In a circle of radius $1$, six arcs of radius $1$ are drawn, which cut the circle as in the figure. Determine the black area.
[img]https://cdn.artofproblemsolving.com/attachments/8/9/0323935be8406ea0c452b3c8417a8148c977e3.jpg[/img]
1. **Identify the structure of the problem**: The circle has a radius of 1, and six arcs of radius 1 are drawn inside it, creating a symmetrical pattern. Each arc subtends an angle of $60^\circ$ at the center of the circle.

2. **Calculate the area of one $60^\circ$ sector**: The area of a sector of a circle is given by:
   \[
   \text{Area of sector} = \frac{\theta}{360^\circ} \times \pi r^2
   \]
   For a $60^\circ$ sector in a circle of radius 1:
   \[
   \text{Area of one sector} = \frac{60^\circ}{360^\circ} \times \pi \times 1^2 = \frac{\pi}{6}
   \]

3. **Calculate the area of the equilateral triangle**: Each $60^\circ$ sector contains an equilateral triangle with side length 1. The area of an equilateral triangle with side length $a$ is given by:
   \[
   \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} a^2
   \]
   For $a = 1$:
   \[
   \text{Area of equilateral triangle} = \frac{\sqrt{3}}{4} \times 1^2 = \frac{\sqrt{3}}{4}
   \]

4. **Calculate the area of the segment**: The area of the segment is the area of the sector minus the area of the triangle:
   \[
   \text{Area of segment} = \frac{\pi}{6} - \frac{\sqrt{3}}{4}
   \]

5. **Calculate the total area of the segments**: There are 6 such segments in the circle:
   \[
   \text{Total area of segments} = 6 \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) = \pi - \frac{3\sqrt{3}}{2}
   \]

6. **Calculate the area of the circle**: The area of the circle with radius 1 is:
   \[
   \text{Area of circle} = \pi \times 1^2 = \pi
   \]

7. **Calculate the black area**: The black area is the area of the circle minus the total area of the segments:
   \[
   \text{Black area} = \pi - \left( \pi - \frac{3\sqrt{3}}{2} \right) = \frac{3\sqrt{3}}{2}
   \]

The final answer is $\boxed{\frac{3\sqrt{3}}{2}}$