Sec 3 G3 Maths Bukit Timah | Composite 2D Shapes

Secondary 3 G3 Mathematics Tutor Bukit Timah | Composite Plane Figures: Area, Perimeter and Cut-Outs — Did you know? Removing a rectangular corner from a shape always reduces its area, but it may leave its perimeter unchanged. That surprises students who think a smaller figure must have a smaller outside boundary. Composite-shape questions become easier when you separate what fills the region from what traces its edge. Here is the quick answer, then checked examples for the most common configurations.

The quick answer: separate area from boundary

Area: Add non-overlapping parts, or subtract the area of a hole or removed region from a larger simple figure.
Perimeter: Trace only the required exposed boundary. Internal shared joins are not exterior edges.
Rectangle: area = length × width; perimeter = 2(length+width).
Triangle: area = ½ × base × perpendicular height, even when the triangle is slanted.
Trapezium: area = ½(sum of parallel bases) × perpendicular height.
Semicircle: area = ½πr²; its curved arc is πr, but a standalone semicircle perimeter also includes its diameter 2r.
Unit check: perimeter has length units, whereas area has squared units. 1 m² = 10,000 cm².
Cut-out check: subtract missing area, but add any new inner boundary only if the question actually counts that boundary.

The official 2027 Singapore–Cambridge SEC G3 Mathematics K310 syllabus includes perimeter and area of composite plane figures, along with trapezium, parallelogram and circular measurements. This article focuses on two-dimensional composite regions, rather than the separate series guide on three-dimensional surface area and composite solids. As with other G3 topics, the school may teach or revisit these skills at different secondary levels and terms.

Jump to: Understand the shapes · 40 worked examples · Common mistakes · Study plan · FAQs.

Why measuring a shape starts with defining the region

Area and perimeter measure different things

Area measures the two-dimensional region inside an outline, usually in cm² or m². Perimeter measures distance along the relevant boundary, usually in cm or m. A large region may have a surprisingly short outline, and a narrow complicated region can have a long outline. The two are related geometrically but are not interchangeable numbers.

Before calculating, ask what the question actually wants: flooring, painting or shaded space suggests area; fencing, edging or walking all around an outline suggests perimeter. In a complicated diagram, label the units beside every length and avoid using a square unit for a boundary.

Addition works when pieces do not overlap

Imagine a rectangle with a triangular roof attached along its upper edge. If the roof and rectangle share only that edge and have no overlapping interiors, total area equals rectangle area plus triangle area. The joined edge has zero area, so it does not alter the area sum.

For perimeter, however, the attached edge is not exposed. A common error is to add the entire rectangle perimeter and all three triangle sides. The shared roof base would then be counted twice despite being inside the composite figure. Different questions require different boundary decisions.

Subtraction is easier when the missing piece is familiar

An L-shaped floor can often be seen as one big rectangle with a smaller rectangle removed from a corner. Rather than splitting the L into several arbitrary strips, subtract the missing rectangle’s area from the bounding rectangle’s area. The same reasoning works for a square frame, a rectangle with holes or a shaded region outside a circle.

State which area you are removing and why. In a valid subtraction model, the missing shape is contained within the original region and there is no unintended overlap. A sketch with the removed piece shaded or crossed out is useful for checking the assumption.

A corner notch is a perimeter surprise

Remove a w-by-h rectangular corner from a W-by-H rectangle. The original outline loses one exterior length w and one exterior length h, but gains an inner horizontal notch of length w and a vertical notch of length h. Their total effects cancel, so the new L-shaped perimeter remains 2(W+H).

The area decreases by wh. This is a good reminder that an area change does not always predict how a perimeter changes. The claim assumes an ordinary notch cut from exactly one corner, not a separate hole in the middle or a slit with a different boundary.

Holes can create additional boundary

Cut an interior rectangular hole through a sheet. The remaining material loses area equal to the hole’s area. If you are asked to measure all exposed edges of the remaining sheet, the hole’s inner boundary counts alongside the outer outline. By contrast, if you are asked for the outside fence only, inner boundaries may be excluded.

Read the wording before adding inner perimeters. A question describing a painted floor area usually needs only the remaining area. An edging task explicitly covering both outside and inside borders needs both sets of boundary lengths. One diagram can therefore support several different correct outputs.

Round parts need careful shared edges

Attach a semicircle to a rectangle along a side equal to the semicircle’s diameter. For area, add the rectangle area and half the circle area. For the exterior perimeter, count the rectangle’s three exposed sides and the semicircular arc. Do not count the diameter twice or add it when it is internal.

Use πr for a semicircle’s curved arc, not πr²/2. The latter is an area. Ask whether the radius or diameter was supplied, and convert correctly before applying the formula. A small radius label inside the drawing can prevent using the diameter by accident.

