The problem involves a square of side length 1, inscribed with four quarter circles, with centers at the corners of the square (A, B, C, D) and radius equal to the side of the square. The question asks for the area of the irregular figure $\triangle DEF$, where $E, F, G, H$ are the intersection points of the arcs.

Step-by-Step Analysis:

1. Understanding the Geometry:

   • The square $ABCD$ has side length 1.
   • The arcs $CAB, BDA, ACD, DCB$ are quarter-circles with radius 1 and are centered at $A, B, C, D$, respectively.
   • Points $E, F, G, H$ are formed by the intersection of these arcs.

2. Key Observations:

   • The arcs divide the square into symmetric regions.
   • The area of $\triangle DEF$ can be derived using symmetry and geometric properties.

3. Finding Areas:

   • The total area of the square is $1 \times 1 = 1$.
   • The area of one quarter-circle is $\frac{1}{4} \pi r^2 = \frac{\pi}{4}$, so the total area covered by the quarter circles is $\pi$ (as there are four quarter-circles).

4. Triangle $\triangle DEF$:

   • $DEF$ lies in the symmetric center, surrounded by arc segments. The calculation involves finding the overlapping regions created by the arcs.

5. Exact Calculation: Using integrals or symmetry, the exact area of $DEF$ is derived to be: $\text{Area of } \triangle DEF = 1 - \pi + 2 \times (\text{area of a lens-shaped region})$ Each lens region's area can be calculated using: $\text{Lens area} = R^2 \cos^{-1}(1/2) - \frac{1}{2} R^2 \sin(\pi/3), \quad R = 1$ Simplifying: $\text{Area of } \triangle DEF = \frac{1}{2} - (\pi/6) + (\sqrt{3}/4)$

Final Result:

The area of the triangle $\triangle DEF$ is approximately: $\frac{1}{2} - \frac{\pi}{6} + \frac{\sqrt{3}}{4}$

Would you like a breakdown of the lens-area calculation or further steps to verify?

Here are 5 related questions:

1. How do you calculate the area of a lens-shaped region between two arcs?
2. What is the total area of the segments surrounding $\triangle DEF$?
3. Why is symmetry important in solving this problem?
4. Can you calculate the area numerically for specific dimensions?
5. How does the formula for a segment area simplify if the radius is 1?

Tip: When dealing with complex geometric figures, breaking them into simpler shapes often simplifies area calculations.