Perimeter

Consider a long, thin rectangle measuring 10 cm by 1 cm. It has an area of 10 cm² and a perimeter of 22 cm. Now consider a square with side length 4 cm. It has an area of 16 cm² and a perimeter of 16 cm. The rectangle has a smaller area (10 cm² vs 16 cm²) but a larger perimeter (22 cm vs 16 cm). This directly counters the “bigger area means bigger perimeter” misconception.

This works because perimeter measures the distance around the outside, while area measures the space inside. Thin, elongated shapes have lots of edge relative to their area. In fact, for a fixed area, the shape with the smallest perimeter is always a circle — so any non-circular shape will have “extra” perimeter for its area.

When we cut a rectangular notch from the corner, we remove two outer edges but replace them with two inner edges of exactly the same lengths. Imagine a 6 cm × 4 cm rectangle with perimeter 2 × (6 + 4) = 20 cm. When we cut a 2 cm × 2 cm notch from a corner, we lose 2 cm of horizontal edge and 2 cm of vertical edge, but we gain 2 cm of new vertical edge going inward and 2 cm of new horizontal edge going inward. The perimeter is still 20 cm.

This surprises students because the shape looks more complicated, so they assume it must have a different (usually larger) perimeter. The key insight is that a corner notch replaces removed edges with new edges of identical total length — the perimeter only changes when a cut doesn’t “swap” edges equally (e.g. cutting from the middle of a side rather than a corner).

A 1 cm × 9 cm rectangle has perimeter 2 × (1 + 9) = 20 cm and area 9 cm². A 5 cm × 5 cm square also has perimeter 2 × (5 + 5) = 20 cm but area 25 cm². Both shapes have a perimeter of 20 cm, yet the square’s area is almost three times that of the thin rectangle.

This demonstrates that perimeter and area are independent measurements. Knowing the perimeter alone does not determine the area. Among all rectangles with a fixed perimeter, the square always has the greatest area — so moving away from a square toward a long, thin shape keeps the perimeter the same while dramatically reducing the area.

A perimeter of 24 cm means the total of all four sides is 24 cm, so \(l + w = 12\). This gives many possibilities: 11 cm and 1 cm, 10 cm and 2 cm, 7 cm and 5 cm, and so on. A 6 cm × 6 cm square is just one of the possibilities — it happens to be the special case where both dimensions are equal.

Students often assume that if the perimeter is 24, each side must be 24 ÷ 4 = 6 cm. This is the “perimeter ÷ 4 = side length” misconception, which only works for squares. For a general rectangle, the correct relationship is that half the perimeter equals the sum of the length and width, so there are infinitely many valid rectangles (including those with non-integer side lengths).

Example: 10 cm × 5 cm (perimeter = 2 × (10 + 5) = 30 cm)

Another: 12 cm × 3 cm (perimeter = 2 × (12 + 3) = 30 cm)

Creative: 14.5 cm × 0.5 cm (perimeter = 2 × (14.5 + 0.5) = 30 cm) — a very long, thin rectangle that students rarely consider because the dimensions are non-integers or extreme.

Trap: 15 cm × 15 cm — a student divides 30 ÷ 2 = 15 and makes both sides 15, but this gives perimeter 2 × (15 + 15) = 60 cm. This is the “halving once instead of twice” misconception. The correct step is 30 ÷ 2 = 15, which gives the semi-perimeter (length + width = 15), and then you choose a pair that sums to 15.

Example: A 5 cm × 5 cm square (perimeter = 4 × 5 = 20 cm)

Another: A 7 cm × 3 cm rectangle (perimeter = 2 × (7 + 3) = 20 cm)

Creative: An equilateral triangle with side length 6⅔ cm (perimeter = 3 × 6⅔ = 20 cm) — students often only think of rectangles when asked about perimeter.

Trap: A 4 cm × 5 cm rectangle — this has an area of 20 cm² but a perimeter of 2 × (4 + 5) = 18 cm. This exploits the “confusing perimeter and area” misconception, where a student sees “20” and produces a shape measured in square centimetres instead of centimetres.

Example: A 6 cm × 4 cm rectangle (perimeter 20 cm) and a 5 cm × 5 cm square (perimeter 20 cm)

Another: A 7 cm × 3 cm rectangle (perimeter 20 cm) and an equilateral triangle with sides of 6⅔ cm (perimeter 20 cm)

Creative: A regular pentagon with sides of 4 cm (perimeter 20 cm) and a 9 cm × 1 cm rectangle (perimeter 20 cm) — different numbers of sides, very different shapes, same perimeter.

Trap: A 6 cm × 4 cm rectangle (perimeter 20 cm, area 24 cm²) and a 3 cm × 8 cm rectangle (perimeter 22 cm, area 24 cm²) — these have the same area (24 cm²) but different perimeters (20 vs 22). This traps students who confuse “same perimeter” with “same area”.

Example: An L-shape where the full dimensions are 8 cm wide and 6 cm tall, with a 3 cm × 2 cm notch cut from one corner. The missing side along the step is 8 − 3 = 5 cm.

Another: A T-shape formed by a 10 cm × 2 cm horizontal bar on top of a 4 cm × 3 cm vertical bar, centred. Missing sides on either side of the step are (10 − 4) ÷ 2 = 3 cm each.

Creative: A staircase shape with two steps — every horizontal missing length can be found by subtracting from the total width, and every vertical missing length from the total height. This tests whether students can apply the “missing side” strategy repeatedly.