Rings and frames are difference-of-areas problems

An annulus is a ring between two concentric circles. Its area is πR²−πr² for outer radius R and inner radius r. A square border can likewise be found as outer-square area minus inner-square area, when the inner square is centred and fully contained.

The amount of border material and the length of all its edges are separate questions. A ring has inner and outer circular boundaries; a square frame has an inner and outer rectangular boundary. If the question specifies edging on both, those lengths must both be counted.

Trapeziums need the perpendicular height

The area formula ½(a+b)h uses the two parallel base lengths a and b and the perpendicular distance h between them. A slanted side is not automatically the height. The formula can be understood as average parallel-base length multiplied by perpendicular separation.

In a combined figure with a rectangle and trapezium, mark which base meets the rectangle. If they share the whole matching side and there is no overlap, areas add. When finding perimeter, the shared base must be excluded from the exposed boundary.

Area conversion squares the length conversion

One metre contains 100 centimetres, so a square metre contains 100×100=10,000 square centimetres. A 250,000 cm² region is therefore 25 m², not 2,500 m². The same squared relationship applies whenever the unit is changed in two dimensions.

Students sometimes know that 100 cm equals 1 m yet divide an area by only 100. Draw a one-metre square and count its 100-by-100 grid to make the multiplier tangible. Unit checks can catch this kind of large mistake immediately.

Composite questions require a diagram with ownership

A shape may be described by several dimensions but not every line belongs to the requested boundary. A useful method is to trace the outside in one colour, shade the area to retain in another and mark cut-out dimensions separately. This makes it easier to choose between addition, subtraction and shared-edge cancellation.

Under the K310 real-world problem-solving emphasis, students should be able to interpret a floor plan or geometric story, choose a sensible decomposition and explain the result. A correct-looking number reached without clarifying the region or boundary is fragile, because the same arithmetic may answer a different question.

40 worked examples: regions, holes and exposed edges

These examples are original teaching exercises, not examination forecasts. Each of ten problem families appears with four changed sizes. Attempt a question with the solution hidden, mark the region or boundary being measured, and then use the check to confirm that the arithmetic fits. Where π occurs, exact expressions are retained; approximations should follow the question’s requested accuracy.

Rectangle with an internal cut-out

This set highlights rectangle with an internal cut-out. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 1. Question: A rectangular sheet is 12 cm by 10 cm. A 4 cm by 3 cm rectangular hole is completely inside it. Find the remaining material area. Method: Outer area=12×10=120 cm². Hole area=4×3=12 cm². Remaining area=120−12=108 cm². Check: The remaining area is positive and smaller than the original 120 cm². Watch for: Do not add the hole’s area or treat its inner edge as another area.

Try a change: Label every dimension with a unit and trace the required region before choosing a formula. Create a second solution by decomposing the region in a different valid way. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 2. Question: A rectangular sheet is 14 cm by 11 cm. A 5 cm by 3 cm rectangular hole is completely inside it. Find the remaining material area. Method: Outer area=14×11=154 cm². Hole area=5×3=15 cm². Remaining area=154−15=139 cm². Check: The remaining area is positive and smaller than the original 154 cm². Watch for: Do not add the hole’s area or treat its inner edge as another area.

Try a change: Ask whether parts overlap in area or only meet along a shared edge. That controls addition and exposed perimeter. Predict the effect of doubling only one linear dimension, rather than doubling all dimensions. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 3. Question: A rectangular sheet is 16 cm by 12 cm. A 6 cm by 3 cm rectangular hole is completely inside it. Find the remaining material area. Method: Outer area=16×12=192 cm². Hole area=6×3=18 cm². Remaining area=192−18=174 cm². Check: The remaining area is positive and smaller than the original 192 cm². Watch for: Do not add the hole’s area or treat its inner edge as another area.

Try a change: Try subtracting a familiar cut-out from a large bounding rectangle rather than dividing a complicated outline arbitrarily. Explain whether a painted-area question and a fencing question would ask for different parts of this diagram. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 4. Question: A rectangular sheet is 18 cm by 13 cm. A 7 cm by 3 cm rectangular hole is completely inside it. Find the remaining material area. Method: Outer area=18×13=234 cm². Hole area=7×3=21 cm². Remaining area=234−21=213 cm². Check: The remaining area is positive and smaller than the original 234 cm². Watch for: Do not add the hole’s area or treat its inner edge as another area.

Try a change: Estimate whether the remaining area should be smaller than the uncut whole; if not, check the subtraction. Check the units on each intermediate multiplication and subtraction. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

L-shape cut from one corner

This set highlights l-shape cut from one corner. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 5. Question: Remove a 3 cm by 2 cm corner rectangle from a 10 cm by 8 cm bounding rectangle. Find the new area and perimeter. Method: Area = 10×8−3×2=74 cm². For perimeter, the lost exterior lengths w and h are replaced by equal inner notch lengths. Hence P=2(10+8)=36 cm. Check: The area decreases, but the simple corner notch changes the outside boundary by −w−h+w+h=0. Watch for: Do not assume a reduced area always has a reduced perimeter.

