can u explain again how u got the length of the square as 4?

I'm also confused. If the origin is the center of the circle, then the radius is 2root2, which squared equals 8! So area of the circle should be 8pi !

I got 6.87, since the area of the square is 32...sorry if i messed something up but i dont seem to be the only one!

1. Find the Side Length of the Square ( $\(s\)$ )

The square has two adjacent vertices at $\(A(0,-2 \sqrt{2})\)$ and $\(B(2 \sqrt{2}, 0)\)$. The distance between these two points is the length of one side:

$$
\(\begin{gathered}
s=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
s=\sqrt{(2 \sqrt{2}-0)^2+(0-(-2 \sqrt{2}))^2} \\
s=\sqrt{(2 \sqrt{2})^2+(2 \sqrt{2})^2}=\sqrt{8+8}=\sqrt{16}=4
\end{gathered}\)
$$


The area of the square ( $\(A_{\text {square }}\)$ ) is:

$$
\(A_{\text {square }}=s^2=4^2=16\)
$$

2. Find the Radius of the Circle ( $r$ )

A circle inscribed in a square has a diameter equal to the square's side length. Therefore, the radius is half the side length:

$$
\(r=\frac{s}{2}=\frac{4}{2}=2\)
$$


The area of the circle ( $\(A_{\text {circle }}\)$ ) is:

$$
\(A_{\text {circle }}=\pi r^2=\pi\left(2^2\right)=4 \pi \approx 12.566\)
$$

3. Calculate the Target Area

The area of the region inside the square but outside the circle is the difference between their areas:

$$
\(\begin{gathered}
\text { Area }=A_{\text {square }}-A_{\text {circle }} \\
\text { Area }=16-4 \pi \approx 16-12.56637 \ldots=3.43362 \ldots
\end{gathered}\)
$$


Rounding to the nearest tenth, we get:
Area $\(\approx 3.4\)$