area, so it follows that the shaded area of the entire diagram is greater than the un-shaded area. The correct answer choice is $\(\mathbf{A}\)$.