Shanti 沿着图 1 所示的虚线剪开一张正方形纸片 ABCD，得到一个面积为 225 cm^2 的小正方形和 8 个完全相同的直角三角形。她移走了其中 4 个这样的三角形，并将它们拼成了一个长方形 PQRS，如图 2 所示。

1. Express the dimensions of the large square and the rectangle in terms of the legs of the right-angled triangle.

Let the legs of the right-angled triangle be $x$ and $y$, where $y$ is the vertical side (QR) and $x$ is the horizontal side of each small rectangle making up PQRS.
The side length of the large square ABCD is $x + y$.
The rectangle PQRS is made of two $x$ by $y$ rectangles side-by-side, so its width is $2x$ and its height is $y$.

2. Form an equation using the given difference in perimeters to find QR.

Perimeter of ABCD $= 4(x + y) = 4x + 4y$
Perimeter of PQRS $= 2(2x + y) = 4x + 2y$
Difference in perimeter $= (4x + 4y) - (4x + 2y) = 2y$
Given the difference is $56$ cm:
$2y = 56$
$y = 28$
Since $y$ is the length of QR, $QR = 28$ cm.