下图所示的长方形 ACDF 由一个正方形和一个长方形组成。长方形 ACDF 的周长为 $(2y + 15)\text{ cm}$。正方形 ABEF 的周长为 16 cm。

1. Find the side length of the square ABEF.

Perimeter of square ABEF = $16\text{ cm}$
Side length of square = $16 \div 4 = 4\text{ cm}$
So, $AB = AF = FE = BE = 4\text{ cm}$.

2. Express the perimeter of rectangle BCDE using the given perimeter of ACDF.

Perimeter of ACDF = $AC + CD + DF + FA = (2y + 15)\text{ cm}$
Notice that $AC = AB + BC = 4 + BC$ and $DF = DE + EF = DE + 4$.
Also, $CD = BE = 4\text{ cm}$ and $FA = 4\text{ cm}$.
So, Perimeter of ACDF = $(4 + BC) + 4 + (DE + 4) + 4 = BC + DE + 16$.
Given $BC + DE + 16 = 2y + 15$, we have $BC + DE = 2y - 1$.

3. Calculate the perimeter of rectangle BCDE.

Perimeter of BCDE = $BC + CD + DE + EB$
We know $CD = 4\text{ cm}$ and $EB = 4\text{ cm}$.
Perimeter of BCDE = $(BC + DE) + 4 + 4 = (2y - 1) + 8 = 2y + 7\text{ cm}$.