When two shapes are joined, some lines stop being part of the outside boundary. They remain useful for showing how the figure was built, but you would not walk along them while travelling around the outside. Adding every visible line can therefore give a neat calculation and the wrong perimeter.
This PSLE Mathematics workshop teaches one specific decision: track which edges remain exposed when shapes are joined, cut or rearranged. Before calculating, identify the boundary the question actually asks you to measure. Then count each part of that boundary once. A formula should summarise the correct path, not replace the decision about which path matters.
The situations below are original practice problems described completely in words. Sketch them on scrap paper where helpful. The descriptions, not the apparent proportions of your sketch, supply the measurements. Use the existing guide to composite figures, missing lengths and boundary tracing for the underlying topic. This workshop concentrates on checking what happens to an edge after a join or cut.
Perimeter follows a boundary; area covers a region
Area and perimeter answer different questions. Area measures the surface covered by a shape. Perimeter measures the length around a specified boundary. Joining two non-overlapping pieces adds their areas, but their perimeters do not usually add unchanged because edges at the join become internal.
Imagine two square cards placed side by side with one full edge touching. Before joining, each touching edge belongs to the outside of its own card. After joining, the two edges meet inside the combined figure. They are not part of the outside path. The amount of card has not disappeared, but some previously exposed edge is no longer exposed.
This distinction explains why a shape can keep the same area while changing perimeter. Rearranging the same tiles does not change how much surface they cover when they do not overlap. It can change how many tile edges are exposed. Do not infer a perimeter change merely from an area change, or assume unchanged area means unchanged perimeter.
Trace first and label the internal join
Choose a starting corner and follow the outside continuously until you return to it. At each turn, identify the length of the next exposed segment. If a line takes you into the middle of the combined region rather than around it, it is probably an internal division rather than part of the requested outside perimeter.
On a sketch, mark the shared edge as internal. This prevents it from returning to the total merely because its measurement is printed clearly. A labelled measurement can be important without being an outside edge. It may tell you how much to remove from the sum of the original perimeters.
Do not rely on the phrase “add all the sides” until you have decided which sides belong to the new figure. The component shapes and the combined shape have different boundaries. The first question is about membership: which segments are on this boundary now?
Worked example 1: two squares become one rectangle
Two squares each have side length 4 cm. Join them along one complete side, without overlap. Find the outside perimeter of the combined figure. A quick sketch shows an 8 cm by 4 cm rectangle, so the perimeter is 8 + 4 + 8 + 4 = 24 cm.
Now examine a common wrong method. Each square has perimeter 16 cm, so a learner adds 16 + 16 and reports 32 cm. That sum describes both separate squares before joining. It includes the touching 4 cm edge of the first square and the touching 4 cm edge of the second square. Both are internal after the join.
Remove those two counted edges: 32 − 4 − 4 = 24 cm. Subtracting the shared length only once would give 28 cm, which still includes one copy of an edge that is no longer outside. The factor of two comes from how the original perimeter sum counted the contact, not from a rule to memorise without explanation.
Check the result through two representations. One treats the new figure as a rectangle. The other starts with the two original perimeters and removes the internal edges. Agreement is useful because the routes identify the same exposed path in different ways. The area is 32 square centimetres, but that number is not a perimeter merely because it also appears in the wrong sum.
Why the shared length is removed twice
For two shapes joined along a boundary segment, without overlap and with no enclosed hole in the examples under discussion, you can use: combined outside perimeter equals the first perimeter plus the second perimeter minus twice the shared length. Every part of the formula has a geometric meaning.
The separate perimeter totals contain every original edge. At the join, one edge from each piece occupies the shared position. Neither belongs to the new outside boundary, so both contributions must be removed. The shared length is not lost from an area calculation; it is a length counted twice in a perimeter calculation.
Check the conditions before using the shortcut. Overlapping shapes require careful tracing of the visible union rather than automatically treating an overlap as a simple edge join. A figure containing a hole also requires you to distinguish the outside boundary from the hole boundary. The formula is useful when you know what its counted edges represent.
Worked example 2: only part of a longer side is shared
A rectangle measures 10 cm horizontally and 6 cm vertically. A second rectangle is 4 cm wide and 3 cm tall. Attach its 4 cm top edge to the middle 4 cm of the larger rectangle’s bottom edge, so the smaller rectangle extends downward. There is no overlap.
