Primary 6 Mathematics tuition for Serangoon North Avenue 4 can help pupils distinguish a circular disc from the four corner regions left inside a square. An inscribed circle in a 28-cm square has radius 14 cm. Using π = 22/7, the square corners together occupy 168 cm².
The weekday-versus-weekend decision depends on whether the child reads diameter as radius, chooses the shaded region correctly or confuses the outer square boundary with the circular curve.
At eduKateSG, focused three-pupil tutorials aim to identify the first geometric misinterpretation and verify it on a changed size. These are original mathematical figures, not physical measurements of local facilities.
A Circle Inside a Square: Shaded Corners
What an inscribed circle means here · Find the square’s area first · Find the circle’s area using radius fourteen · Worked example: all four square corners
What an inscribed circle means here
The fictional drawing contains a square with each side twenty-eight centimetres long and a circle centred inside so that it touches the middle of all four sides.
The circle’s full diameter is therefore twenty-eight centimetres, equal to the square side. The radius is half of that, or 14 cm.
The four outer corner pieces are the regions within the square but outside the circle. They together make a composite shaded region, not a separate larger circle.
Ask the learner to shade only those four corner pieces. A pupil who shades the circle itself has changed the requested region before any arithmetic begins.
Find the square’s area first
The square’s length and breadth both measure twenty-eight centimetres. Its area is 28 × 28 = 784 cm².
This complete area includes the central circular disc as well as the four corners. It is the appropriate starting whole when the final question asks for the area not occupied by the circle.
A quick dimensional check uses squared units. The square’s perimeter is separately four times twenty-eight, or 112 cm.
Do not subtract a circumference from an area. Both numbers can be calculated from the diagram but measure incompatible quantities.
Find the circle’s area using radius fourteen
The inscribed circle has radius fourteen, not twenty-eight. With π = 22/7, its area is (22/7) × 14 × 14 = 616 cm².
The fraction cancels neatly because 14 is a multiple of seven. Multiplying with the full diameter as though it were the radius would give a greatly inflated disc area.
Check the size: a circle entirely inside the 784-cm² square must have area less than 784. The calculated 616 passes that bound.
The square and circle share a centre, but their area calculations use different formulas because their shapes are different.
Worked example: all four square corners
Subtract the inscribed disc from the complete square: 784 − 616 = 168 cm² for all four corner regions combined.
The four corners are equal under the specified centred arrangement. Each therefore occupies 168 ÷ 4 = 42 cm².
Check independently by taking a quarter of the square: 784 ÷ 4 = 196. A quarter of the disc is 616 ÷ 4 = 154. The difference is 196 − 154 = forty-two per corner.
Multiply forty-two by four to recover 168. That independent quarter-region route verifies the same shaded material.
A learner who writes 168 centimetres has calculated a region with the wrong unit. Area must use cm².
Further worked examples and independent checks
Trace the square boundary and the circle boundary separately · Worked example: area and full boundary of one corner · Worked variation: a circle inside a fourteen-centimetre square · Worked variation: a circle inside a forty-two-centimetre square · Recover a square side from the circle circumference · What if only two opposite corners are shaded? · Follow the problem’s approximation for π · Recover the circle area from the corner area
Trace the square boundary and the circle boundary separately
The original twenty-eight-centimetre square has an outside boundary of four equal sides. Its perimeter is 4 × 28 = 112 cm.
The inscribed radius-fourteen circle has circumference 2 × 22/7 × 14 = 88 cm under the given π approximation.
The four corner regions are separated from one another by the circle’s tangent points. If edging is required around every one of the four corner pieces, the combined boundary length includes the complete square perimeter and the full circular boundary: 200 cm.
This 200-cm answer describes the sum of boundaries of four separate regions, not their combined area. Ask the learner what lines are being traced before adding.
A pupil who reports only 112 ignores the curved edges, while one who reports only eighty-eight ignores the square edges. Both are correct numbers attached to incomplete boundary targets.
Worked example: area and full boundary of one corner
The centred circle divides the unshaded part of the square into four congruent corner regions. The combined corner area is 168 cm², so each has area 42 cm².
One corner contains two straight edge segments, each running from a square corner to a circle-tangency point. Each straight segment has length fourteen centimetres, half the square side.
The curved boundary is one quarter of the circle’s eighty-eight-centimetre circumference, or 22 cm.
Thus one corner’s full boundary length is 14 + 14 + 22 = 50 cm. Four separate corners have combined boundary length two hundred centimetres.
