BTT Mathematics / Primary Mathematics Learning Hub / Geometric Dissections and Area Proofs
A geometric dissection cuts a shape into pieces and rearranges those pieces without stretching them. If every piece is preserved and the new arrangement has no gaps or overlaps, total area is unchanged. This simple conservation law can explain why formulas work, not merely how to use them.
Cut an 8 cm by 6 cm rectangle along a diagonal. The two pieces are congruent right triangles. The rectangle has area 48 cm², so each triangle has area 24 cm². The familiar triangle formula one-half × base × height is therefore visible as half of a rectangle in this case.
This guide is Primary Mathematics enrichment. It extends Measurement, Units, Perimeter, Area and Volume, Triangle Area and Composite Figures and Tessellations, Tiling and Spatial Construction Puzzles.
Some proof-style rearrangements below are optional enrichment rather than required Primary syllabus content. Use the MOE Primary curriculum page and the learner’s school programme for required scope.
Area conservation · Triangle formula · Parallelogram rearrangement · Composite shapes · Area versus perimeter · Check a rearrangement proof · 24 questions · Worked answers
1. Cutting changes boundaries, not the amount of surface
Imagine a paper rectangle. Cutting it creates more exposed edges, but it does not create or destroy paper. If the pieces are translated, rotated or reflected and then fitted without overlap, their areas still add to the original area.
Worked example A: Diagonal cut
An 8×6 rectangle has area 48. A diagonal divides it into two congruent triangles, so each has area 24 square units.
Worked example B: Two triangles re-form a rectangle
Two congruent right triangles each have perpendicular legs 3 and 4. Each triangle has area 1/2×3×4=6. Together they can form a 3×4 rectangle with area 12. The numerical equality checks the dissection.
Worked example C: Rearrange strips
A 7×5 rectangle has area thirty-five. Cut it into strips and rearrange every strip with no gaps or overlaps. Whatever new outline is formed, the total area remains 35 square units.
What can go wrong?
If pieces overlap, area is counted twice in the overlap. If a hidden gap appears, some area is missing from the new outline. If a piece is stretched, the transformation is no longer a rigid rearrangement. A valid dissection proof must rule out all three failures.
2. Triangle area can be proved by duplication
Take a triangle with base b and perpendicular height h. A congruent copy can be rotated and joined to form a parallelogram with the same base b and height h. The parallelogram area is bh, and the original triangle is half of the two-triangle figure. Therefore triangle area is 1/2 bh.
Worked example D: Base 12, height 7
Two copies form a parallelogram area 12×7=84. One triangle has area 42.
Worked example E: Reverse the formula
A triangle has area 36 and base 9. Since 1/2×9×h=36, 9h=72 and h=8.
Worked example F: Perpendicular height, not sloping side
A triangle may have a side of length ten but perpendicular height six to a base of eight. Its area is 1/2×8×6=24. The sloping side is irrelevant unless it is the required perpendicular height.
Worked example G: Same base and height
Two different triangles with base ten and perpendicular height four each have area twenty, even if their top vertices are horizontally shifted. The outline changes while base-height area stays fixed.
3. A parallelogram can be cut and shifted into a rectangle
Cut a right triangle from one sloping side of a parallelogram and move it to the opposite side. When the pieces fit exactly, the result is a rectangle with the same base and perpendicular height. Hence parallelogram area is base×height.
Worked example H: Base 9, height 5
The rearranged rectangle is 9×5, so parallelogram area=45.
Worked example I: Base 13, height 4
Area=52. The sloping side length does not replace the perpendicular height.
Worked example J: Why the cut works
The triangle removed from one side is translated to the other. No piece changes area. The new rectangle covers exactly the same total surface, so its easier formula transfers back to the parallelogram.
Optional extension: trapezium by pairing
Two congruent trapezia with parallel sides a and b and height h can be paired to make a parallelogram with base a+b and height h. Therefore one trapezium has area 1/2(a+b)h.
Worked example K: Parallel sides 6 and 10, height 4
Two copies form a parallelogram with base sixteen and height four, area sixty-four. One trapezium has area 32.
4. Composite-area problems are dissections in reverse
Instead of physically cutting a shape, we can imagine a large simple shape with a smaller piece removed or a set of simpler pieces joined.