The large rectangle has perimeter 32 cm and the small one 14 cm. The shared length is 4 cm, not the full 10 cm bottom side of the large rectangle. The combined perimeter is therefore 32 + 14 − 2 × 4 = 38 cm.
Trace it directly to check. Begin at the top left of the large rectangle: move 10 cm right, 6 cm down, 3 cm left, 3 cm down, 4 cm left, 3 cm up, 3 cm left and 6 cm up. The lengths total 38 cm. The two 3 cm exposed portions of the large rectangle’s bottom edge remain on the boundary.
A learner who subtracts twice 10 cm removes six centimetres of legitimate exposed edge twice. A learner who adds the small rectangle’s entire perimeter to 32 cm leaves the shared segment counted twice. Both errors come from failing to identify the actual contact length. The picture’s general appearance is not enough; locate precisely which parts touch.
Worked example 3: rotating the attached piece changes the contact
A large rectangle measures 12 cm by 8 cm. A smaller rectangle measures 5 cm by 3 cm. Join them without overlap. In Arrangement A, the smaller rectangle touches the larger along its complete 3 cm side. In Arrangement B, it touches along its complete 5 cm side. Both arrangements are possible on the stated larger rectangle.
The original perimeters are 40 cm and 16 cm, giving a separate total of 56 cm. Arrangement A has perimeter 56 − 6 = 50 cm. Arrangement B has perimeter 56 − 10 = 46 cm. The longer contact removes more exposed boundary from the separate perimeter total.
Both combined figures have area 96 + 15 = 111 square centimetres. The pieces are the same, and neither arrangement overlaps them. Their areas agree, but their perimeters differ by 4 cm. This is a concrete counterexample to the claim that the same two pieces must always make the same perimeter.
Do not identify the shared length by automatically choosing the shorter side of the smaller rectangle. That would work for Arrangement A and fail for B. The stated orientation decides the contact. A method that gives one correct answer by coincidence needs to be tested on the rotated version before you trust it.
Worked example 4: long and compact arrangements use the same material
Take two identical rectangles measuring 8 cm by 3 cm. Join their 3 cm ends to make a long rectangle. The resulting shape measures 16 cm by 3 cm and has perimeter 38 cm. Alternatively, join their 8 cm sides to make an 8 cm by 6 cm rectangle with perimeter 28 cm.
The two separate perimeters total 44 cm. Removing twice a 3 cm contact gives 38; removing twice an 8 cm contact gives 28. The two arrangements use the same 48 square centimetres of material. Their outside lengths differ because the length hidden inside the join differs.
Suppose the question asks which arrangement needs less edging. The compact arrangement needs 10 cm less. If it asks which covers more surface, neither does: their areas are equal. Read the target before selecting which comparison matters. A correct perimeter calculation cannot answer an area question on its own.
This example is also useful for explanation. Instead of saying “the compact one looks smaller”, state that its longer shared edge leaves less total boundary exposed. The reason is measurable and follows the construction. It does not depend on how accurately the sketch was drawn.
Worked example 5: a corner cut can leave perimeter unchanged
Begin with a 12 cm by 8 cm rectangle. Remove a 3 cm by 2 cm rectangle from its top-right corner. The removed piece uses 3 cm of the original top edge and 2 cm of the original right edge. Find the perimeter of the remaining L-shaped region.
The original perimeter is 40 cm. The cut removes two outside segments totalling 5 cm, but exposes two new segments of lengths 3 cm and 2 cm. Those replace the removed lengths. The new perimeter remains 40 cm, even though the area decreases from 96 to 90 square centimetres.
Trace the new path to verify: 9 cm across the remaining top, 2 cm down the cut, 3 cm across the inward ledge, 6 cm down the remaining right edge, 12 cm along the bottom and 8 cm up the left edge. The sum is 40 cm. Every listed segment belongs to the remaining region’s outside path.
The trap is assuming that removing material must reduce perimeter. That is an area-based intuition applied to a boundary question. The right check compares the outside lengths removed with the new lengths exposed. In this corner-cut construction they are equal. Do not turn that result into a rule that every cut leaves perimeter unchanged.
Worked example 6: an inward notch adds two new depths
Use another 12 cm by 8 cm rectangle. This time remove a 3 cm wide, 2 cm deep rectangular notch from the middle of the top edge. The notch does not reach either side or the bottom. Its position differs from the corner cut even though the removed piece has the same dimensions.
