17. From a square with a side length of 20, remove a rectangle with an area of 36, where the two side lengths of the rectangle are integers, and one side of the rectangle is part of one side of the square. The maximum perimeter of the remaining shape is $\qquad$.
【Answer】116.
【Analysis】Key point: Skillfully finding the perimeter. To maximize the perimeter, a rectangle should be cut out from the middle of the square (allowing only one side of the rectangle to coincide with the square). According to the translation method, the perimeter of the remaining shape is the perimeter of the square plus two remaining line segments. To maximize the perimeter, the remaining line segments must be as long as possible. $36=2 \times 18=3 \times 12=4 \times 9=6 \times 6$. The remaining line segments can be as long as two 18s, so the maximum perimeter is $20 \times 4+2 \times 18=116$.