Are Rectangles with Perimeter 20 cm and Equal Area Congruent?

Rectangle Congruence with Constraint Conditions

The perimeter of A is 20 cm.

The perimeter of B is also 20 cm.

The area of them is identical.

Are the rectangles congruent?

P=18P=18P=18P=18P=18P=18ab

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Step-by-step written solution

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1

Understand the problem

The perimeter of A is 20 cm.

The perimeter of B is also 20 cm.

The area of them is identical.

Are the rectangles congruent?

P=18P=18P=18P=18P=18P=18ab

2

Step-by-step solution

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) P = 2(l + w)
Area of a rectangle: A=l×w A = l \times w

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

The identical area condition gives us:
lA×wA=lB×wB l_A \times w_A = l_B \times w_B .

Given that both the sums of l l and w w (using perimeter) and their products (using area) are equal, this enforces that lA=lB l_A = l_B and wA=wB 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.

3

Final Answer

Yes

Key Points to Remember

Essential concepts to master this topic
  • Congruent Rectangles: Identical length and width dimensions make rectangles congruent
  • Constraint Analysis: Same perimeter (20 cm) and area forces identical dimensions
  • Verification: Check lA=lB l_A = l_B and wA=wB w_A = w_B from given conditions ✓

Common Mistakes

Avoid these frequent errors
  • Assuming different rectangles can't be congruent with same constraints
    Don't think rectangles with identical perimeter and area must be different = wrong conclusion! This ignores how constraints work together mathematically. Always analyze how multiple conditions like l+w=10 l + w = 10 and l×w=constant l \times w = \text{constant} force unique solutions.

Practice Quiz

Test your knowledge with interactive questions

Are the rectangles congruent?

A=20A=20A=20A=24A=24A=24

FAQ

Everything you need to know about this question

How can two different rectangles have the same perimeter AND area?

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They can't! When rectangles have both identical perimeter and area, the mathematical constraints force them to have exactly the same dimensions, making them congruent.

What if rectangles only have the same perimeter?

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Then they could be different! A 6×4 rectangle and a 7×3 rectangle both have perimeter 20, but different areas (24 vs 21). You need both conditions to guarantee congruence.

How do I prove rectangles are congruent mathematically?

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Show that their corresponding sides are equal. Use the constraints: lA+wA=lB+wB l_A + w_A = l_B + w_B and lA×wA=lB×wB l_A \times w_A = l_B \times w_B to prove lA=lB l_A = l_B and wA=wB w_A = w_B .

Could there be multiple rectangle solutions with perimeter 20?

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Yes for just perimeter! But when you add the identical area constraint, it becomes a unique solution. The two conditions together eliminate all other possibilities.

What does congruent actually mean for rectangles?

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Congruent rectangles are exactly the same size and shape - they have identical length and width. You could place one on top of the other and they'd match perfectly.

Why is this problem asking about congruence instead of just similarity?

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Because similarity only requires same proportions, while congruence requires identical actual dimensions. The specific perimeter and area values determine exact size, not just shape.

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