Worked example L: Rectangle with corner removed
An 8×6 rectangle has area 48. Remove a 3×2 corner rectangle of area six. Remaining L-shape area=42.
Worked example M: Another subtraction
A 10×7 rectangle has area seventy. Remove a 4×3 rectangle of area twelve. Remaining area=58.
Worked example N: Add two triangles
Two non-overlapping triangles have areas eighteen and twenty-seven. If joined without overlap, total area=45, regardless of the new outer outline.
Worked example O: Rearrange an L-shape
If an L-shape is cut into two rectangles and those same rectangles are rearranged into another shape, the sum of their areas is invariant. The easiest proof is to compute the two piece areas before and after rather than trust appearance.
5. Area can stay fixed while perimeter changes
Cutting and rearranging can change which edges lie on the outside. Therefore area conservation does not imply perimeter conservation.
Worked example P: Same area, different perimeter
A 2×6 rectangle and a 3×4 rectangle both have area twelve. Their perimeters are 16 and 14. Equal area does not force equal perimeter.
Worked example Q: Twelve unit squares
Arrange twelve unit squares as a 1×12 strip: perimeter 26. Arrange them as a 3×4 rectangle: perimeter 14. Same twelve square units of area, very different boundary.
Worked example R: Why compactness matters
When unit squares share more edges internally, fewer unit edges remain exposed on the perimeter. Rearrangement can therefore reduce perimeter while leaving the number of unit squares unchanged.
Worked example S: A corner removal can preserve perimeter
From a 4×4 square, remove one corner unit square. Two original outer unit edges disappear, but two previously internal unit edges become exposed. The area falls from 16 to 15 while perimeter stays 16. Area and perimeter respond differently to the same cut.
6. A rearrangement picture needs mathematical checks
Some visual puzzles appear to create or lose area. The usual cause is a hidden gap, overlap, changed slope, or non-identical pieces. A valid area proof needs an invariant that can be audited.
Worked example T: Check total piece area
Suppose four pieces have areas 6,8,11 and15. Their total is forty. Any exact rearrangement of those same pieces without overlap must also have area 40. If a drawn new outline claims area 41, then the outline or fit is not exact.
Check corresponding lengths and slopes
Two slanted segments that look collinear may have different slopes. If they do, the apparent straight boundary is slightly bent and may hide a thin gap or overlap. Measurement or coordinate comparison can expose the difference.
Check every piece is used exactly once
A rearrangement is not valid if one piece is duplicated, omitted or mirrored when reflection is forbidden by the puzzle. The piece inventory is part of the proof.
Check the new figure’s dimensions independently
If pieces from a 12×8 rectangle are rearranged, total area is 96. A claimed exact rectangle of width 10 must therefore have height 9.6. If the drawing labels height ten, something is inconsistent.
7. Practice: 24 original questions
Questions 1–8: Conservation and formulas
1. An 8×6 rectangle is cut along a diagonal. Find the area of each triangle.
2. A 10×4 rectangle is cut along a diagonal. Find the area of each triangle.
3. A parallelogram has base 9 and perpendicular height 5. Find its area.
4. A triangle has base 12 and perpendicular height 7. Find its area.
5. A square of side 8 is cut along a diagonal. Find each triangle’s area.
6. A 7×5 rectangle is cut into pieces and exactly rearranged with no gaps or overlaps. What is the total area after rearrangement?
