BTT Mathematics / Primary Mathematics Learning Hub / Tessellations and Tiling
A tessellation covers a region with repeated shapes without gaps or overlaps. A spatial construction puzzle asks whether pieces can be fitted under stated rules, and if so, how. The central questions are geometric: what angles meet, which edges coincide, what area is covered and which constraints remain after pieces are moved.
Four squares can meet around one vertex because four right angles total 360°. Six equilateral triangles can meet because 6×60°=360°. Three regular hexagons can meet because 3×120°=360°. A regular pentagon has interior angle 108°, and no whole number of those angles totals 360°, so regular pentagons alone cannot form an edge-to-edge regular tessellation around a point.
This guide develops angle conditions, repeated translations and rotations, unit-square constructions, domino tilings and simple colouring arguments. It is Primary Mathematics enrichment built on Geometry, Angles, Symmetry and Coordinates, Measurement, Units, Perimeter, Area and Volume and Nets and 3D Visualisation.
It is not a claim that every parity or colouring proof below is required in the Primary syllabus. Use the MOE Primary curriculum page and the learner’s actual school programme to separate required work from extension.
Angles around a point · Translate, rotate and reflect · Unit-square constructions · Domino tilings · Colouring arguments · 24 questions · Worked answers · Teaching and transfer
1. Tiles meeting at a vertex must fill 360°
For an edge-to-edge tessellation, the angles that meet around an interior point must total exactly 360°. A smaller total leaves a gap; a larger total forces overlap.
Worked example A: Squares
A square contributes 90° at a vertex. Four squares give 4×90°=360°, so they can meet exactly around a point.
Worked example B: Equilateral triangles
An equilateral triangle contributes 60°. Six such angles make 360°, so six triangles can meet around one point.
Worked example C: Regular hexagons
A regular hexagon has interior angle 120°. Three meet exactly because 3×120°=360°.
Worked example D: Regular pentagons
A regular pentagon contributes 108°. Three give 324°, leaving a 36° gap. Four give 432°, which overlaps. Therefore identical regular pentagons cannot meet edge-to-edge around an interior point with no gap.
Mixed polygons
A triangle, square, regular hexagon and another square can meet around one point because 60+90+120+90=360. A local angle fit is necessary, though a full infinite tessellation still requires the edge lengths and repeated arrangement to remain compatible.
2. Moving a tile can preserve its size and shape
A translation slides a shape without turning it. A rotation turns it about a point. A reflection flips it across a line. These rigid transformations preserve lengths, angles and area.
Translation builds repeating rows
A square tile shifted one side length horizontally can cover the neighbouring square position exactly. Repeating that translation constructs a row. Repeating vertically produces the familiar square-grid tessellation.
Rotation changes orientation, not congruence
A square rotated 90° about its centre coincides with itself. An equilateral triangle needs a 120° rotation to coincide with itself. A regular hexagon needs 60°.
Reflection can reverse handedness
An L-shaped triomino reflected across a vertical line becomes a mirror L. The piece remains congruent, but a puzzle may or may not allow flipping. Always read whether pieces may be rotated, reflected or translated only.
A construction can fail under “rotation only” rules but succeed if reflection is permitted. The allowed transformations are part of the puzzle, not decoration.
3. Shared edges change perimeter but not covered area
Joining unit squares edge to edge adds their areas but removes shared edges from the outside boundary. Two unit squares contain area two square units but perimeter six units, not eight, because the shared edge is internal.
Worked example E: Three in a row
Three unit squares in a 1×3 rectangle have area three and perimeter 2(1+3)=8 units.
Worked example F: L-triomino
Three unit squares arranged as a 2×2 square with one corner missing have area three. Their outside boundary has eight unit edges, so perimeter is 8 units.
Worked example G: Six-square rectangle
A 2×3 rectangle contains six unit squares. Its area is six square units and perimeter is 2(2+3)=10 units.
