PSLE Mathematics Learning Guide · Guide 6
Return to the PSLE Learning Guide · Explore the Mathematics Learning Hub
A composite figure is not difficult because it contains many formulas. It is difficult because the learner must decide which pieces belong together, which lengths are missing, which lines are internal and which edges actually form the required boundary.
The safest habit is not “memorise a composite-shape formula”. There is no single useful formula for every composite figure. The reliable method is: decompose or recombine the figure into familiar shapes, mark the lengths that are known or can be found, calculate each required part, then audit whether every square unit or boundary segment has been counted exactly once.
The examples here use squares, rectangles, triangles, semicircles and quarter circles because these appear in the Primary 6 Measurement and Geometry outcomes in the MOE Primary Mathematics syllabus updated October 2025. The page also links the current SEAB 2026 PSLE examination formats. All worked tasks below are original eduKate teaching material.
Start with the question: area, perimeter or both?
Area measures the amount of surface inside a figure. Perimeter measures the distance around a boundary. A composite diagram can use the same lengths for two very different jobs.
If a rectangular lawn has a semicircular end, finding the grass needed is an area problem. Finding the fence around the outside is a perimeter problem. The interior line where the rectangle joins the semicircle may help with area decomposition, but it is not part of the exposed outside boundary.
A learner should therefore write one word before calculating: AREA or PERIMETER. That small label reduces formula drift.
Two main strategies: add pieces or subtract a cut-out
Composite area can usually be approached in one of two ways.
Add pieces: Split the figure into familiar non-overlapping shapes, calculate each area and add.
Subtract a cut-out: Imagine a larger familiar figure containing the target shape. Calculate the larger area, calculate the missing or removed part, then subtract.
Neither method is automatically better. Choose the one that creates fewer unknown lengths and fewer opportunities to double-count.
Example: An L-shaped region can be split into two rectangles, or treated as one large rectangle minus a smaller missing rectangle. If the outer dimensions and cut-out dimensions are clearly labelled, subtraction is often shorter. If the two arms have direct dimensions, addition may be clearer.
Find missing lengths before applying formulas
Many composite problems hide one side length. The missing length often comes from a total length minus known parts.
If the full top edge of a rectangle is 18 cm and a cut-out uses 7 cm of that width, the remaining horizontal segment is 18 − 7 = 11 cm.
If the full height is 14 cm and one vertical section is 5 cm, the other aligned vertical section may be 9 cm, provided the diagram shows they partition the same total height.
Do not infer equality merely because two lines look similar in a sketch. Use labels, right-angle structure, parallel sides, or the geometry of rectangles and squares to justify the missing length.
Trace the perimeter instead of adding every visible line
Composite diagrams often contain internal lines used to show how shapes join. Those lines are not automatically part of the perimeter.
Use a boundary walk. Start at one corner, move around the outside edge and list each segment once until you return to the starting point. If you cross through the interior, you have left the perimeter.
For a rectangle with a semicircle attached along one side, the shared diameter is internal. The perimeter uses the three exposed rectangle sides plus the semicircular arc, not the diameter twice.
Avoid overlap when adding areas
If two component shapes overlap, adding both full areas counts the overlap twice. Either partition the figure into non-overlapping pieces or subtract the repeated overlap once.
This matters when a diagram is described as “a square with a quarter circle inside” or “two rectangles overlapping”. The drawing method must preserve the actual region being measured.
Curved pieces need their own boundary audit
For semicircles and quarter circles, distinguish the arc from the whole perimeter of the sector-shaped region.
A semicircle attached to another shape along its diameter contributes only its curved arc to the outer perimeter if the diameter is fully internal. A detached semicircle standing alone contributes both arc and diameter.
A quarter circle attached along one or both radii may contribute its arc and whichever radii remain exposed. Trace the actual outer edge instead of automatically adding two radii.
For circle fundamentals, see Guide 5: Circle Area and Circumference.
Triangles are often the cleanest missing piece
A triangle’s area is 1/2 × base × height. The height must be perpendicular to the chosen base.
In a composite figure, a triangular corner may be easier to subtract from a rectangle than to treat the irregular figure directly.