7. Can such a rearrangement change the perimeter? Answer yes or no and explain.
8. Two congruent right triangles have legs 3 and 4. What is their combined area?
Questions 9–16: Composite shapes and reverse problems
9. An 8×6 rectangle has a 3×2 corner removed. Find the remaining area.
10. A 10×7 rectangle has a 4×3 rectangle removed. Find the remaining area.
11. A parallelogram has base 13 and perpendicular height 4. Find its area.
12. A triangle has base 15 and perpendicular height 8. Find its area.
13. A triangle has area 36 and base 9. Find its perpendicular height.
14. A triangle has area 54 and base 12. Find its perpendicular height.
15. Two non-overlapping pieces have areas 18 and 27. They are joined without overlap. Find the combined area.
16. Two congruent triangles each have area 18 and are paired to form a parallelogram. Find the parallelogram area.
Questions 17–24: Proof and perimeter
17. A trapezium has parallel sides 6 and 10 and height 4. Use the two-copy rearrangement to find its area.
18. Explain why duplicating a triangle and joining the two copies can justify the formula 1/2×base×height.
19. Compare a 2×6 rectangle and a 3×4 rectangle. Find each area and perimeter.
20. Twelve unit squares form a 1×12 strip. Find its perimeter.
21. The same twelve unit squares form a 3×4 rectangle. Find its perimeter.
22. A 12×8 rectangle is cut and exactly rearranged. What total area must the new figure have?
23. Four pieces have areas 6,8,11 and15. An alleged gap-free, overlap-free rearrangement claims total area 41. What can you conclude?
24. Why must a “missing area” dissection puzzle check for hidden gaps, overlaps or non-collinear boundaries before claiming area has changed?
8. Worked answers
Answers 1–8
1. 24 square units each. Rectangle area 48, divided into two congruent triangles.
2. 20 square units each. Rectangle area 40.
3. 45 square units. 9×5.
4. 42 square units. 1/2×12×7.
5. 32 square units each. Square area 64, halved.
6. 35 square units. Exact cut-and-rearrangement preserves area.
7. Yes. Rearrangement can change which edges are exposed, so perimeter may change even when area does not.
8. 12 square units. Each triangle has area six.
Answers 9–16
9. 42 square units. 48−6.
10. 58 square units. 70−12.
11. 52 square units. 13×4.
12. 60 square units. 1/2×15×8.
13. 8 units. 1/2×9×h=36.
14. 9 units. 1/2×12×h=54 gives 6h=54.
15. 45 square units. Add the non-overlapping areas.
16. 36 square units. Two copies of area eighteen.
Answers 17–24
17. 32 square units. Two copies form a parallelogram base sixteen, height four, area sixty-four; halve it.
18. Two congruent copies make a parallelogram with area base×height. One original triangle occupies exactly half that area.
19. Both areas are 12. Perimeters: 2×6 gives 16; 3×4 gives 14.
20. 26 units. 2(1+12).
21. 14 units. 2(3+4).
22. 96 square units. 12×8; exact rearrangement preserves total area.
23. The claim is inconsistent. The pieces total 6+8+11+15=40, so a genuine gap-free, overlap-free rearrangement of those same pieces cannot have area 41.
24. Rigid pieces cannot create or destroy area. An apparent change means one assumption of exact rearrangement has failed, commonly through a hidden gap, overlap or slightly bent boundary.
9. Teaching and transfer
If a learner remembers formulas but cannot explain them, use paper shapes. Cut a parallelogram and slide one triangular piece; duplicate a triangle and pair the copies. The formula then records a conserved-area transformation.
When area and perimeter are fused
Give twelve unit squares and build a 1×12 strip and a 3×4 rectangle. Count squares for area and exposed edges for perimeter. The two measures respond differently to rearrangement.
When a composite shape feels complicated
Ask whether it is easier to add known pieces or subtract a missing piece from a larger rectangle. Both are virtual dissections; choose the one with fewer uncertain dimensions.
When a picture is trusted more than the arithmetic
Compute the sum of piece areas before rearrangement. That sum becomes an invariant. If the claimed new outline has a different calculated area, inspect the fit rather than accepting a visual paradox.
When height is taken along a sloping side
Physically or visually drop a perpendicular to the base. Rearranging a parallelogram into a rectangle makes clear why perpendicular height is the distance controlling area.
Connect to invariants
Area conservation is an invariant: rigid cutting and exact rearrangement changes position and boundary but not total surface. This connects directly to Invariants, Before-and-After and Unchanged Quantities.
Continue through this enrichment collection
For guarantee proofs, use Pigeonhole Principle, Guaranteed Repetition and Distribution. For structured counting, use Pascal’s Triangle, Combinations and Path Counting. For fraction-preserving decomposition, use Egyptian Fractions, Unit Fractions and Decomposition.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Exact rearrangement means the same pieces are used once each, without stretching, gaps or overlaps.