The figure below was made by gluing together 5 non-overlapping congruent squares. The figure has area 45. Find the perimeter of the figure.

[center][img]https://snag.gy/ZeKf4q.jpg[/center][/img]
1. **Determine the area of each square:**
   The total area of the figure is given as \(45\) square units. Since the figure is composed of 5 congruent squares, the area of each square is:
   \[
   \text{Area of each square} = \frac{45}{5} = 9 \text{ square units}
   \]

2. **Find the side length of each square:**
   The area of a square is given by the square of its side length. Let \(s\) be the side length of each square. Then:
   \[
   s^2 = 9 \implies s = \sqrt{9} = 3 \text{ units}
   \]

3. **Identify the perimeter of the figure:**
   To find the perimeter, we need to sum the lengths of all the outer edges of the figure. The figure is composed of 5 squares arranged in a specific pattern. We will label the vertices as given in the problem and calculate the perimeter by summing the lengths of the outer edges.

4. **Calculate the lengths of the outer edges:**
   - The horizontal segments are \(AB\), \(BC\), \(CD\), \(DE\), \(EF\), \(FG\), \(GH\), \(HI\), \(IJ\), \(JK\), \(KL\), \(LM\), and \(MA\).
   - Each of these segments corresponds to the side length of a square, which is \(3\) units.

5. **Sum the lengths of the outer edges:**
   - The segments \(AB\), \(BC\), \(CD\), \(DE\), \(EF\), \(FG\), \(GH\), \(IJ\), \(JK\), \(KL\), and \(LM\) each have a length of \(3\) units.
   - The segments \(HI\) and \(MA\) are vertical segments that connect the top and bottom of the figure. Since the figure is composed of 5 squares, the total height of the figure is \(3 \times 3 = 9\) units. However, the segments \(HI\) and \(MA\) are not full heights but rather the difference between the total height and the height of one square, which is \(6\) units.

6. **Adjust for overlapping segments:**
   - The segments \(HI\) and \(MA\) are not full heights but rather the difference between the total height and the height of one square, which is \(6\) units.
   - Therefore, the total length of the outer edges is:
   \[
   \mathcal{P} = 11 \times 3 + 3 = 33 + 3 = 36 \text{ units}
   \]

The final answer is \(\boxed{36}\) units.