Example: A 12 cm by 8 cm rectangle has a right triangular corner of base 4 cm and height 3 cm removed. Rectangle area = 96 cm². Triangle area = 1/2 × 4 × 3 = 6 cm². Remaining area = 90 cm².
Main worked workshop: twelve original composite-figure problems
Problem 1: L-shape by subtraction
A 15 cm by 12 cm rectangle has a 5 cm by 4 cm rectangular corner removed. Find the remaining area.
Reasoning: Large rectangle = 180 cm². Cut-out = 20 cm². Remaining area = 160 cm².
Problem 2: L-shape by addition
The same L-shape can be split into a 10 cm by 12 cm rectangle and a 5 cm by 8 cm rectangle.
Reasoning: 120 + 40 = 160 cm². The agreement with Problem 1 checks the decomposition.
Problem 3: perimeter of an L-shape
Use the 15 cm by 12 cm rectangle with a 5 cm by 4 cm corner removed from the top-right corner. Find the perimeter.
Reasoning: Boundary segments are 10, 4, 5, 8, 15 and 12 cm. Total = 54 cm. Notice that removing the corner changed the route but, in this specific corner-cut case, the total perimeter equals the original rectangle perimeter 2(15 + 12) = 54 cm.
Problem 4: rectangle plus semicircle area
A rectangle is 14 cm long and 10 cm wide. A semicircle of diameter 10 cm is attached to one short side. Take π = 3.14. Find the total area.
Reasoning: Rectangle = 140 cm². Semicircle radius = 5 cm. Semicircle area = 1/2 × 3.14 × 25 = 39.25 cm². Total = 179.25 cm².
Problem 5: rectangle plus semicircle perimeter
Use the same figure.
Reasoning: The joined diameter is internal. Exposed rectangle edges = 14 + 14 + 10 = 38 cm. Semicircle arc = πr = 15.7 cm. Perimeter = 53.7 cm.
Problem 6: square minus quarter circle
A 14 cm by 14 cm square contains a quarter-circle cut-out of radius 14 cm at one corner. Take π = 22/7. Find the area remaining.
Reasoning: Square = 196 cm². Quarter circle = 1/4 × 22/7 × 196 = 154 cm². Remaining = 42 cm².
Problem 7: rectangle minus triangle
A 20 cm by 12 cm rectangle has a right triangle of base 6 cm and perpendicular height 8 cm removed. Find the remaining area.
Reasoning: Rectangle = 240 cm². Triangle = 24 cm². Remaining = 216 cm².
Problem 8: find missing width before area
A composite figure is made of two adjoining rectangles. The total width is 18 cm. One rectangle is 7 cm wide. Find the width of the other rectangle.
Answer: 11 cm. The missing measure must be resolved before area calculations that use it.
Problem 9: two rectangles with different heights
Rectangle A is 8 cm by 6 cm. Rectangle B is 5 cm by 4 cm and is attached to the right side of A along a 4 cm segment. Find the total area.
Reasoning: No overlap. Total = 48 + 20 = 68 cm².
Problem 10: perimeter with shared edge
Use the figure in Problem 9. Find the perimeter.
Reasoning: One 4 cm joining segment is internal. Either trace the outside boundary, or use P(A) + P(B) − 2(shared edge). P(A)=28, P(B)=18. Perimeter = 28 + 18 − 8 = 38 cm.
Problem 11: two semicircles forming a full circle in area
Two non-overlapping semicircular regions each have radius 7 cm. Take π = 22/7. Find their combined area.
Reasoning: Together they equal one full circle of radius 7 cm. Area = 154 cm².
Problem 12: compare area methods
An irregular rectilinear shape fits exactly inside a 20 cm by 15 cm rectangle. Two missing rectangular corners measure 4 cm by 3 cm and 5 cm by 2 cm. Find the irregular area.
Reasoning: Large rectangle = 300 cm². Missing areas = 12 + 10 = 22 cm². Target area = 278 cm².
Run a double-count audit
After an area calculation, ask:
- Did any piece overlap another piece I also counted?
- Did I leave a gap?
- Did I subtract a cut-out that was never inside the original region?
After a perimeter calculation, ask:
- Did I include any internal joining line?
- Did I miss an indentation edge?
- Did I count a shared edge twice?
