PRIMARY 4 MATHEMATICS LEARNING GUIDE · BATCH 2 · GUIDE 8
Composite figures are not difficult because the formulas are difficult. They are difficult because some of the information is hidden inside the shape. Primary 4 students must learn to reconstruct missing lengths, decide whether to split or subtract regions, trace only the outside boundary for perimeter and preserve units throughout the solution.
This guide develops composite rectangles and squares as a reasoning system: reconstruct, decompose, calculate, trace, verify. The goal is to help the learner see the figure as connected constraints rather than a collection of unrelated sides.
Series route: return to the Primary 4 Mathematics Learning Hub. For the wider geometry foundation, see Area, Perimeter, Angles, Symmetry and Nets.
1. Start by Naming the Quantity
Before touching a formula, ask: is the question asking for a boundary length or a covered region?
Perimeter measures the outside boundary and uses length units such as cm or m.
Area measures the covered region and uses square units such as cm² or m².
This distinction is the first control point in every composite-figure problem.
2. A Composite Figure Is Built From Simpler Figures
An L-shape may be understood as:
- one large rectangle with a smaller rectangle removed; or
- two smaller rectangles joined without overlap.
Both decompositions can produce the same area when applied correctly.
The best decomposition is the one that exposes known dimensions and keeps every region counted exactly once.
3. Reconstructing a Missing Horizontal Length
Suppose the full width of a composite figure is 14 cm. One horizontal segment is 6 cm and another aligned segment completes the same total width.
Missing horizontal length = 14−6 = 8 cm.
This works because aligned horizontal segments partition the same total width.
The learner should see the equality of total horizontal spans before subtracting.
4. Reconstructing a Missing Vertical Length
Suppose the full height is 11 cm and an upper vertical section is 4 cm. The remaining aligned vertical section is:
11−4 = 7 cm.
Missing-length questions often require no new formula. They require recognising that smaller aligned parts combine to make a known total.
5. Area by Subtraction
An L-shape fits inside a 12 cm by 9 cm rectangle. A 5 cm by 3 cm rectangle is removed from one corner.
Outer rectangle area = 12×9 = 108 cm².
Removed area = 5×3 = 15 cm².
Composite area = 108−15 = 93 cm².
This route is efficient when the missing corner is clearly rectangular.
6. Area by Splitting
The same L-shape can often be split into two rectangles.
Suppose one rectangle is 7 cm by 9 cm and another is 5 cm by 6 cm, without overlap.
Total area = 7×9 + 5×6 = 63+30 = 93 cm².
The two routes agree. Agreement between different decompositions is a strong verification method.
7. Why Overlap Matters
If two rectangles used in a decomposition overlap, adding both areas counts the overlap twice.
If there is a gap, some region is not counted.
A valid split must cover the original figure exactly once.
Every square unit belongs to one counted region—not zero regions and not two.
8. Perimeter Is an Exterior Journey
For perimeter, imagine walking around the outside edge of the figure.
Every exterior segment is counted exactly once. Internal lines used to split the figure for area are not automatically part of the perimeter.
This “boundary journey” model is safer than adding the perimeters of component rectangles.
9. Why Adding Sub-Perimeters Fails
Suppose two rectangles share an edge. If you add their individual perimeters, the shared edge is counted twice even though it lies inside the combined figure.
The combined perimeter excludes that internal boundary.
This is why area decomposition and perimeter decomposition obey different counting rules.
10. Tracing an L-Shaped Perimeter
Begin at one corner and move clockwise. Record each side length. If a side is missing, reconstruct it from aligned totals before adding.
Example boundary lengths: 12, 4, 5, 3, 7, 7 cm.
Perimeter = 12+4+5+3+7+7 = 38 cm.
After tracing, check that the total horizontal movement to the right equals the total horizontal movement to the left, and similarly for vertical movement. A closed shape must return to its starting point.
11. The Closure Check
A closed rectilinear figure must balance directions.
Total rightward horizontal length = total leftward horizontal length.
Total upward vertical length = total downward vertical length.
This gives students a powerful way to reconstruct missing edges and verify a perimeter diagram without relying only on visual intuition.
12. Missing Length From Area
A rectangle has area 96 cm² and width 8 cm. Its length is 96÷8 = 12 cm.
In a composite problem, such an inverse step may be needed before the missing rectangle can be reconstructed.
Area can therefore supply a side length that later becomes part of a perimeter calculation.
13. Missing Length From Perimeter
A rectangle has perimeter 34 cm and length 10 cm.
Half-perimeter = 17 cm.
Width = 17−10 = 7 cm.
In a composite problem, this reconstructed width may then determine a missing region or aligned edge.
14. A Two-Stage Composite Problem
A rectangular board is 15 cm by 10 cm. A rectangular corner 6 cm by 4 cm is removed. Find the remaining area and describe how to find the perimeter.
Area = 15×10−6×4 = 150−24 = 126 cm².
For perimeter, do not subtract the removed rectangle’s perimeter. Trace the new boundary. Removing a corner removes some old exterior edges but creates new interior-cut edges that become exterior.
This is why perimeter must be reconstructed from the boundary itself.
15. Corner Removal and Perimeter
When a rectangle is removed exactly from a corner of a larger rectangle, the total perimeter may remain unchanged if the lengths removed from the original outer edges are replaced by equal new cut edges.
This surprising result is a useful reasoning question. It should be checked from actual dimensions rather than memorised as a universal trick.