Try a change: Keep exact π expressions when appropriate and round only according to the problem’s instructions. Describe which line segment is shared internally and why it should not appear in exterior perimeter. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 6. Question: Remove a 4 cm by 3 cm corner rectangle from a 12 cm by 9 cm bounding rectangle. Find the new area and perimeter. Method: Area = 12×9−4×3=96 cm². For perimeter, the lost exterior lengths w and h are replaced by equal inner notch lengths. Hence P=2(12+9)=42 cm. Check: The area decreases, but the simple corner notch changes the outside boundary by −w−h+w+h=0. Watch for: Do not assume a reduced area always has a reduced perimeter.

Try a change: For perimeter, read ‘outside edge’ or ‘all exposed edges’ carefully, because inner holes can change the boundary counted. Try a reverse question using the area to recover one missing side length. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 7. Question: Remove a 5 cm by 4 cm corner rectangle from a 14 cm by 10 cm bounding rectangle. Find the new area and perimeter. Method: Area = 14×10−5×4=120 cm². For perimeter, the lost exterior lengths w and h are replaced by equal inner notch lengths. Hence P=2(14+10)=48 cm. Check: The area decreases, but the simple corner notch changes the outside boundary by −w−h+w+h=0. Watch for: Do not assume a reduced area always has a reduced perimeter.

Try a change: A composite answer needs a unit: centimetres for lengths and square centimetres for areas. Create a second solution by decomposing the region in a different valid way. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 8. Question: Remove a 6 cm by 5 cm corner rectangle from a 16 cm by 11 cm bounding rectangle. Find the new area and perimeter. Method: Area = 16×11−6×5=146 cm². For perimeter, the lost exterior lengths w and h are replaced by equal inner notch lengths. Hence P=2(16+11)=54 cm. Check: The area decreases, but the simple corner notch changes the outside boundary by −w−h+w+h=0. Watch for: Do not assume a reduced area always has a reduced perimeter.

Try a change: Draw an alternative decomposition of the same region and check that the area total agrees. Predict the effect of doubling only one linear dimension, rather than doubling all dimensions. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Rectangle with an attached semicircle

This set highlights rectangle with an attached semicircle. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 9. Question: A rectangle measures 8 cm long by 6 cm high. A semicircle of diameter 6 cm is attached along its full right-hand side. Give total area and exposed perimeter exactly. Method: Radius=3 cm. Total area=48 + (9/2)π cm². Its exposed perimeter is the top and bottom (2×8), left side (6), and semicircle arc (π×3): 22+3π cm. Check: The shared diameter of 6 cm is internal and is not counted in the exterior perimeter. Watch for: Do not add the semicircle diameter and the rectangle’s joined edge to the outside length.

Try a change: Mark shared internal joints with an X when calculating outside perimeter so they are not counted twice. Explain whether a painted-area question and a fencing question would ask for different parts of this diagram. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 10. Question: A rectangle measures 10 cm long by 8 cm high. A semicircle of diameter 8 cm is attached along its full right-hand side. Give total area and exposed perimeter exactly. Method: Radius=4 cm. Total area=80 + (16/2)π cm². Its exposed perimeter is the top and bottom (2×10), left side (8), and semicircle arc (π×4): 28+4π cm. Check: The shared diameter of 8 cm is internal and is not counted in the exterior perimeter. Watch for: Do not add the semicircle diameter and the rectangle’s joined edge to the outside length.

Try a change: Return after two days to an L-shape or ring with different measurements and explain the first step unaided. Check the units on each intermediate multiplication and subtraction. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 11. Question: A rectangle measures 12 cm long by 10 cm high. A semicircle of diameter 10 cm is attached along its full right-hand side. Give total area and exposed perimeter exactly. Method: Radius=5 cm. Total area=120 + (25/2)π cm². Its exposed perimeter is the top and bottom (2×12), left side (10), and semicircle arc (π×5): 34+5π cm. Check: The shared diameter of 10 cm is internal and is not counted in the exterior perimeter. Watch for: Do not add the semicircle diameter and the rectangle’s joined edge to the outside length.

Try a change: Label every dimension with a unit and trace the required region before choosing a formula. Describe which line segment is shared internally and why it should not appear in exterior perimeter. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 12. Question: A rectangle measures 14 cm long by 12 cm high. A semicircle of diameter 12 cm is attached along its full right-hand side. Give total area and exposed perimeter exactly. Method: Radius=6 cm. Total area=168 + (36/2)π cm². Its exposed perimeter is the top and bottom (2×14), left side (12), and semicircle arc (π×6): 40+6π cm. Check: The shared diameter of 12 cm is internal and is not counted in the exterior perimeter. Watch for: Do not add the semicircle diameter and the rectangle’s joined edge to the outside length.