Trap: A shape where a student simply adds all the labelled dimensions without realising a side is missing. For instance, if an L-shape labels five of its six sides, the student adds only those five and gets the wrong answer. This targets the “only add the numbers you can see” misconception — students must recognise that unmarked sides still contribute to the perimeter.

Example: A rectangle with side lengths \(2x\) and \(6\) (perimeter = \(2(2x + 6) = 4x + 12\)).

Another: A rectangle with side lengths \(x + 4\) and \(x + 2\).

Creative: A rectangle with side lengths \(2x + 5.5\) and \(0.5\) — students rarely mix algebraic expressions with decimal constants.

Trap: A rectangle with sides \(4x\) and \(12\) — the student forgets to halve the perimeter before choosing the lengths, resulting in a perimeter of \(8x + 24\).

A regular hexagon with side length 1 cm has perimeter 6 cm (6 sides), while a rectangle measuring 10 cm × 5 cm has perimeter 30 cm (4 sides). Here the shape with fewer sides has the much larger perimeter. But a hexagon with side length 10 cm has perimeter 60 cm compared to a triangle with side length 3 cm (perimeter 9 cm) — here the shape with more sides has the larger perimeter.

The perimeter depends on both the number of sides and the length of each side, so the number of sides alone doesn’t determine which perimeter is greater. This targets the “more sides means bigger perimeter” misconception.

A rectangle has four sides. Even the thinnest possible rectangle has two long sides and two short sides (which must have positive length). The perimeter = 2 × length + 2 × width. Since both length and width are positive, the perimeter is at least 2 × length + something positive, which is always more than the longest side.

For example, a 100 cm × 0.1 cm rectangle has a longest side of 100 cm and a perimeter of 200.2 cm. Students might try to “break” this with an extremely thin rectangle, but the perimeter will always exceed the longest side because there are always at least three additional side lengths to add.

If you cut a corner notch from a rectangle (cut in from a corner at right angles), the perimeter stays the same — the removed outer edges are replaced by new inner edges of the same total length. If you cut a thin strip entirely off one end of a rectangle, the perimeter gets smaller — you’ve genuinely removed edge. But if you make a V-shaped cut into one side of a shape, the perimeter actually increases — you remove a small section of edge and replace it with two longer diagonal edges.

So cutting can make the perimeter smaller, the same, or even larger depending on the type of cut. The key misconception is that “removing area always reduces perimeter” — area and perimeter do not always change in the same direction.

If the original rectangle has length l and width w, the perimeter is 2(l + w). Doubling gives new dimensions 2l and 2w, so the new perimeter is 2(2l + 2w) = 2 × 2(l + w) — exactly double the original. For example, a 5 × 3 rectangle has perimeter 16 cm; a 10 × 6 rectangle has perimeter 32 cm = 2 × 16.

Students who confuse perimeter with area often fall for the “doubling sides quadruples everything” misconception. Doubling sides does quadruple the area (from lw to 4lw), but perimeter is a linear measure, so it only doubles. This is a powerful way to test whether students understand the difference between one-dimensional and two-dimensional scaling.

The student has made the “adding length + width only” misconception error — they found half the perimeter instead of the full perimeter. A rectangle has four sides: two lengths and two widths. The student only counted one of each.

The correct calculation is perimeter = 2 × (9 + 4) = 2 × 13 = 26 cm (or equivalently: 9 + 4 + 9 + 4 = 26 cm). This error often arises when students learn “add the length and width” and forget the crucial “then double” step.

The answer happens to be correct — the perimeter of a 6 cm × 3 cm rectangle is indeed 2 × (6 + 3) = 18 cm. But the student used multiplication (the area formula) instead of the perimeter formula. They got lucky because 6 × 3 happens to equal 2 × (6 + 3) for this specific rectangle.

This is the “perimeter means multiply” misconception that coincidentally produces the correct answer here. If the dimensions were 8 cm × 3 cm, the student would get 24 instead of the correct 22 — the method fails for almost every other rectangle. The correct answer doesn’t validate the faulty reasoning.

The student has made the “only add the labelled numbers” misconception error. The L-shape has six sides, but only four measurements were given on the diagram. The right side of the shape measures 8 − 3 = 5 cm, and the top of the shape measures 10 − 4 = 6 cm. The student needed to calculate these missing sides before adding.

The correct perimeter is 10 + 5 + 4 + 3 + 6 + 8 = 36 cm. Interestingly, this equals the perimeter of the full 10 × 8 rectangle (2 × (10 + 8) = 36 cm), because the removed corner swaps equal lengths of edge — but students shouldn’t rely on this shortcut until they understand why it works.

The student has applied the “perimeter = 4 × side” misconception — using the square/rectangle formula for a shape that isn’t a quadrilateral. A regular hexagon has 6 equal sides, not 4. The correct perimeter is 6 × 4 = 24 cm.

This error comes from over-generalising the formula for a square (perimeter = 4 × side) to all shapes. For any regular polygon, perimeter = number of sides × side length. Students need to identify how many sides the shape has before choosing a formula.

The student has made the “ignoring mixed units” misconception. They added cm and mm together directly without converting.

Before any calculations can happen, the units must be matched. 20 mm is equal to 2 cm. Using the correct matching units, the calculation is perimeter = 2 × (5 + 2) = 14 cm (or alternatively, 140 mm).