We call an isosceles trapezoid $PQRS$ [i]interesting[/i], if it is inscribed in the unit square $ABCD$ in such a way, that on every side of the square lies exactly one vertex of the trapezoid and that the lines connecting the midpoints of two adjacent sides of the trapezoid are parallel to the sides of the square.

Find all interesting isosceles trapezoids and their areas.
1. **Define the vertices of the unit square and the trapezoid:**
   Let the unit square \(ABCD\) have vertices \(A(0,0)\), \(B(1,0)\), \(C(1,1)\), and \(D(0,1)\). Let the vertices of the isosceles trapezoid \(PQRS\) be \(P\), \(Q\), \(R\), and \(S\) such that \(P \in AB\), \(Q \in BC\), \(R \in CD\), and \(S \in DA\).

2. **Assign coordinates to the vertices of the trapezoid:**
   Let \(P = (x_1, 0)\), \(Q = (1, y_1)\), \(R = (x_2, 1)\), and \(S = (0, y_2)\).

3. **Use the condition that the midpoints of adjacent sides are parallel to the sides of the square:**
   - The midpoint of \(PQ\) is \(\left(\frac{x_1 + 1}{2}, \frac{y_1}{2}\right)\).
   - The midpoint of \(QR\) is \(\left(\frac{1 + x_2}{2}, \frac{y_1 + 1}{2}\right)\).
   - The midpoint of \(RS\) is \(\left(\frac{x_2}{2}, \frac{1 + y_2}{2}\right)\).
   - The midpoint of \(SP\) is \(\left(\frac{x_1}{2}, \frac{y_2}{2}\right)\).

   For the midpoints of \(PQ\) and \(RS\) to be parallel to the sides of the square, the y-coordinates must be equal:
   \[
   \frac{y_1}{2} = \frac{1 + y_2}{2} \implies y_1 = 1 + y_2
   \]

   For the midpoints of \(QR\) and \(SP\) to be parallel to the sides of the square, the x-coordinates must be equal:
   \[
   \frac{1 + x_2}{2} = \frac{x_1}{2} \implies 1 + x_2 = x_1
   \]

4. **Solve the system of equations:**
   - From \(y_1 = 1 + y_2\), we get \(y_2 = y_1 - 1\).
   - From \(1 + x_2 = x_1\), we get \(x_2 = x_1 - 1\).

5. **Determine the possible values for the coordinates:**
   Since \(P\) and \(S\) must lie on the sides of the unit square, \(0 \leq x_1 \leq 1\) and \(0 \leq y_1 \leq 1\). Similarly, \(0 \leq x_2 \leq 1\) and \(0 \leq y_2 \leq 1\).

   - For \(x_2 = x_1 - 1\) to be valid, \(x_1\) must be greater than or equal to 1. Thus, \(x_1 = 1\) and \(x_2 = 0\).
   - For \(y_2 = y_1 - 1\) to be valid, \(y_1\) must be greater than or equal to 1. Thus, \(y_1 = 1\) and \(y_2 = 0\).

6. **Verify the coordinates:**
   - \(P = (1, 0)\)
   - \(Q = (1, 1)\)
   - \(R = (0, 1)\)
   - \(S = (0, 0)\)

   These coordinates form a square, not an isosceles trapezoid. Therefore, we need to re-evaluate the conditions.

7. **Re-evaluate the conditions:**
   Consider the possibility that the trapezoid is symmetric about the center of the square. Let the midpoints of \(PQ\) and \(RS\) be \((\frac{1}{2}, y)\) and the midpoints of \(QR\) and \(SP\) be \((x, \frac{1}{2})\).

   - For the midpoints of \(PQ\) and \(RS\) to be parallel to the sides of the square, \(y\) must be \(\frac{1}{2}\).
   - For the midpoints of \(QR\) and \(SP\) to be parallel to the sides of the square, \(x\) must be \(\frac{1}{2}\).

   This implies that the trapezoid is symmetric about the center of the square, and the vertices are:
   - \(P = (a, 0)\)
   - \(Q = (1, a)\)
   - \(R = (1-a, 1)\)
   - \(S = (0, 1-a)\)

8. **Calculate the area of the trapezoid:**
   The bases of the trapezoid are \(PQ\) and \(RS\), both of length \(1 - 2a\). The height of the trapezoid is \(1\).

   The area \(A\) of the trapezoid is given by:
   \[
   A = \frac{1}{2} \times (PQ + RS) \times \text{height} = \frac{1}{2} \times (1 - 2a + 1 - 2a) \times 1 = \frac{1}{2} \times (2 - 4a) = 1 - 2a
   \]

   Since \(0 \leq a \leq \frac{1}{2}\), the area ranges from \(1\) to \(0\).

The final answer is \(\boxed{1 - 2a}\).