Perimeter of polygons — regular and irregular: Worked examples

This page is the "battle drill" for the parent topic. We are going to hunt down every kind of question perimeter can throw at you and solve one of each — regular, irregular, mixed-unit, "find the missing side", degenerate shapes, a word problem, and a nasty exam twist. Before each solution you make a Forecast (a guess), because guessing first is how you find out what you actually understand.

The scenario matrix

Every perimeter question falls into one of these case classes. The last column names the worked example that lands in that cell.

Cell	Case class	What makes it tricky	Covered by
A	Regular polygon, side given	none — pure shortcut $P = n \cdot s$	Example 1
B	Irregular polygon, all sides given	must add each one, no shortcut	Example 2
C	Mixed units	convert BEFORE adding	Example 3
D	Missing side found by Pythagorean theorem	perimeter needs a length you must compute	Example 4
E	Reverse question (perimeter → side)	undo the formula, divide	Example 5
F	Degenerate / zero input	a "side" of length 0, or a flattened shape	Example 6
G	Composite / real-world word problem	strip the story down to boundary steps	Example 7
H	Exam twist (regular polygon, perimeter given, find $n$ )	two unknowns hiding in one formula	Example 8

The tools we lean on and why each is chosen:

Addition — because perimeter is defined as a sum of side lengths (Cell B, C, G).
Multiplication — because "add the same length $n$ times" is multiplication (Cell A, H).
Division — because it undoes multiplication, so it answers "which side/how many sides gives this perimeter?" (Cell E, H).
Pythagorean theorem — because it is the one tool that turns two known legs of a right triangle into the unknown third side, which we then feed into the sum (Cell D).

Cell A — Regular polygon, side given

Worked example Example 1 — Regular octagon

A regular octagon has each side $s = 9$ cm. Find the perimeter.

Forecast: An octagon has 8 sides. Multiply — do you expect roughly 70? 80? Write your guess.

Step 1 — Count the sides. $n = 8$ . Why this step? The word "octagon" means 8 sides — "oct" like octopus. In the picture all 8 edges are equal white segments, so we are allowed to use the shortcut.

Step 2 — Confirm "regular". The problem says regular, so every side is the same length $s$ . Why this step? The multiplication shortcut $P = n \cdot s$ is only legal when all sides are equal. Checking this is not optional.

Step 3 — Apply the formula. $P = n \cdot s = 8 \times 9 = 72 cm$ Why this step? Multiplying by 8 is a compact way of writing $9 + 9 + 9 + 9 + 9 + 9 + 9 + 9$ .

Verify: Add the long way: $9 \times 8$ . Split as $9 \times 8 = (10 - 1) \times 8 = 80 - 8 = 72$ . ✓ Units stay cm (a length), not cm². ✓

Cell B — Irregular polygon, all sides given

Worked example Example 2 — Irregular pentagon

A five-sided plot has sides $12$ m, $7$ m, $15$ m, $9$ m, $11$ m. Find its perimeter.

Forecast: Five different lengths, none equal. Can you multiply by 5? Guess yes or no, then a rough total.

Step 1 — Reject the shortcut. These are five different numbers. Why this step? $P = n \cdot s$ needs a single shared $s$ . There is no such $s$ here, so multiplication would be nonsense (multiply by which side?).

Step 2 — Add each side once, in order around the boundary. $P = 12 + 7 + 15 + 9 + 11$ Why this step? Walking the boundary passes each side exactly once; the sum is the total walk.

Step 3 — Group to add safely. $(12 + 15 + 11) + (7 + 9) = 38 + 16 = 54$ m. Why this step? Pairing numbers into round-ish chunks lowers the chance of an arithmetic slip.

Verify: Re-add in a different order: $12 + 7 = 19$ , $19 + 15 = 34$ , $34 + 9 = 43$ , $43 + 11 = 54$ . Same answer $54$ m. ✓

Cell C — Mixed units

Worked example Example 3 — Rectangle with mixed units

A rectangle has length $1.2$ m and width $80$ cm. Find its perimeter.

Forecast: Danger — one number is in metres, one in centimetres. Guess: what goes wrong if you just add $1.2 + 80$ ?

Step 1 — Spot the unit clash. $1.2$ m and $80$ cm are different rulers. Why this step? Adding $1.2 + 80$ would mean "1.2 metre-steps plus 80 cm-steps", mixing two different step sizes — meaningless. See Units of measurement.

Step 2 — Convert to one unit. $1$ m $= 100$ cm, so $1.2$ m $= 120$ cm. Why this step? Now everything is measured in the same centimetre-ruler, so addition is honest.

Step 3 — A rectangle is regular in pairs: two lengths, two widths. $P = 2 \times (length) + 2 \times (width) = 2 (120) + 2 (80)$ Why this step? Opposite sides of a rectangle are equal (see Quadrilaterals), so we can multiply each pair by 2 instead of adding four separate sides.

Step 4 — Compute. $240 + 160 = 400$ cm $= 4$ m. Why this step? Finishing the arithmetic; giving the answer in metres too makes it human-readable.

Verify: $2 (120 + 80) = 2 (200) = 400$ cm. ✓ And $400$ cm $\div 100 = 4$ m. ✓

Cell D — Missing side via the Pythagorean theorem

Worked example Example 4 — Right triangle with a hidden side

A right-angled triangle has legs $6$ cm and $8$ cm. Find its perimeter.

Forecast: You are only given TWO sides but a triangle has three. Guess the third side, then the perimeter.

Step 1 — Realise a side is missing. Perimeter needs all three sides; we have only two. Why this step? You cannot add what you do not have. The third side (the hypotenuse) must be found first.