Try a change: Ask whether parts overlap in area or only meet along a shared edge. That controls addition and exposed perimeter. Try a reverse question using the area to recover one missing side length. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Square courtyard frame with a clear border

This set highlights square courtyard frame with a clear border. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 13. Question: A square tiled border has outer side length 12 m and uniform width 2 m. Find its tiled area and the total length of its outer and inner edges. Method: Inner square side=12−2×2=8 m. Border area=12²−8²=80 m². Outer edge=4×12=48 m and inner edge=4×8=32 m, total=80 m. Check: Both inner and outer edges are included because the question explicitly requests both. Watch for: Do not subtract a single border width from the outer side; the border lies on both sides.

Try a change: Try subtracting a familiar cut-out from a large bounding rectangle rather than dividing a complicated outline arbitrarily. Create a second solution by decomposing the region in a different valid way. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 14. Question: A square tiled border has outer side length 14 m and uniform width 3 m. Find its tiled area and the total length of its outer and inner edges. Method: Inner square side=14−2×3=8 m. Border area=14²−8²=132 m². Outer edge=4×14=56 m and inner edge=4×8=32 m, total=88 m. Check: Both inner and outer edges are included because the question explicitly requests both. Watch for: Do not subtract a single border width from the outer side; the border lies on both sides.

Try a change: Estimate whether the remaining area should be smaller than the uncut whole; if not, check the subtraction. Predict the effect of doubling only one linear dimension, rather than doubling all dimensions. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 15. Question: A square tiled border has outer side length 16 m and uniform width 4 m. Find its tiled area and the total length of its outer and inner edges. Method: Inner square side=16−2×4=8 m. Border area=16²−8²=192 m². Outer edge=4×16=64 m and inner edge=4×8=32 m, total=96 m. Check: Both inner and outer edges are included because the question explicitly requests both. Watch for: Do not subtract a single border width from the outer side; the border lies on both sides.

Try a change: Keep exact π expressions when appropriate and round only according to the problem’s instructions. Explain whether a painted-area question and a fencing question would ask for different parts of this diagram. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 16. Question: A square tiled border has outer side length 18 m and uniform width 5 m. Find its tiled area and the total length of its outer and inner edges. Method: Inner square side=18−2×5=8 m. Border area=18²−8²=260 m². Outer edge=4×18=72 m and inner edge=4×8=32 m, total=104 m. Check: Both inner and outer edges are included because the question explicitly requests both. Watch for: Do not subtract a single border width from the outer side; the border lies on both sides.

Try a change: For perimeter, read ‘outside edge’ or ‘all exposed edges’ carefully, because inner holes can change the boundary counted. Check the units on each intermediate multiplication and subtraction. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Annulus: area and both circular boundaries

This set highlights annulus: area and both circular boundaries. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 17. Question: An annular ring has outer radius 6 cm and inner radius 3 cm. Find its material area and the combined length of its two circular boundaries. Method: Ring area=π(6²−3²)=27π cm². Boundary length=2π6+2π3=18π cm. Check: The inner radius is smaller, so material area is positive; both circles form ring boundaries. Watch for: Do not confuse π(R²−r²), an area, with 2π(R+r), a boundary length.

Try a change: A composite answer needs a unit: centimetres for lengths and square centimetres for areas. Describe which line segment is shared internally and why it should not appear in exterior perimeter. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 18. Question: An annular ring has outer radius 8 cm and inner radius 4 cm. Find its material area and the combined length of its two circular boundaries. Method: Ring area=π(8²−4²)=48π cm². Boundary length=2π8+2π4=24π cm. Check: The inner radius is smaller, so material area is positive; both circles form ring boundaries. Watch for: Do not confuse π(R²−r²), an area, with 2π(R+r), a boundary length.

Try a change: Draw an alternative decomposition of the same region and check that the area total agrees. Try a reverse question using the area to recover one missing side length. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 19. Question: An annular ring has outer radius 10 cm and inner radius 5 cm. Find its material area and the combined length of its two circular boundaries. Method: Ring area=π(10²−5²)=75π cm². Boundary length=2π10+2π5=30π cm. Check: The inner radius is smaller, so material area is positive; both circles form ring boundaries. Watch for: Do not confuse π(R²−r²), an area, with 2π(R+r), a boundary length.

Try a change: Mark shared internal joints with an X when calculating outside perimeter so they are not counted twice. Create a second solution by decomposing the region in a different valid way. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 20. Question: An annular ring has outer radius 12 cm and inner radius 6 cm. Find its material area and the combined length of its two circular boundaries. Method: Ring area=π(12²−6²)=108π cm². Boundary length=2π12+2π6=36π cm. Check: The inner radius is smaller, so material area is positive; both circles form ring boundaries. Watch for: Do not confuse π(R²−r²), an area, with 2π(R+r), a boundary length.

