Probablility of a dart landing closer to the center than the edge of a square dartboard?

The question is: There is a square dart board with a side length of $2$m, and a dart has equal probability to land anywhere on the board. What is the probability that the dart will land closer to the center than to the edges? So I know that if I center the square at the origin, I have to find $f(x)$ such that every point on $f(x)$ is equidistant between the origin and the line $x=1$ or $y=1$ (I'm only working with the first quadrant since by symmetry the probability should be the same). Another reasonable assumption that I made is that $f(x)$ is reflected over $y=x$, so I only have to work with the first $45$ degrees of the first quadrant. So to find $f(x)$ I did the following: $\sqrt{x^2+[f(x)]^2}=1-x$, so $f(x)=\sqrt{1-2x}$. Now I know that I just find the area under this curve and divide it by $1/8$ the area of the square to find the probability. I just wanted to know if my $f(x)$ was okay? Thanks!

Your thought process looks good, but the integration is actually a bit tricky. Here is your dartboard:

We'd like to calculate the area of the center shape, and as you say, we can split this into eight slices of equal area. Here we see the two slices in the first quadrant:

Let's compute the area of the slice from $45^{\circ}$ to $90^{\circ}$:

$$\int_{0}^{\sqrt{2}-1}(\frac{1-x^2}{2}-x)dx=\frac{1}{6}(4\sqrt{2}-5)$$

Multiplying this by $8$ gives us the area of the center shape. Then, we divide by $4$, the area of the dartboard. The final probability is then:

$$\frac{1}{3}(4\sqrt{2}-5)\approx.2190$$

Judging by our picture above, this seems right. The center shape is about $1/5$ the area of the dartboard.