Are Rectangles with Perimeter 20 cm and Equal Area Congruent?

To determine if the two rectangles are congruent, we start by understanding that two rectangles are congruent if they have identical lengths and widths. In this problem, both rectangles have a perimeter of 20 cm and identical areas, which suggests they could potentially be congruent.

Let's recall the formulas:
Perimeter of a rectangle: $P = 2(l + w)$
Area of a rectangle: $A = l \times w$

Given that each rectangle has a perimeter $P = 20$, we can write:
$2(l_A + w_A) = 20$ for Rectangle A,
$2(l_B + w_B) = 20$ for Rectangle B,
which simplifies to:
$l_A + w_A = 10$,
$l_B + w_B = 10$.

The identical area condition gives us:
$l_A \times w_A = l_B \times w_B$.

Given that both the sums of $l$ and $w$ (using perimeter) and their products (using area) are equal, this enforces that $l_A = l_B$ and $w_A = w_B$.

This implies that the rectangles are congruent (i.e., have identical lengths and widths).

Therefore, the solution to the problem is Yes.