Try a change: Return after two days to an L-shape or ring with different measurements and explain the first step unaided. Predict the effect of doubling only one linear dimension, rather than doubling all dimensions. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

A rectangular house shape with a triangle roof

This set highlights a rectangular house shape with a triangle roof. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 21. Question: A 6 cm wide, 5 cm tall rectangle has a triangle roof attached along its full 6 cm top. The roof’s perpendicular height is 4 cm. Find total area. Method: Rectangle area=6×5=30 cm². Roof area=½×6×4=12 cm². Combined non-overlapping area=42 cm². Check: The shared roof base has zero area, so the two region areas add without subtraction. Watch for: A sloping roof side cannot replace the stated perpendicular height in the triangle-area formula.

Try a change: Label every dimension with a unit and trace the required region before choosing a formula. Explain whether a painted-area question and a fencing question would ask for different parts of this diagram. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 22. Question: A 8 cm wide, 6 cm tall rectangle has a triangle roof attached along its full 8 cm top. The roof’s perpendicular height is 5 cm. Find total area. Method: Rectangle area=8×6=48 cm². Roof area=½×8×5=20 cm². Combined non-overlapping area=68 cm². Check: The shared roof base has zero area, so the two region areas add without subtraction. Watch for: A sloping roof side cannot replace the stated perpendicular height in the triangle-area formula.

Try a change: Ask whether parts overlap in area or only meet along a shared edge. That controls addition and exposed perimeter. Check the units on each intermediate multiplication and subtraction. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 23. Question: A 10 cm wide, 7 cm tall rectangle has a triangle roof attached along its full 10 cm top. The roof’s perpendicular height is 6 cm. Find total area. Method: Rectangle area=10×7=70 cm². Roof area=½×10×6=30 cm². Combined non-overlapping area=100 cm². Check: The shared roof base has zero area, so the two region areas add without subtraction. Watch for: A sloping roof side cannot replace the stated perpendicular height in the triangle-area formula.

Try a change: Try subtracting a familiar cut-out from a large bounding rectangle rather than dividing a complicated outline arbitrarily. Describe which line segment is shared internally and why it should not appear in exterior perimeter. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 24. Question: A 12 cm wide, 8 cm tall rectangle has a triangle roof attached along its full 12 cm top. The roof’s perpendicular height is 7 cm. Find total area. Method: Rectangle area=12×8=96 cm². Roof area=½×12×7=42 cm². Combined non-overlapping area=138 cm². Check: The shared roof base has zero area, so the two region areas add without subtraction. Watch for: A sloping roof side cannot replace the stated perpendicular height in the triangle-area formula.

Try a change: Estimate whether the remaining area should be smaller than the uncut whole; if not, check the subtraction. Try a reverse question using the area to recover one missing side length. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Two separate holes in a rectangular panel

This set highlights two separate holes in a rectangular panel. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 25. Question: A panel is 16 cm by 12 cm. Two separate interior square holes have sides 2 cm and 3 cm. Find its remaining area. Method: Outer area=16×12=192 cm². Two holes have area 2²+3²=4+9=13 cm². Material area=192−13=179 cm². Check: The removed squares are distinct and contained, so subtracting their areas does not double-count any region. Watch for: Subtract square AREAS (4 and 9), not the square side lengths (2 and 3).

Try a change: Keep exact π expressions when appropriate and round only according to the problem’s instructions. Create a second solution by decomposing the region in a different valid way. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 26. Question: A panel is 18 cm by 13 cm. Two separate interior square holes have sides 2 cm and 3 cm. Find its remaining area. Method: Outer area=18×13=234 cm². Two holes have area 2²+3²=4+9=13 cm². Material area=234−13=221 cm². Check: The removed squares are distinct and contained, so subtracting their areas does not double-count any region. Watch for: Subtract square AREAS (4 and 9), not the square side lengths (2 and 3).

Try a change: For perimeter, read ‘outside edge’ or ‘all exposed edges’ carefully, because inner holes can change the boundary counted. Predict the effect of doubling only one linear dimension, rather than doubling all dimensions. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 27. Question: A panel is 20 cm by 14 cm. Two separate interior square holes have sides 2 cm and 3 cm. Find its remaining area. Method: Outer area=20×14=280 cm². Two holes have area 2²+3²=4+9=13 cm². Material area=280−13=267 cm². Check: The removed squares are distinct and contained, so subtracting their areas does not double-count any region. Watch for: Subtract square AREAS (4 and 9), not the square side lengths (2 and 3).

