To solve this problem, we begin by analyzing the unit square \(ABCD\) with vertices \(A(0, 0)\), \(B(1, 0)\), \(C(1, 1)\), and \(D(0, 1)\).
The division of the unit square into two congruent triangles and two congruent trapezoids is essential. The most common way to achieve this with geometrical symmetry is by drawing a diagonal, say from \(A\) to \(C\). This diagonal splits the square into two congruent right triangles, \( \triangle ABC \) and \( \triangle CDA \).
Now we have the two congruent triangles. The remaining area will be two congruent trapezoids that make up the rest of the unit square. Since the problem specifies that the trapezoids are formed and are also congruent, it implies that we can visualize these trapezoids as being created by creating midpoints or intersections along the edges of the triangles.
For simplicity, we can take each triangle and conceptualize that the two trapezoids are formed by lines parallel to the bases of the triangles:
- We observe that to form trapezoids, if we take points along segments \(AB\) and \(CD\), moving inwards or outwards while maintaining parallel lines, we can shape the trapezoids appropriately.
The trapezoids would then have their top and bottom sides parallel to the bases of the triangles. Let's denote the heights of these trapezoids in a simpler fashion.
- Let us denote the dimensions of rectangle \(PQRS\) that incorporates both triangles and trapezoids.
- Since each triangle has a base of \(1\) and a height of \(1\), and since they are combined to form a rectangle while maintaining congruence, we can think of \(PQRS\) as the rectangle obtained by aligning both bases (in this case, the bases of trapezoids would be the sides of triangle).
Next step is to find the dimensions of rectangle \(PQRS\). The rectangle \(PQRS\) that results from these triangles will span the same horizontal distance as the unit square, thus having a width of \(1\). In regard to the height, considering congruence and rearrangement of equal areas from triangles and trapezoids, it compensates.
Assuming no scaling applies to the newly formed rectangle from the rearrangement and leveraging from these congruences:
- The width of rectangle \(PQRS\) is \(1\),
- The total height will span the maximum alignment when standing side by side or in full depth maintaining the shape dynamics of rectangle, which also results in \(1\).
Thus, \(PQRS\) must also be a rectangle with dimensions \(1\) by \(1\).
Finally, the perimeter \(P\) of rectangle \(PQRS\) is calculated as follows:
\[ P = 2 \times (\text{length} + \text{width}) = 2 \times (1 + 1) = 2 \times 2 = 4. \]
Therefore, the perimeter of rectangle \(PQRS\) is:
\[ \boxed{4}. \]