The intersection of two squares with perimeter $8$ is a rectangle with diagonal length $1$. Given that the distance between the centers of the two squares is $2$, the  perimeter of the rectangle can be expressed as $P$. Find $10P$.
1. **Determine the side length of the squares:**
   The perimeter of each square is given as \(8\). Since the perimeter \(P\) of a square with side length \(s\) is \(P = 4s\), we have:
   \[
   4s = 8 \implies s = 2
   \]

2. **Set up the equations for the dimensions of the rectangle:**
   Let \(x\) and \(y\) be the dimensions of the rectangle formed by the intersection of the two squares. The diagonal of this rectangle is given as \(1\). Using the Pythagorean theorem, we have:
   \[
   x^2 + y^2 = 1
   \]

3. **Use the distance between the centers of the squares:**
   The distance between the centers of the two squares is given as \(2\). Construct a right triangle where the hypotenuse is the segment connecting the centers of the two squares, and the legs are \(2 - x\) and \(2 - y\). Applying the Pythagorean theorem to this triangle, we get:
   \[
   (2 - x)^2 + (2 - y)^2 = 2^2
   \]
   Simplifying the left-hand side:
   \[
   (2 - x)^2 + (2 - y)^2 = 4 - 4x + x^2 + 4 - 4y + y^2 = x^2 + y^2 - 4x - 4y + 8
   \]
   Since \(x^2 + y^2 = 1\), substitute this into the equation:
   \[
   1 - 4x - 4y + 8 = 4
   \]
   Simplify to find \(x + y\):
   \[
   9 - 4x - 4y = 4 \implies 4x + 4y = 5 \implies x + y = \frac{5}{4}
   \]

4. **Calculate the perimeter of the rectangle:**
   The perimeter \(P\) of a rectangle is given by:
   \[
   P = 2(x + y)
   \]
   Substituting \(x + y = \frac{5}{4}\):
   \[
   P = 2 \left( \frac{5}{4} \right) = \frac{10}{4} = \frac{5}{2}
   \]

5. **Find \(10P\):**
   \[
   10P = 10 \left( \frac{5}{2} \right) = 25
   \]

The final answer is \(\boxed{25}\).