Try a change: A composite answer needs a unit: centimetres for lengths and square centimetres for areas. Explain whether a painted-area question and a fencing question would ask for different parts of this diagram. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 28. Question: A panel is 22 cm by 15 cm. Two separate interior square holes have sides 2 cm and 3 cm. Find its remaining area. Method: Outer area=22×15=330 cm². Two holes have area 2²+3²=4+9=13 cm². Material area=330−13=317 cm². Check: The removed squares are distinct and contained, so subtracting their areas does not double-count any region. Watch for: Subtract square AREAS (4 and 9), not the square side lengths (2 and 3).

Try a change: Draw an alternative decomposition of the same region and check that the area total agrees. Check the units on each intermediate multiplication and subtraction. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Rectangle joined to a trapezium

This set highlights rectangle joined to a trapezium. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 29. Question: A rectangle 8 cm by 5 cm has a trapezium attached along its entire 8-cm upper side. The trapezium’s parallel bases are 8 cm and 12 cm; perpendicular height is 3 cm. Find total area. Method: Rectangle area=40 cm². Trapezium area=½(8+12)×3=30 cm². Joined along a line with no interior overlap, total area=70 cm². Check: The trapezium height used is perpendicular to its bases; the attachment is a shared line, not overlapping area. Watch for: Do not use one slanted side as the trapezium height or subtract the shared base from area.

Try a change: Mark shared internal joints with an X when calculating outside perimeter so they are not counted twice. Describe which line segment is shared internally and why it should not appear in exterior perimeter. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 30. Question: A rectangle 10 cm by 6 cm has a trapezium attached along its entire 10-cm upper side. The trapezium’s parallel bases are 10 cm and 14 cm; perpendicular height is 4 cm. Find total area. Method: Rectangle area=60 cm². Trapezium area=½(10+14)×4=48 cm². Joined along a line with no interior overlap, total area=108 cm². Check: The trapezium height used is perpendicular to its bases; the attachment is a shared line, not overlapping area. Watch for: Do not use one slanted side as the trapezium height or subtract the shared base from area.

Try a change: Return after two days to an L-shape or ring with different measurements and explain the first step unaided. Try a reverse question using the area to recover one missing side length. A useful correction is one the student can independently apply to a new outline without being told whether addition or subtraction will work.

Worked example 31. Question: A rectangle 12 cm by 7 cm has a trapezium attached along its entire 12-cm upper side. The trapezium’s parallel bases are 12 cm and 16 cm; perpendicular height is 5 cm. Find total area. Method: Rectangle area=84 cm². Trapezium area=½(12+16)×5=70 cm². Joined along a line with no interior overlap, total area=154 cm². Check: The trapezium height used is perpendicular to its bases; the attachment is a shared line, not overlapping area. Watch for: Do not use one slanted side as the trapezium height or subtract the shared base from area.

Worked example 32. Question: A rectangle 14 cm by 8 cm has a trapezium attached along its entire 14-cm upper side. The trapezium’s parallel bases are 14 cm and 18 cm; perpendicular height is 6 cm. Find total area. Method: Rectangle area=112 cm². Trapezium area=½(14+18)×6=96 cm². Joined along a line with no interior overlap, total area=208 cm². Check: The trapezium height used is perpendicular to its bases; the attachment is a shared line, not overlapping area. Watch for: Do not use one slanted side as the trapezium height or subtract the shared base from area.

Paintable wall with a door opening

This set highlights paintable wall with a door opening. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 33. Question: A wall is 5 m wide and 3 m high, with an unpainted door opening 2 m by 1 m. Two coats each cost $4 per m² of wall covered. Find painted area per coat and total coating cost. Method: Wall area=5×3=15 m². Door area=2 m², so painted area per coat=13 m². Two coats cost 2×13×$4=$104. Check: Total coating cost divided by two gives a single-coat cost of $52 for 13 m². Watch for: Do not subtract the door for each coat twice; calculate area once and then apply the coat count.

Worked example 34. Question: A wall is 6 m wide and 4 m high, with an unpainted door opening 2 m by 1 m. Two coats each cost $5 per m² of wall covered. Find painted area per coat and total coating cost. Method: Wall area=6×4=24 m². Door area=2 m², so painted area per coat=22 m². Two coats cost 2×22×$5=$220. Check: Total coating cost divided by two gives a single-coat cost of $110 for 22 m². Watch for: Do not subtract the door for each coat twice; calculate area once and then apply the coat count.

Worked example 35. Question: A wall is 7 m wide and 5 m high, with an unpainted door opening 2 m by 1 m. Two coats each cost $6 per m² of wall covered. Find painted area per coat and total coating cost. Method: Wall area=7×5=35 m². Door area=2 m², so painted area per coat=33 m². Two coats cost 2×33×$6=$396. Check: Total coating cost divided by two gives a single-coat cost of $198 for 33 m². Watch for: Do not subtract the door for each coat twice; calculate area once and then apply the coat count.

