Two rectangles have the same area and the same perimeter. Does this mean that they are congruent to each other?
Answer: Yes. If the perimeter is 2p and the area is A, the sides satisfy l + b = p and lb = A. Then (l − b)2 = p2 − 4A is fixed, so l − b is fixed, and l, b are fixed. Same sides ⇒ congruent rectangles.
Step-by-step solution
Idea: A rectangle is fixed (up to position) by its two side lengths. Knowing the sum and the product of two numbers fixes the two numbers.
- Let one rectangle have sides l ≥ b. Same perimeter: l + b = p (half the perimeter). Same area: lb = A. The other rectangle’s sides satisfy the same two equations.1 mark
- (l − b)2 = (l + b)2 − 4lb = p2 − 4A, which is the same for both rectangles. So l − b = √(p2 − 4A) is the same for both.1 mark
- Knowing l + b and l − b fixes l = (l + b) + (l − b)2 and b. So both rectangles have the same length and the same breadth, hence they are congruent. (This is not true for other shapes: e.g. two different triangles can share area and perimeter.)1 mark
Yes. Equal perimeter and equal area fix the sum and the product of the sides, which fix the sides themselves; so the two rectangles are congruent.
Check: Area 24 and perimeter 20: l + b = 10, lb = 24 forces {6, 4} only ✓.
Answer to write in the exam
l + b = p, lb = A (same for both)
(l − b)2 = (l + b)2 − 4lb = p2 − 4A (same for both)
⇒ l − b same ⇒ l, b same
∴ Yes, the rectangles are congruent.
Common mistakes that cost marks
- Answering ‘no’ by thinking of a square and a rectangle; those have different perimeters or areas.
- Showing it for one example only. A general argument (sum and product) is needed.
How this can come in the exam
MCQ (1 mark)
A rectangle has perimeter 26 cm and area 40 cm2. Its sides are
- 10 cm and 4 cm
- 13 cm and 3 cm
- 20 cm and 2 cm
- 8 cm and 5 cm
Show answer
(D) 8 cm and 5 cm
l + b = 13 and lb = 40 give 8 and 5.
Try one yourself
A rectangle has perimeter 34 cm and area 60 cm2. Find its sides.
Show answer
l + b = 17, lb = 60: (l − b)2 = 289 − 240 = 49, l − b = 7: 12 cm and 5 cm.
More questions like this
- You know that the area of a parallelogram is base × height. Using this and the figure, show that the area of a trapezium is half the sum of the parallel sides × height, i.e., 12(a + b)h.
- By dividing a trapezium into two triangles show that its area is, half the sum of the parallel sides multiplied by the height (the same formula as the one given above).
- Show how we can use two identical copies of a trapezium to make a parallelogram. How will this give us the formula for the area of a trapezium?
- Show that the area of a kite is half the product of its diagonals. Show this:
- Three problems about fitting congruent shapes together: