Congruent Rectangles - Examples, Exercises and Solutions

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

$8\times4=32$

Therefore, the rectangles are 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.

By definition congruent rectangles are rectangles that have the same area and the same perimeter. 

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.

Since JI intersects AH and divides it into two identical parts, we can claim that:

$AI=IH=BJ=JG=3$

And since it is given that:

$AB=HG=5$

Rectangles ABJI and IJGH are equal and overlapping.

Let's look at rectangle ADEH where AH equals 6, therefore:

$AH=DE=6$

From this it follows that:

$AH=CF=DE=6$

It is also given that:

$BC=CD=GF=FE=2.5$

Therefore we can claim that rectangles BCFG and CDEF are equal and overlapping.

Since side BG divides sides HE and AD into two equal parts:

$AB=BD=HG=GE=5$

And since:

$AH=BG=DE=6$ we can claim that rectangles BDEG and ABGH are equal and overlapping.