Worked example 36. Question: A wall is 8 m wide and 6 m high, with an unpainted door opening 2 m by 1 m. Two coats each cost $7 per m² of wall covered. Find painted area per coat and total coating cost. Method: Wall area=8×6=48 m². Door area=2 m², so painted area per coat=46 m². Two coats cost 2×46×$7=$644. Check: Total coating cost divided by two gives a single-coat cost of $322 for 46 m². Watch for: Do not subtract the door for each coat twice; calculate area once and then apply the coat count.

A T-shaped joined region and its outer edge

This set highlights a t-shaped joined region and its outer edge. Before solving, sketch the shape and mark its included, excluded and shared parts. Make the first decision about the region or boundary, then use the formula for each simple piece. Four number variations make the method easier to retain.

Worked example 37. Question: A T is formed by a horizontal rectangle 10 cm wide and 3 cm tall, with a 4-cm wide vertical stem extending 5 cm below it. They meet along the stem’s whole top edge without overlap. Find area and outer perimeter. Method: Area=bar 10×3 plus stem 4×5=50 cm². For perimeter, adding the rectangle outlines then removing the two copies of their shared width gives 2(10+3)+2(4+5)−24=36 cm. Check: The stem’s attachment is an internal seam. Cancelling its two copies keeps only the outside edge. Watch for: Counting both parts’ full perimeters without removing the shared seam overstates the exterior length.

Worked example 38. Question: A T is formed by a horizontal rectangle 12 cm wide and 4 cm tall, with a 5-cm wide vertical stem extending 7 cm below it. They meet along the stem’s whole top edge without overlap. Find area and outer perimeter. Method: Area=bar 12×4 plus stem 5×7=83 cm². For perimeter, adding the rectangle outlines then removing the two copies of their shared width gives 2(12+4)+2(5+7)−25=46 cm. Check: The stem’s attachment is an internal seam. Cancelling its two copies keeps only the outside edge. Watch for: Counting both parts’ full perimeters without removing the shared seam overstates the exterior length.

Worked example 39. Question: A T is formed by a horizontal rectangle 14 cm wide and 5 cm tall, with a 6-cm wide vertical stem extending 9 cm below it. They meet along the stem’s whole top edge without overlap. Find area and outer perimeter. Method: Area=bar 14×5 plus stem 6×9=124 cm². For perimeter, adding the rectangle outlines then removing the two copies of their shared width gives 2(14+5)+2(6+9)−26=56 cm. Check: The stem’s attachment is an internal seam. Cancelling its two copies keeps only the outside edge. Watch for: Counting both parts’ full perimeters without removing the shared seam overstates the exterior length.

Worked example 40. Question: A T is formed by a horizontal rectangle 16 cm wide and 6 cm tall, with a 7-cm wide vertical stem extending 11 cm below it. They meet along the stem’s whole top edge without overlap. Find area and outer perimeter. Method: Area=bar 16×6 plus stem 7×11=173 cm². For perimeter, adding the rectangle outlines then removing the two copies of their shared width gives 2(16+6)+2(7+11)−27=66 cm. Check: The stem’s attachment is an internal seam. Cancelling its two copies keeps only the outside edge. Watch for: Counting both parts’ full perimeters without removing the shared seam overstates the exterior length.

Six common composite-shape mistakes

Using perimeter formulas when asked for shaded area

Identify the requested mathematical object first. Shaded material uses squared units; fencing or edging uses length units. Locate the first line in the student’s original attempt where a region, edge or unit was misidentified. Draw that line correctly and solve one changed diagram without hints. A repair should transfer beyond the first example.

Counting shared internal seams as exterior edges

Trace the outer boundary and exclude joins between adjacent non-overlapping pieces. For two perimeters added together, subtract both copies of a shared edge. Locate the first line in the student’s original attempt where a region, edge or unit was misidentified. Draw that line correctly and solve one changed diagram without hints. A repair should transfer beyond the first example.

Forgetting inner holes when the question counts all boundaries

An internal opening creates a new inner edge. Include it only when the task measures every exposed edge, not when it asks solely for the external fence. Locate the first line in the student’s original attempt where a region, edge or unit was misidentified. Draw that line correctly and solve one changed diagram without hints. A repair should transfer beyond the first example.

Treating a semicircle’s diameter as its radius

Read whether the given length spans the entire circle and divide a diameter by two before using r in πr² or πr. Locate the first line in the student’s original attempt where a region, edge or unit was misidentified. Draw that line correctly and solve one changed diagram without hints. A repair should transfer beyond the first example.

Using a slant length as a perpendicular height

The triangle and trapezium area formulas use perpendicular height. Mark a right angle before substituting an apparent side length. Locate the first line in the student’s original attempt where a region, edge or unit was misidentified. Draw that line correctly and solve one changed diagram without hints. A repair should transfer beyond the first example.

