1. **Stating the problem:** We have a square ABCD with area 3. Inside it, two circles are drawn touching the sides, and some regions inside the circles and corners are shaded black. We need to find the total area of the black parts (the four shaded corners). 2. **Given:** - Area of square ABCD = 3 - The square has side length $s$ such that $s^2 = 3$, so $s = \sqrt{3}$. 3. **Understanding the figure:** - The black parts are the four corners of the square outside the two circles. - The two circles are tangent to the sides of the square and arranged vertically. 4. **Key insight:** - The two circles together cover the central part of the square, leaving the four corners shaded. - The total area of the black parts = Area of square - Area covered by the two circles. 5. **Finding the radius of the circles:** - Each circle touches the top and bottom sides of the square, so the diameter of each circle equals half the side length of the square. - Since the two circles are stacked vertically and tangent, their combined height equals the side length $s$. - Let radius of each circle be $r$, then $2r + 2r = s \Rightarrow 4r = s \Rightarrow r = \frac{s}{4} = \frac{\sqrt{3}}{4}$. 6. **Area of each circle:** $$\text{Area} = \pi r^2 = \pi \left(\frac{\sqrt{3}}{4}\right)^2 = \pi \frac{3}{16} = \frac{3\pi}{16}$$ 7. **Total area of two circles:** $$2 \times \frac{3\pi}{16} = \frac{3\pi}{8}$$ 8. **Area of black parts (four corners):** $$\text{Area}_{black} = \text{Area}_{square} - \text{Area}_{circles} = 3 - \frac{3\pi}{8} = 3\left(1 - \frac{\pi}{8}\right)$$ **Final answer:** $$\boxed{3\left(1 - \frac{\pi}{8}\right)}$$