Removing a corner may preserve perimeter
Start with a 4×4 square of sixteen unit cells, perimeter sixteen. Remove one corner unit square. Two outside unit edges disappear, but two previously internal unit edges are exposed. The new fifteen-cell shape still has perimeter 16.
Compact arrangements hide more shared edges
Twelve unit squares in a 3×4 rectangle have perimeter fourteen. The same twelve squares in a 1×12 strip have perimeter twenty-six. Same area does not imply same perimeter.
4. A domino covers exactly two adjacent unit cells
A 1×2 domino covers two unit squares sharing an edge. A board can be tiled by dominoes only if its area contains an even number of unit cells, but even area alone does not guarantee a tiling.
Worked example H: A 4×4 board
A 4×4 board contains sixteen cells, so any domino tiling uses exactly eight dominoes. One construction places four horizontal dominoes in each pair of rows.
Worked example I: A 3×3 board
A 3×3 board has nine cells. Since every domino covers two, an exact tiling is impossible by area parity alone.
Worked example J: A 3×4 board
A 3×4 board has twelve cells and needs six dominoes. A construction exists: split the board into three 1×4 rows, each tiled by two horizontal dominoes.
Even area is necessary, not sufficient
A damaged board can have an even number of cells but still be impossible to tile because the remaining cells are arranged incompatibly. A colouring argument can reveal such a hidden obstruction.
5. Colouring can expose an invariant that every tile must obey
Colour a rectangular grid like a chessboard, alternating black and white so edge-adjacent cells have opposite colours. Every domino covers two adjacent cells and therefore covers exactly one black and one white cell.
Worked example K: Remove two same-colour corners
A 4×4 board has eight black and eight white cells. Opposite corners have the same colour. Remove both opposite corners of that colour. The remaining board has six cells of one colour and eight of the other.
If dominoes tiled the board, every domino would cover one of each colour, so the numbers covered would have to be equal. They are not. Therefore the damaged board cannot be domino-tiled, even though fourteen cells is an even area.
The colouring is a proof tool, not a physical requirement
The real board does not need to be painted. The imagined colouring reveals a property preserved by every legal domino placement.
A successful colour count does not prove a tiling exists
Equal numbers of black and white cells remove this particular obstruction, but another geometric obstruction may remain. A colouring argument can prove impossibility when the counts disagree; agreement is not automatically a construction.
Build a construction when claiming possibility
To show a 3×4 board is tileable, give an explicit arrangement of six dominoes. “The area is even” is only a necessary check.
6. Practice: 24 original questions
All regular polygons below have equal side lengths where they are intended to meet edge-to-edge. Unit-square boards are made from edge-adjacent unit cells.
Questions 1–8: Angles and transformations
1. A square contributes 90° at a vertex. How many squares fit around an interior point with no gap?
2. How many equilateral triangles of angle 60° fit around a point?
3. How many regular hexagons of angle 120° fit around a point?
4. Explain why regular pentagons of angle 108° cannot alone meet edge-to-edge around an interior point without gaps.
5. Three square angles meet around a point. What gap remains?
6. Two regular hexagon angles meet around a point. What angle remains to fill?
7. Can one equilateral triangle, two squares and one regular hexagon meet around a point? Check the angle sum.
8. Which smallest positive rotation maps a square onto itself: 45°, 90°, 120° or 180°?
Questions 9–16: Unit-square constructions and dominoes
9. Two unit squares share one full edge. Find the area and perimeter of the combined shape.
10. Three unit squares form a 1×3 rectangle. Find its perimeter.
11. Three unit squares form an L-triomino. Find its perimeter.
12. A 2×3 rectangle is tiled by unit squares. Find its area and perimeter.
13. A 4×4 unit-square board has one corner cell removed. Find the remaining area and perimeter.
14. How many 1×2 dominoes tile a complete 4×4 board?
15. Can a 3×3 board be tiled by dominoes? Give the quickest reason.
16. How many dominoes are needed for a 3×4 board, and is a tiling possible?
Questions 17–24: Coverage, symmetry and colouring
17. Twelve unit squares form a 3×4 rectangle. Find the perimeter.
18. The same twelve unit squares form a 1×12 strip. Find the perimeter and compare it with Question 17.
19. If every side length of a square tile is doubled, by what factor does its area change?
20. What is the rotational symmetry order of an equilateral triangle?
21. What is the rotational symmetry order of a regular hexagon?
22. A 4×4 checkerboard has two opposite same-colour corners removed. Can the remaining fourteen cells be tiled by dominoes? Explain using colours.
