Think of various rectangles with perimeter 40 units (the sides do not have to be integers).
- 1. How many such rectangles are there?
- 2. Among them, is there one whose area is the largest? What are its dimensions?
- 3. Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?
Answer: 1. Infinitely many (length + breadth = 20). 2. Yes: the 10 × 10 square, area 100. 3. No smallest: thin rectangles have areas as close to 0 as we like, but never 0.
Step-by-step solution
Idea: Perimeter 40 means length + breadth = 20. Write the area as 100 − (l − 10)2 to see where it is biggest.
1. How many such rectangles are there?
- Length + breadth = 40 ÷ 2 = 20. The length can be any number between 0 and 20 (e.g. 1 and 19, 2.5 and 17.5, 7.25 and 12.75), so there are infinitely many such rectangles.1 mark
Infinitely many
2. Among them, is there one whose area is the largest? What are its dimensions?
- Area = l(20 − l) = 100 − (l − 10)2, because (l − 10)2 = l2 − 20l + 100.1 mark
- (l − 10)2 ≥ 0, so the area is at most 100, and it equals 100 only when l = 10. Yes: the 10 × 10 square, area 100 sq. units.½ mark
Yes: the square 10 × 10 (area 100)
3. Among all these rectangles, is there one whose area is the smallest? What are its dimensions? Do either of these answers come as a surprise to you?
- No smallest one. 19 × 1 has area 19, 19.9 × 0.1 has area 1.99, 19.99 × 0.01 has 0.1999 The area can be made as small as we like, but a breadth of 0 is not a rectangle, so no rectangle has the least area.1 mark
- Surprises: the biggest is a square (it is the ‘most balanced’ rectangle), and the same perimeter can enclose almost no area at all.½ mark
No smallest area (it can be as close to 0 as we like)
1. Infinitely many. 2. The largest area is 100 sq. units, for the 10 × 10 square. 3. There is no smallest area: it can be made as close to 0 as we like. It is surprising that the best rectangle is a square and that the same perimeter can enclose almost nothing.
Check: 10 × 10 = 100 beats nearby rectangles: 9 × 11 = 99 and 9.5 × 10.5 = 99.75 ✓.
Answer to write in the exam
1.
l + b = 20, 0 < l < 20
∴ Infinitely many rectangles
2.
A = l(20 − l) = 100 − (l − 10)2
(l − 10)2 ≥ 0 ⇒ A ≤ 100
∴ Largest area 100 when l = b = 10 (a square)
3.
19 × 1 ⇒ 19; 19.9 × 0.1 ⇒ 1.99; 19.99 × 0.01 ⇒ 0.1999
Area → 0 but b = 0 is not allowed
∴ No rectangle has the smallest area.
Common mistakes that cost marks
- Counting only whole-number rectangles (19 of them). The sides need not be integers.
- Thinking all rectangles with the same perimeter have the same area.
- Saying 20 × 0 is the smallest rectangle: a side of 0 does not make a rectangle.
How this can come in the exam
MCQ (1 mark)
Among all rectangles with perimeter 36 cm, the greatest area is
- 72 cm2
- 80 cm2
- 324 cm2
- 81 cm2
Show answer
(D) 81 cm2
Length + breadth = 18; the square 9 × 9 gives the greatest area, 81 cm2.
Try one yourself
A farmer has 60 m of fencing for a rectangular pen. What are the dimensions of the pen with the largest area, and what is that area?
Show answer
l + b = 30; the square 15 m × 15 m gives 225 m2.
More questions like this
- Let us test Heron’s formula against some known cases: an equilateral triangle with side a units.
- Let us test Heron’s formula against some known cases: an isosceles triangle with equal sides a units and base 2b units.
- Let us test Heron’s formula against some known cases: a triangle with sides 3 units, 4 units and 5 units.
- In the same way we ask: can we find the area of a 4-gon if we only know the lengths of its sides? The figures below reveal the answer to this question. The problem (see the figure) is about a 4-gon whose sides are known to be 3, 3, 3, 3 (it is a ‘rhombus’). As you can see, the areas of the three figures are different. (We drew the figures using GeoGebra and found the areas using the ‘Area’ tool. Please try this exercise yourself, or by using four rods joined together at their ends.)
- Verify Brahmagupta’s formula for the case of a rectangle.