Step 2 — Choose the right tool. The two given sides meet at a right angle, so they are the legs of a right triangle → use the Pythagorean theorem $c^{2} = a^{2} + b^{2}$ . Why THIS tool? Pythagoras is the unique rule connecting the two legs to the slanted third side. No other tool converts two legs into a hypotenuse.

Step 3 — Compute the hypotenuse. $c = 6^{2} + 8^{2} = 36 + 64 = 100 = 10 cm$ Why this step? Squaring, adding, then square-rooting undoes the squaring to leave the raw length $c$ .

Step 4 — Now add all three sides. $P = 6 + 8 + 10 = 24 cm$ Why this step? With every side known, perimeter is the ordinary sum.

Verify: $6$ – $8$ – $10$ is just the famous $3$ – $4$ – $5$ triangle doubled, so it is a valid right triangle. ✓ $6 + 8 + 10 = 24$ cm. ✓

Cell E — Reverse question (perimeter → side)

Cell F — Degenerate / zero input

Worked example Example 6 — When a side shrinks to zero

A quadrilateral has sides $5$ cm, $0$ cm, $5$ cm, $6$ cm (one side has collapsed to nothing). What is its perimeter, and what has the shape become?

Forecast: A side of length $0$ — is that even a shape? Guess the perimeter and the resulting figure.

Step 1 — Add honestly, including the zero. $P = 5 + 0 + 5 + 6 = 16 cm$ Why this step? Adding $0$ changes nothing, but writing it keeps the "walk each side once" rule intact.

Step 2 — Interpret the geometry. A side of length $0$ means two corners have merged into one point. The four-sided figure has collapsed into a triangle with sides $5$ , $5$ , $6$ . Why this step? Understanding the limiting behaviour stops you from over-counting corners. This is the boundary case between a quadrilateral and a triangle.

Step 3 — Cross-check as a triangle. As a triangle: $5 + 5 + 6 = 16$ cm — identical. Why this step? The perimeter must not jump when a side smoothly shrinks to $0$ ; it doesn't, which confirms the answer is consistent.

Verify: $5 + 0 + 5 + 6 = 16$ and $5 + 5 + 6 = 16$ . Both give $16$ cm. ✓

Cell G — Real-world composite word problem

Worked example Example 7 — L-shaped garden fence

A garden is L-shaped. Reading around the boundary the sides are $10$ m, $4$ m, $6$ m, $3$ m, $4$ m, $7$ m. Fencing costs ₹ $50$ per metre. Find the total fencing cost.

Forecast: Strip away the money for a second — how many metres of fence, roughly? Then multiply by ₹50.

Step 1 — Translate the story into "boundary steps". Fencing surrounds the edge → this is a perimeter problem, not an area problem. Why this step? Word problems disguise the maths; naming the quantity ("perimeter") tells us to add sides, not multiply length×width.

Step 2 — Add every side once, walking the L. $P = 10 + 4 + 6 + 3 + 4 + 7 = 34 m$ Why this step? The L-shape is an irregular hexagon; no shortcut, so we sum all six edges.

Step 3 — Convert length to cost. $Cost = 34 m \times ₹50/ m = ₹1700$ Why this step? Cost scales with length, so multiply metres by rupees-per-metre; the "m" cancels leaving ₹.

Verify: Re-sum: $(10 + 7) + (6 + 4) + (4 + 3) = 17 + 10 + 7 = 34$ m. ✓ $34 \times 50 = 34 \times 5 \times 10 = 170 \times 10 = ₹1700$ . ✓

Cell H — Exam twist (find the number of sides)

Worked example Example 8 — Two unknowns, one formula

A regular polygon has perimeter $60$ cm and each side is $12$ cm long. How many sides does it have, and name the polygon.

Forecast: This time $n$ is the mystery. Which operation pulls $n$ out of $P = n \cdot s$ ?

Step 1 — Write the formula with the unknown. $P = n \cdot s$ , with $P = 60$ , $s = 12$ . Why this step? Placing knowns into the formula reveals $n$ as the single unknown.

Step 2 — Isolate $n$ with division. $n = \frac{P}{s} = \frac{60}{12} = 5$ Why THIS tool? $n$ is multiplied by $s$ , so dividing by $s$ strips $s$ away, leaving $n$ alone. Division is again the inverse of multiplication.

Step 3 — Name it. $5$ equal sides → a regular pentagon (see Triangles/Quadrilaterals for the smaller cases). Why this step? The count $n$ names the polygon; interpreting the number completes the answer.

Sanity guard: $n$ must be a whole number $\geq 3$ (you cannot have 2.5 sides). Here $n = 5$ ✓. Had division given a fraction, the data would be impossible.

Verify: Rebuild: $5 \times 12 = 60$ cm, matching the given perimeter. ✓

Recall Quick self-test across the matrix

Which cell needs the Pythagorean theorem, and why? ::: Cell D — the third side of a right triangle is unknown, and Pythagoras is the only tool that turns two legs into the hypotenuse before we can add. A regular polygon has $P = 60$ and $s = 12$ . How many sides? ::: $n = 60 \div 12 = 5$ (a pentagon) — divide because division undoes the multiplication in $P = n \cdot s$ . Why can't you add $1.2$ m and $80$ cm directly? ::: They use different rulers; convert to one unit first ( $1.2$ m $= 120$ cm) so the "steps" are the same size. When one side of a quadrilateral shrinks to $0$ , what happens to the shape? ::: Two corners merge and it becomes a triangle; the perimeter stays consistent ( $5 + 0 + 5 + 6 = 5 + 5 + 6 = 16$ ).