Answer: 29.

Solution. The side of the largest square (with vertex $A$) is greater than the side of the second largest square (with vertex $C$) by the length of segment $A B$, which is 11. Similarly, the side of the second largest square is greater than the side of the third largest square (with vertex $E$) by the length of segment $C D$, which is 5. And its side is greater than the side of the smallest square by the length of segment $E F$, which is 13. In total, the side of the largest square is greater than the side of the smallest square by the length of segment $G H$, which is $11+5+13=29$.
Problem 6.5. The figure shows 4 squares. It is known that the length of segment $A B$ is 11, the length of segment $F E$ is 13, and the length of segment $C D$ is 5. What is the length of segment $G H$?

![](https://cdn.mathpix.com/cropped/2024_05_06_c49ecf27446af41a8bcdg-17.jpg?height=500&width=464&top_left_y=927&top_left_x=495)