Converting square units with a one-dimensional factor

Square the length conversion: 1 m² equals 10,000 cm². Always state converted area units clearly. Locate the first line in the student’s original attempt where a region, edge or unit was misidentified. Draw that line correctly and solve one changed diagram without hints. A repair should transfer beyond the first example.

A small-group tutor can build area reasoning, not just formula recall

A focused Secondary 3 G3 Mathematics tutorial in Bukit Timah can give each student the same L-shaped diagram and three different questions: area remaining, outside perimeter and length of all cut-out edges. If they use one familiar rectangle formula for all three, the misunderstanding becomes visible. The tutor can trace the requested measurement, then ask pupils to justify which pieces belong in their answer.

Next, introduce an attached semicircle or an interior hole. Each learner should label the shared edge before adding areas or perimeters. A delayed unseen question tests whether the student retains the habit once the tutorial’s example is no longer in view. Class size may support feedback, but genuine progress depends on independent reasoning and corrections.

The unchanged Clementi Secondary 1 Mathematics small-group guide describes eduKateSG’s foundation approach, while the Secondary 3 G3 Mathematics tuition Bukit Timah overview provides year-specific context. The school chapter schedule and actual class arrangements should be confirmed for an individual learner.

A practical six-week composite-area routine

Week 1 — separate area and perimeter

Use simple rectangles, triangles and half-circles to identify output units and which lines count as boundaries. Keep a dated original attempt and a later unseen attempt. Watch whether the student chooses the correct region or boundary before performing the calculation.

Week 2 — add adjacent shapes

Join rectangles with triangles or trapeziums, and justify why their areas add while shared seams do not belong to external perimeters. Keep a dated original attempt and a later unseen attempt. Watch whether the student chooses the correct region or boundary before performing the calculation.

Week 3 — practise subtraction and cut-outs

Take rectangles with rectangular holes and one-corner notches. Label the region removed and explain the effect on the outside outline. Keep a dated original attempt and a later unseen attempt. Watch whether the student chooses the correct region or boundary before performing the calculation.

Week 4 — work with curved boundaries

Combine rectangles and semicircles, distinguish arc length from area, and retain π correctly until the final reported answer. Keep a dated original attempt and a later unseen attempt. Watch whether the student chooses the correct region or boundary before performing the calculation.

Week 5 — introduce frames, rings and units

Compare inside and outside edges, then change between cm² and m² with the squared conversion factor. Keep a dated original attempt and a later unseen attempt. Watch whether the student chooses the correct region or boundary before performing the calculation.

Week 6 — solve mixed floor-plan questions

Use fresh diagrams without a method heading. Check whether the learner independently decides which shapes to add, subtract or trace. Keep a dated original attempt and a later unseen attempt. Watch whether the student chooses the correct region or boundary before performing the calculation.

Frequently asked questions

How do I find the area of an L-shaped figure?

Often subtract a smaller missing rectangle from a complete outer rectangle, or add non-overlapping rectangular parts. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

Does removing a corner always reduce perimeter?

No. A rectangular notch cut from one corner can leave the overall perimeter unchanged even though area decreases. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

Should I count a shared edge?

For external perimeter, no. A shared join between attached pieces lies inside the composite outline. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

Does a hole count toward total boundary length?

Only if the question includes the inner exposed edge or edging around that opening. It does not count toward an outside-only fence. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

What is the arc length of a semicircle?

πr, half a circle’s full circumference. A standalone semicircle’s full perimeter also includes the diameter 2r. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

How do I calculate a square frame area?

Subtract the smaller internal square area from the outer square area. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

Is a trapezium’s slanted side its height?

Not generally. The formula uses perpendicular distance between the parallel bases. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

How many square centimetres are in one square metre?

10,000 cm², because both sides of the unit square are 100 cm. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

Do I have to calculate every side separately?

For perimeter, trace and total the exposed sides. Symmetry or cancelling equal cut-out lengths can simplify the calculation. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

Are composite 2D shapes in the G3 syllabus?

Yes. The 2027 K310 Geometry and Measurement mensuration section explicitly includes perimeter and area of composite plane figures. Try a fresh example and ask the learner to identify which lines or regions belong to the requested quantity before substituting numbers.

The official course reference is SEAB 2027 SEC G3 Mathematics syllabus K310, Mensuration G5. This resource offers original explanations and practice rather than reproduced examination papers.

The takeaway

Did you know? The hardest part of a composite shape is often deciding what belongs to the figure. Shade the included region, cross out the removed part, trace the external edge and label the units. When those decisions are correct, the arithmetic usually reduces to familiar rectangles, triangles and circles. A tutor’s job is to make that reasoning independent.

Secondary 3 G3 Mathematics Tutor Bukit Timah | Composite Plane Figures: Area, Perimeter and Cut-Outs