The cut removes a 3 cm segment from the old top boundary. It exposes a new 3 cm horizontal segment at the bottom of the notch and two vertical segments, each 2 cm deep. The horizontal lengths replace one another, while the two depths add 4 cm. The new perimeter is 44 cm.
Both the corner-cut shape and the notched shape have area 90 square centimetres. Their perimeters are 40 cm and 44 cm respectively. The amount removed is not enough information to determine the boundary change. Where the piece was removed matters because it determines which old edges disappear and which new edges appear.
For a direct check, place the notch so the remaining top segments are 4 cm and 5 cm. Their total is 9 cm. Add the 3 cm bottom of the notch, the 12 cm bottom of the large shape, the two 8 cm outer sides and the two 2 cm notch sides. The total is 44 cm. The internal-looking notch sides are actually exposed boundary because the cut opens to the outside.
Worked example 7: an enclosed hole requires a precise question
Cut a 2 cm by 2 cm square hole entirely inside a 12 cm by 8 cm rectangle. The hole does not touch any outer edge. If the question asks for the outside perimeter, it remains 40 cm. The outer rectangle has not changed.
If the question asks for the total length of ribbon needed along every exposed edge, including the hole, add the 8 cm boundary of the square hole. The total ribbon length is 48 cm. The remaining area is 96 − 4 = 92 square centimetres, a separate quantity again.
There is no need to argue from an ambiguous everyday use of perimeter when the task can state which boundary it wants. Read the actual wording. Outside boundary excludes the hole; every exposed edge including the hole includes it. Keep both paths distinct on your sketch.
This example marks a limit on the simple join shortcut. Adding component perimeters and removing shared segments tracks exposed edges, but a configuration with an enclosed gap can have more than one boundary. Before reporting an outside perimeter, make sure an inner boundary has not been included accidentally. Trace the requested path rather than treating every remaining edge as one continuous outside loop.
Worked example 8: count shared tile edges, then attach the unit
Six squares, each with side length 1 cm, are joined in a straight row. Their separate perimeters total 24 cm. There are five joins between neighbouring squares, each 1 cm long. Each join removes two edge contributions, so the outside perimeter is 24 − 10 = 14 cm.
Arrange the same six squares as a rectangle with three columns and two rows. There are four horizontal-neighbour joins, two within each row, and three vertical-neighbour joins between the rows. That is seven shared edges. The perimeter is 24 − 14 = 10 cm, matching a 3 cm by 2 cm rectangle.
The number of tiles remains six, but the number of joins changes from five to seven. A formula based only on tile count would miss that difference. The internal arrangement matters. Count each shared side once in the join count, then remove its two copies from the separate-perimeter total.
If each square instead has side length 3 cm, the edge lengths scale while the join counts remain five and seven. The row perimeter becomes 42 cm and the compact rectangle perimeter 30 cm. Do not subtract “five” from a length total without deciding whether five means joins or centimetres. A count and a measured length are not interchangeable.
Worked example 9: a printed line may show construction, not perimeter
Four squares of side 3 cm form a 2-by-2 block. A diagram leaves the dividing lines visible, so you can see the original squares. The combined shape is a 6 cm by 6 cm square with perimeter 24 cm. The internal cross does not add to that outside perimeter.
The separate perimeters total 48 cm. Four shared sides, each 3 cm long, account for a total contact length of 12 cm. Removing twice that length gives 48 − 24 = 24 cm. This agrees with tracing the four 6 cm outer edges.
Now suppose the question instead asks for the total length of all lines drawn, including the internal cross, with each visible segment counted once. The outer lines total 24 cm and the internal cross consists of a 6 cm vertical line and a 6 cm horizontal line, giving 36 cm altogether. That is a different job from measuring perimeter.
The same picture can support different quantities. You must identify whether the question asks for outer edging, all drawn lines, material used, or a path travelled. A clear target protects you from applying a correct perimeter method to a non-perimeter question.
Worked example 10: an opening changes fencing, not the geometric outline
A rectangular garden is 12 m by 8 m. Its geometric perimeter is 40 m. A 4 m entrance is left without fencing. The amount of fencing required is 40 − 4 = 36 m, assuming all the remaining outer boundary is fenced once.
Do not say the rectangle now has perimeter 36 m. The shape still has the same outer dimensions. The practical construction omits a portion of fencing. Perimeter and fencing length happen to be equal only when the whole requested boundary is fenced without an opening.
