Two $5\times1$ rectangles have 2 vertices in common as on the picture. 
(a) Determine the area of overlap
(b) Determine the length of the segment between the other 2 points of intersection, $A$ and $B$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=290[/img]
### Part (a): Determine the area of overlap

1. **Identify the variables and relationships:**
   - Let \( m \) be the length of the segments \( BQ = AK = AR = BP \).
   - Let \( n \) be the length of the segments \( AL = BL = BN = AN \).
   - Given that \( m + n = 5 \).

2. **Apply the Law of Cosines:**
   - Consider the triangle \( \triangle LAN \):
     \[
     LN^2 = AN^2 + AL^2 - 2 \cdot AN \cdot AL \cdot \cos(\angle LAN)
     \]
   - Since \( \angle LAN = 180^\circ - \angle LAK \), we have:
     \[
     \cos(180^\circ - \angle LAK) = -\cos(\angle LAK)
     \]
   - Substituting \( \cos(\angle LAK) = \frac{m}{n} \):
     \[
     LN^2 = 2n^2 - 2n^2 \cdot \left(-\frac{m}{n}\right) = 2n^2 + 2mn
     \]

3. **Solve for \( n \):**
   - Given \( LN = 5 \):
     \[
     5^2 = 2n^2 + 2mn
     \]
     \[
     25 = 2n^2 + 2mn
     \]
   - Using \( m + n = 5 \):
     \[
     n = \frac{-m + \sqrt{m^2 + 52}}{2}
     \]
   - Substitute \( n \) back into the equation:
     \[
     \frac{-m + \sqrt{m^2 + 52}}{2} + m = 5
     \]
     \[
     m^2 + 52 = m^2 - 20m + 100
     \]
     \[
     20m = 48 \implies m = 2.4
     \]

4. **Calculate the area of overlap:**
   - The overlap area is the sum of the areas of triangles \( \triangle QLB \) and \( \triangle RAN \):
     \[
     \text{Area} = \frac{1}{2} \cdot BQ \cdot AL + \frac{1}{2} \cdot AR \cdot BN = m = 2.4
     \]

### Part (b): Determine the length of the segment between the other 2 points of intersection, \( A \) and \( B \)

1. **Calculate \( n \):**
   - Using \( n = \sqrt{m^2 + 1} \):
     \[
     n = \sqrt{2.4^2 + 1} = \sqrt{5.76 + 1} = \sqrt{6.76} \approx 2.6
     \]

2. **Apply the Law of Cosines to find \( AB \):**
   - Consider the triangle \( \triangle ALB \):
     \[
     AB^2 = AL^2 + BQ^2 - 2 \cdot AL \cdot BQ \cdot \cos(\angle ALB)
     \]
   - Since \( \angle ALB = \angle LAK \):
     \[
     AB^2 = 2n^2 - 2n^2 \cdot \cos(\angle LAK)
     \]
     \[
     AB^2 = 2n^2 - 2n^2 \cdot \frac{m}{n}
     \]
     \[
     AB^2 = 2n^2 - 2mn
     \]
   - Substitute \( n \) and \( m \):
     \[
     AB^2 = 2 \cdot 2.6^2 - 2 \cdot 2.6 \cdot 2.4
     \]
     \[
     AB^2 = 2 \cdot 6.76 - 2 \cdot 6.24
     \]
     \[
     AB^2 = 13.52 - 12.48 = 1.04
     \]
     \[
     AB = \sqrt{1.04} = \frac{\sqrt{26}}{5}
     \]

The final answer is \( \boxed{\frac{\sqrt{26}}{5}} \)