The lesson is that perimeter depends on boundary length, not simply on “how much shape is missing”.
16. Composite Figures With Squares
A square of side 8 cm has area 64 cm². If a 3 cm by 2 cm rectangle is attached without overlap, total area becomes:
64+6 = 70 cm².
For perimeter, identify which square edge length is covered by the attachment and therefore becomes internal.
Attachments change the boundary differently from the area.
17. Multiple Rectangles: Build a Map
When a shape contains three or more rectangles, label them A, B and C. Write each rectangle’s dimensions before calculating.
Then choose one of two area strategies:
- sum non-overlapping component areas; or
- outer bounding rectangle minus missing rectangular regions.
A labelled map prevents double-counting and makes missing dimensions easier to track.
18. Units as a Diagnostic
If the final calculation adds 40 cm² and 12 cm, something has gone wrong because area and length are different quantities.
You may add two areas or two lengths when they represent compatible quantities. You cannot directly add an area to a perimeter.
Units reveal structural mistakes that arithmetic alone may hide.
19. Estimating Composite Area
If a composite figure fits inside a 10 cm by 10 cm square, its area cannot exceed 100 cm².
If the figure visibly covers most of that square, an answer of 12 cm² is suspicious.
Bounding rectangles provide quick plausibility checks even before exact decomposition.
20. Estimating Perimeter
If a figure is roughly 12 cm wide and 8 cm tall, a perimeter near a few tens of centimetres is plausible. A result of 300 cm would require unusually long boundary detours that should be visible.
Estimation cannot prove the exact perimeter, but it can expose major scale errors.
21. Comparing Two Decompositions
Suppose Method A uses outer area minus a cut-out. Method B splits the same shape into two rectangles.
If the answers disagree, do not average them. Inspect the decompositions:
- Was a missing length reconstructed correctly?
- Was there overlap?
- Was a gap omitted?
- Did both methods cover the same original figure?
Two methods are valuable because disagreement creates evidence of an error.
22. Common Composite-Figure Errors
| Error | Weak link | Repair question |
|---|---|---|
| Adds sub-rectangle perimeters | Internal edges counted | Can I walk along this edge from outside? |
| Uses full length for both sub-rectangles | Missing length not reconstructed | Which aligned parts make the total? |
| Area subtraction gives negative value | Removed region larger than container | Does the cut-out actually fit inside? |
| cm used for area | Quantity/unit mismatch | Am I covering or tracing? |
| Two decompositions disagree | Overlap or omission | Is every region counted exactly once? |
23. Practice Laboratory
- A rectangle is 12 cm by 7 cm. Find area and perimeter.
- A rectangle has area 108 cm² and width 9 cm. Find length.
- A rectangle has perimeter 42 cm and length 13 cm. Find width.
- A 14 cm total width is split into 5 cm and an unknown aligned segment. Find the unknown.
- A 12 cm total height is split into 7 cm and an unknown aligned segment. Find the unknown.
- A 12×9 rectangle has a 4×3 corner removed. Find remaining area.
- Explain why adding component perimeters can overcount.
- A composite figure has exterior sides 10, 4, 3, 2, 7 and 6 cm. Find perimeter.
- Why must total rightward and leftward horizontal distances match around a closed rectilinear figure?
- A square has side 9 cm. Find its area and perimeter.
- A 2×5 rectangle is attached to a square without overlap. What area is added?
- A figure fits inside a 10×8 rectangle. State an upper bound for its area.
- A rectangular field has area 180 m² and length 15 m. Find width and perimeter.
- An L-shape can be solved by subtraction or splitting. What should be true of both correct area answers?
- Why is 25 cm² + 8 cm not a meaningful single area or perimeter calculation?
24. Explained Answers
1. Area=84 cm²; perimeter=38 cm.
2. 108÷9=12 cm.
3. Half-perimeter=21; width=21−13=8 cm.
4. 14−5=9 cm.
5. 12−7=5 cm.
6. 108−12=96 cm².
7. Shared internal edges appear in the perimeter of each component but are not part of the combined outside boundary.
8. 10+4+3+2+7+6=32 cm.
9. The path returns to its starting horizontal position, so net horizontal movement must be zero.
10. Area=81 cm²; perimeter=36 cm.
11. 10 square units.
12. At most 80 square units.
13. Width=12 m; perimeter=54 m.
14. They should be equal.
15. Square centimetres measure area while centimetres measure length; they are different kinds of quantities.
25. Teaching Routine: Reconstruct Before Formula
When a student sees a composite figure, delay the formula. First ask them to highlight full horizontal spans, full vertical spans, missing aligned parts and the exterior boundary.
Then ask for two possible area decompositions. For perimeter, trace the outside separately. Finally, verify with units and a bounding estimate.
The reusable sequence is: label → reconstruct → decompose → calculate → trace → verify.
26. Why Composite Figures Matter Later
Later geometry and mensuration rely on exactly this ability to decompose complicated objects into known parts. Algebra will formalise missing lengths. Coordinate geometry will place boundaries on a grid. Volume and surface area will extend decomposition into three dimensions.
Primary 4 composite figures therefore teach more than area and perimeter. They teach how to find hidden information from constraints.
Return to the Primary 4 Mathematics Learning Hub →
Sources and Boundaries
Curriculum scope is aligned with the MOE Primary Mathematics Syllabus, updated October 2025. All examples and teaching routines are independently written by eduKate Publishing.
Editorial control: Wintour House V1.0 · CivDJ · eduKate Publishing.