Suppose a second 2 m opening is added elsewhere. The fencing length becomes 34 m. If both openings are distinct, subtract both. If the question describes the same entrance twice in different words, do not subtract it twice. As with shared edges, you need to identify physical segments rather than simply react to every number in the text.
This distinction is especially useful in word problems about borders, ribbons, rails and paths. The material required follows the construction instructions. A final label such as “fencing required” keeps that practical quantity separate from the shape’s full boundary length.
Worked example 11: recover the hidden contact length
A 9 cm by 5 cm rectangle and a 7 cm by 4 cm rectangle are joined without overlap along one straight segment. The combined figure has no hole and an outside perimeter of 42 cm. Find the length of the shared segment.
The original perimeters are 28 cm and 22 cm, totalling 50 cm. Joining reduces that total by 50 − 42 = 8 cm. This reduction contains two copies of the shared length, so the contact is 8 ÷ 2 = 4 cm.
A learner who reports 8 cm has found the total length removed from the separate perimeter sum, not the physical length of the join. Label the intermediate quantity. The word “reduction” and the word “contact” refer to related but different lengths in this calculation.
Check forward: 28 + 22 − 2 × 4 = 42 cm. Also check that a 4 cm contact can fit on the available sides. The smaller rectangle has a 4 cm side, and the larger has sides long enough to receive it. The numerical answer and a feasible construction agree.
Worked example 12: sometimes the join has not been specified
You are told that a 12 cm by 8 cm rectangle is joined to a 7 cm by 4 cm rectangle, with no overlap, but nothing says which sides touch or how long the contact is. Can you find one unique combined perimeter from those dimensions alone?
No. The separate perimeter total is 40 + 22 = 62 cm. A full 4 cm contact produces 62 − 8 = 54 cm. A full 7 cm contact produces 62 − 14 = 48 cm. Both contacts can fit on the larger rectangle, so the same supplied dimensions allow different outside lengths.
This is not a failure to calculate. It is a recognition that one necessary relationship is missing. A labelled diagram or a sentence identifying the contact could settle it. An unlabelled sketch that merely looks as though a full side is shared may not justify a precise length unless the task establishes that relationship.
In ordinary examination work, read all the given information before deciding that a value is missing. The relationship may be stated through equal-length marks or a clear construction instruction. The lesson is to use those facts, not to invent them from visual appearance.
A boundary audit before your final line
First, name the target: outside perimeter, all exposed edges, fencing excluding an entrance, or all lines in a drawing. Second, identify internal joins and any holes. Third, trace the requested path or explain how the original perimeters are being adjusted. Finally, attach the correct length unit.
Check whether every segment appears once in the final path. Repeating one short side and omitting another of equal length can accidentally preserve the total, so a matching number alone does not prove that the route is correct. The working should correspond to actual boundary segments.
When using a second method, make it genuinely different. Compare direct tracing with the shared-edge adjustment, or recognise an outer rectangle where that is justified. Repeating the same mistaken list of sides in reverse order will not expose a wrong decision about which edges belong outside.
Practice questions: draw, decide and calculate
Question one: join two squares of side 5 cm along one full side. Find the combined perimeter. Explain why subtracting only one shared side from the separate perimeter total leaves an error.
Question two: attach a 3 cm by 2 cm rectangle to an 8 cm by 5 cm rectangle along the smaller rectangle’s complete 3 cm side, without overlap. Find the combined perimeter. Then find it if the shared side is 2 cm instead.
Question three: remove a 2 cm by 3 cm piece from a corner of a 10 cm by 7 cm rectangle. Find the remaining perimeter. Then move the same cut to form a notch 2 cm wide and 3 cm deep in the middle of a 10 cm side. Find the new perimeter and explain the difference.
Question four: arrange four squares of side 2 cm in a straight row. Find the outside perimeter. Rearrange them as a 2-by-2 square block and find the perimeter again. State which quantity remains unchanged.
Question five: cut a 1 cm square hole entirely inside a 9 cm by 6 cm rectangle. Find the outside perimeter and the total length of all exposed edges including the hole. Explain why those are different answers to different questions.
Question six: two shapes have separate perimeters of 30 cm and 18 cm. They join without overlap along one segment, forming a figure with no hole and outside perimeter 38 cm. Find the contact length. Identify what the subtraction 48 − 38 represents before dividing.
