What fraction of the diagram is black?

Let the side of the big square be $x$ units. Hence it's area $=x^2$ .

The radius of the circle $=\dfrac x2$ , because $x$ is the diameter. Hence the area of the circle $=\pi \dfrac {x^2}4$ .

We can observe that the diagonal of the smaller square is the diameter of the circle, therefore each side of the smaller square $=\sqrt{\dfrac{x^2}2}=\dfrac{x\sqrt 2}2$ . Hence it's area $=\left(\dfrac{x\sqrt 2}2\right)^2=\dfrac{2x^2}4$ .

Therefore the area of the black region $=\text{ar(circle)-ar(small square)}=\pi \dfrac {x^2}4-\dfrac{2x^2}4=\dfrac{x^2}2\left[\dfrac{\pi}2-1\right]$

Hence the ratio of the black region to the whole square $=\dfrac{\dfrac{x^2}2\left[\dfrac{\pi}2-1\right]}{x^2}=\dfrac{\pi}4 - \dfrac 12 \approx 0.29$

If you imagine the small square rotated 45 degrees it becomes obvious that it is half the area of the large one.

So the shapes have areas: 4, Pi, 2 — from outer to inner — expressed as multiples of the square of the radius of the circle.

As a fraction of the outer square the black area is: (Pi - 2) / 4 = .285…