Congruent Rectangles: Are the rectangles congruent?

We can see that the length is identical in both rectangles: 3=3.

However their widths are not equal, as one is 2 while the other is 4.

Therefore, the rectangles are not congruent.

To determine whether the rectangles are congruent, we need to understand what congruence means for geometric figures.

Definition of Congruent Rectangles:
Two rectangles are congruent if and only if they have exactly the same dimensions. This means they must have the same length and the same width. Congruent figures can be placed on top of each other through rigid motions (translation, rotation, reflection) and match perfectly.

Key Observation:
An important property of congruent figures is that they must have equal areas. While equal areas don't guarantee congruence for rectangles, different areas guarantee that the rectangles are NOT congruent.

Analysis of the Given Rectangles:
From the diagram, we can see:

• Rectangle 1 has area $A = 24$
• Rectangle 2 has area $A = 20$

Conclusion:
Since the two rectangles have different areas ($24 \neq 20$), they cannot possibly have the same dimensions. Therefore, the rectangles are not congruent.

The answer is: No

Since there are two pairs of sides that are equal, they also have the same area:

$8\times4=32$

Therefore, the rectangles are congruent.

Note that DC divides AE into two unequal parts.

AC=5 while CE=4

The area of rectangle ABDC is equal to:

$5\times2=10$

The area of rectangle CDGE is equal to:

$4\times2=8$

Therefore, the rectangles do not overlap.

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.