- Did I include all exposed curved and straight pieces?
Area and perimeter must finish in different unit families
Perimeter is a length: cm, m, km. Area is measured in square units: cm², m², km².
If a composite calculation finishes with 160 cm for an area, the unit itself reveals a problem. Conversely, a perimeter in cm² has the wrong dimension.
For mixed-unit diagrams, convert lengths before combining them. Guide 2 covers this in detail: Make Units Agree Before You Calculate.
Choose the decomposition that minimises unknowns
Two correct decompositions can differ greatly in difficulty. Suppose a shape can be split into three small rectangles, but can also be treated as one large rectangle minus one obvious cut-out. The second method may be safer because it needs fewer intermediate lengths.
Good mathematical strategy is not simply “can I solve it this way?” It also asks “which valid representation makes the relationships easiest to verify?”
Independent transfer check: the activity courtyard
A school courtyard is a 20 m by 14 m rectangle with a 6 m by 4 m rectangular storage corner removed. A semicircular stage of diameter 14 m is attached along the full 14 m left side of the original rectangle. The removed storage corner is on the opposite side. Take π = 22/7.
- Find the area of the original rectangle.
- Find the storage cut-out area.
- Find the area of the rectangular courtyard after the cut-out.
- Find the semicircular stage area.
- Find the total usable area including the stage.
- Explain why the 14 m diameter shared by the rectangle and semicircle is not part of the outside perimeter.
- Find the curved arc length of the semicircular stage.
- Describe the correct perimeter method without assuming the removed corner lies on a particular side not specified by a diagram.
Independent-check answers
1. 20 × 14 = 280 m².
2. 6 × 4 = 24 m².
3. 280 − 24 = 256 m².
4. Radius = 7 m. Semicircle area = 1/2 × 22/7 × 49 = 77 m².
5. 256 + 77 = 333 m².
6. The diameter lies where the two regions join; it is internal rather than exposed outside boundary.
7. Semicircle arc = πr = 22 m.
8. Trace the actual outer boundary shown in the diagram, adding every exposed straight segment and the semicircular arc once. The exact numerical perimeter cannot be determined from the text alone unless the storage corner’s precise position on the boundary is specified. Preserving that uncertainty is better than inventing a diagram.
Repair the smallest reasoning defect
Draft: Add the areas of two overlapping rectangles.
Problem: The overlap is counted twice.
Repair: Partition into non-overlapping pieces or subtract the overlap once.
Draft: Perimeter of rectangle-plus-semicircle includes the shared diameter.
Problem: The diameter is internal after attachment.
Repair: Trace only the exposed outside boundary.
Draft: Area of an L-shape = multiply all visible side lengths.
Problem: No valid area relationship is represented.
Repair: Split into familiar areas or subtract a cut-out.
Draft: Use a missing length because it “looks equal” to another segment.
Problem: Diagram appearance is not proof.
Repair: Derive the length from labelled totals and known shape properties.
Parent and tutor guide: ask the learner to colour the region and trace the edge
Two simple prompts separate area from perimeter. For area, ask the learner to shade the exact region being measured. For perimeter, ask the learner to trace the exact boundary.
If the learner cannot find a missing side, ask which full length the small segments make together. If two decompositions are possible, compare them and choose the one with fewer uncertain lengths.
Do not reward a complicated solution merely because it has more steps. A shorter decomposition that preserves the geometry is often stronger.
The learner’s final card
What region am I measuring? What boundary am I tracing? Which familiar shapes can I add or subtract? Which lengths must be found first? Have I counted every area or boundary piece exactly once?
Continue through the PSLE Mathematics Learning Guide
Return to Guide 5: Circle Area and Circumference. Continue with Guide 7: Rebuild Cube and Cuboid Volume From Base Area and Height and Guide 8: Find Unknown Angles by Naming the Geometric Rule First.
Return to the PSLE Learning Guide for the wider pathway.
Sources and boundaries
Official references: SEAB 2026 PSLE formats and MOE Primary Mathematics syllabus, updated October 2025.
Teaching boundary: All diagrams described in words, examples and suggested solutions are original teaching material. Where a numerical perimeter depends on a diagram not fully specified in text, the guide states that limitation rather than inventing geometry.