Practice answers with the edge decision shown
For question one, the separate perimeters total 40 cm. Remove two copies of the 5 cm contact to obtain 30 cm. The combined shape is a 10 cm by 5 cm rectangle, providing a direct check. Removing only 5 cm would leave one internal edge contribution in the total.
For question two, the original perimeters are 26 cm and 10 cm. A 3 cm contact gives 36 − 6 = 30 cm. A 2 cm contact gives 36 − 4 = 32 cm. The smaller contact leaves a longer outside boundary even though the area of material is unchanged.
For question three, the original perimeter is 34 cm. The corner cut replaces the removed 2 cm and 3 cm edges with equal new lengths, so 34 cm remains. The middle notch replaces its 2 cm opening with a 2 cm inner horizontal edge and adds two 3 cm depths, giving 40 cm.
For question four, the row is 8 cm by 2 cm and has perimeter 20 cm. The square block is 4 cm by 4 cm and has perimeter 16 cm. Both have area 16 square centimetres. The square-centimetre result describes coverage, not exposed length.
For question five, the outside perimeter is 30 cm. The square hole adds 4 cm only when the task asks for every exposed edge, giving 34 cm. For question six, the reduction of 10 cm contains two copies of the shared segment, so the contact length is 5 cm.
Delayed transfer: change the join before changing the numbers
After learning one worked example, keep its numbers but change how the shapes touch. This tests whether you understood the boundary decision or merely remembered an arithmetic result. Rotating the smaller rectangle in Example 3 changes the contact from 3 cm to 5 cm and therefore changes the answer.
Next, keep the construction but change the dimensions. The method should survive. Finally, present a mixture of joins, corner cuts, notches and holes without naming the method first. Ask the learner to explain which edges disappear and which become exposed before calculating.
Include a case where the answer does not change. The corner cut is useful for this purpose. If every modified picture in practice produces a different total, a learner may infer that every visible change must change perimeter. Good transfer practice tests both sensitivity to relevant changes and resistance to irrelevant ones.
Guidance for parents and tutors
When an answer is wrong, ask the learner to trace the path that their calculation represents. Do not begin by supplying the formula. If the learner includes an internal join, the main difficulty is boundary selection. If the path is right but the addition is wrong, the difficulty is execution. Those need different corrections.
Ask why twice the shared length is removed. A learner who can point to the two original edge contributions has a reason for the operation. A learner who only repeats “times two” may fail when the figure has several joins or a different requested boundary.
Use simple paper sketches or squared paper if helpful, but do not let drawing quality become the main assessment. The learner should use the stated dimensions and construction relationships. A rough but correctly labelled sketch can support better reasoning than a beautiful diagram whose proportions have been guessed.
Frequently asked questions
Why not add both original perimeters?
That total includes edges that become internal when the shapes touch. Remove the contributions from those shared edges or trace the new outside boundary directly. The original shapes and the combined shape do not have the same set of exposed sides.
Is the shared side always the shorter side?
No. The construction or diagram determines the contact. A rectangle can be attached along its longer side, shorter side or only part of a side. Find the actual contact length before calculating the adjustment.
Does removing a piece always reduce perimeter?
No. A corner cut can leave perimeter unchanged, and a middle notch can increase it. Compare the old boundary removed with the new boundary exposed. Area and perimeter must be considered separately.
Should a hole be included?
Read the requested quantity. Outside perimeter excludes the hole’s boundary; total edging around every exposed edge includes it when the task says so. Keep the outer and inner boundaries separate instead of making an unstated assumption.
Can I use a diagram that is not to scale?
Yes. Use its labels, equal-length markings and stated relationships. Do not measure the picture or assume a missing equality merely because two segments look similar. The diagram represents relationships, not necessarily actual proportions.
How do I check without solving everything twice?
Check the risky boundary decision. Trace one disputed segment, identify the shared length or compare the result with an obvious outer rectangle. Choose a second method that could reveal a different error rather than repeating the same side list mechanically.
Official reference and next route
The 2026 PSLE Mathematics syllabus includes interpreting information, applying concepts and choosing appropriate problem-solving strategies. These original boundary exercises practise those decisions; they are not a prediction of particular examination questions or a substitute for the full syllabus.
Return to the Primary 6 Mathematics Learning Hub and the PSLE Learning Guide. For a different subject using the same discipline of updating only what changed, try Vol 0053 on revised listening plans. The practical Mathematics rule is to decide which edges still belong to the requested boundary before adding their lengths.