23. A board has equal numbers of black and white cells under checkerboard colouring. Does that alone prove it can be tiled by dominoes? Explain.
24. At one vertex, a regular triangle, square, regular hexagon and square meet in that order. Verify the angle condition for a local tessellation fit.
7. Worked answers
Answers 1–8
1. 4 squares. 360÷90=4.
2. 6 triangles. 360÷60=6.
3. 3 hexagons. 360÷120=3.
4. They do not fit exactly. Three angles total 324°, leaving 36°; four total 432°, causing overlap.
5. 90°. Three squares contribute 270°, so 360−270=90.
6. 120°. Two regular hexagons contribute 240°.
7. Yes locally. 60+90+90+120=360°.
8. 90°. A quarter-turn maps a square onto itself.
Answers 9–16
9. Area 2 square units; perimeter 6 units. The shared edge becomes internal.
10. 8 units. A 1×3 rectangle has perimeter 2(1+3).
11. 8 units. Trace the outside boundary of the L-triomino.
12. Area 6 square units; perimeter 10 units. Use 2×3 and 2(2+3).
13. Area 15; perimeter 16. Removing the corner loses two outside edges and exposes two new boundary edges.
14. 8 dominoes. Sixteen cells divided by two cells per domino.
15. No. Nine cells is odd, while every domino covers two.
16. 6 dominoes, and yes. Twelve cells need six; for example, tile each row with horizontal dominoes.
Answers 17–24
17. 14 units. Perimeter of a 3×4 rectangle is 2(3+4).
18. 26 units, which is 12 more. A 1×12 strip has perimeter 2(1+12)=26.
19. ×4. Doubling both perpendicular dimensions multiplies area by 2×2.
20. 3. Rotations of 120°, 240° and 360° return it to coincidence.
21. 6. The smallest positive rotational symmetry is 60°.
22. No. Removing two same-colour corners leaves unequal black and white counts, but every domino covers one of each.
23. No. Equal colour counts remove one possible obstruction but do not construct a tiling or rule out every other obstruction.
24. 360°. 60+90+120+90=360, so the angles fit locally around the point.
8. Teaching and transfer
If a learner says a shape tessellates because “it looks like it should,” ask what happens at one interior vertex. Angle sums provide a necessary local check that can be explained without relying on visual confidence.
Use physical tiles before abstract impossibility
Let a learner arrange squares, triangles or paper dominoes. Then record the angle or colour property that remained true in every arrangement. The proof grows from a visible invariant rather than appearing as a trick.
Separate area from boundary
Use two and three unit squares. Count area first, then trace perimeter. Shared edges explain why adding a tile increases area by one but may increase perimeter by only two or even leave it unchanged after a corner removal.
When “even area means tileable” appears
Use the damaged 4×4 checkerboard. Its area fourteen is even, yet the colouring proof shows impossibility. A necessary condition is not sufficient.
When a colouring proof is treated as a construction
Reverse the lesson: give a board with equal black and white counts and ask the learner to actually place dominoes. Passing the invariant test means “not ruled out by this test,” not “proved possible.”
Continue through the puzzle-enrichment collection
For digit constraints, use Cryptarithms, Alphametics and Digit Puzzles. For balanced numerical structures, use Magic Squares, Number Grids and Balanced Sums. For best-possible spatial and numerical constructions, continue to Optimisation, Minimum Moves and Efficient Constructions.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Colourings and transformations are proof tools under the stated tile rules. No official examination requirement is implied.