1. The problem is to find the dimensions of a rectangle given that its perimeter is 56. 2. The formula for the perimeter $P$ of a rectangle with length $l$ and width $w$ is: $$P = 2(l + w)$$ 3. Given $P = 56$, substitute into the formula: $$56 = 2(l + w)$$ 4. Divide both sides by 2 to isolate $l + w$: $$\cancel{2}(l + w) = \cancel{2}28$$ $$l + w = 28$$ 5. This means the sum of the length and width is 28. Without additional information, the dimensions can be any pair $(l, w)$ such that $l + w = 28$. 6. For example, if $l = 15$, then $w = 28 - 15 = 13$. 7. Therefore, the rectangle's length and width satisfy $l + w = 28$ with perimeter 56.