Check area independently through quarter shapes: ¼ × 784 − ¼ × 616 = 196 − 154 = 42 cm². This verifies the area separately from the boundary calculation.
Worked variation: a circle inside a fourteen-centimetre square
A second fictional diagram contains a square of side fourteen centimetres and an inscribed circle of radius seven.
The square area is 14 × 14 = 196 cm². The circular disc area with π = 22/7 is 154 cm².
The four corners together occupy 196 − 154 = 42 cm², and each corner occupies 10½ cm².
The square perimeter is fifty-six centimetres, while the circle circumference is forty-four. The combined boundaries of all four corner regions have total length 100 cm.
For one small corner, the straight edge lengths are seven and seven while the curved quarter-circle arc is eleven, yielding 25 cm of perimeter. Compare how every result changes with the square size.
Worked variation: a circle inside a forty-two-centimetre square
Now consider a square with side forty-two centimetres and a centred inscribed circle of radius twenty-one.
The square area is 42 × 42 = 1,764 cm². The circle area is 22/7 × 21 × 21 = 1,386 cm².
Subtract to obtain 378 cm² across all four square corners. Each congruent corner has area 378 ÷ 4 = 94½ cm².
The square perimeter is 168 cm and the circle circumference 132 cm. The total boundary of all four corners is therefore 300 cm.
Each corner has two straight segments twenty-one centimetres long and a quarter-circle arc thirty-three centimetres long, for a perimeter of 21 + 21 + 33 = 75 cm.
These changed measurements test whether the pupil understands the square–disc decomposition rather than remembering 168 square centimetres from the main example.
Recover a square side from the circle circumference
Suppose a different question states that a circle inscribed within an unknown square has circumference eighty-eight centimetres, with π taken as 22/7.
Use circumference = π × diameter. Divide eighty-eight by 22/7 to obtain 28 cm for the diameter.
An inscribed circle’s diameter equals its containing square side, so the square side is also twenty-eight centimetres and its radius is fourteen.
Check forwards: 22/7 × 28 = 88 centimetres of circumference. The reconstructed square area is 784 cm² and the disc area is 616 cm².
This inverse route is useful for pupils who can use the formulas forward but struggle to identify which length must be recovered first.
What if only two opposite corners are shaded?
Keep the main twenty-eight-centimetre square with its inscribed radius-fourteen circle. Suppose only the top-left and bottom-right corner regions are shaded.
Each corner has area forty-two square centimetres, so those two together have 84 cm² of shading.
If the question instead asks for the perimeter of both separate shaded pieces, each has boundary length fifty centimetres and their total is 100 cm.
These two shaded regions do not share a boundary segment, so their combined boundary total is simply the sum of two separate perimeters.
A pupil who uses the entire 168-cm² four-corner area despite the instruction that only two are shaded has answered a different region question.
Follow the problem’s approximation for π
The principal example specifies π = 22/7, giving the radius-fourteen circle an area of 616 cm² and the combined corner area of 168 cm².
If a changed problem instead specifies π = 3.14, the circle area is 3.14 × 196 = 615.44 cm². The corners then total 784 − 615.44 = 168.56 cm².
The circle circumference using 3.14 would be 3.14 × 28 = 87.92 cm, so both boundary types together total 112 + 87.92 = 199.92 cm.
The small differences arise from the specified numerical approximations. Neither 22/7 nor 3.14 is mathematically identical to the exact constant π.
Do not mix approximations across an area subtraction, or round an intermediate value without a question instructing that change.
Recover the circle area from the corner area
Imagine an unknown inscribed circle in a square with area 784 cm². The four corner regions together have area 168 cm².
The central disc must therefore have area 784 − 168 = 616 cm². This calculation reverses the shading subtraction.
If π = 22/7, divide 616 by 22/7 to obtain radius squared equal to 196. The positive radius is 14 cm.
Check that the diameter twenty-eight fits exactly across a square of side twenty-eight, as required by an inscribed arrangement.
The pupil should distinguish the circle’s radius from the area of a corner. Finding the radius is a separate inverse operation after recovering the missing region.
Circle and square area extensions with independent checks
A square boundary and circular boundary measure different things · Worked example: perimeter around one shaded corner region · Why all four separate corner perimeters total 200 cm · What percentage of the square lies outside the circle? · A corner area as a fraction of the complete square · Verify one corner with a quarter-square minus a quarter-disc · Changed example: a circle inside a 56 cm square · Why the total corner area scales quadratically · Different prescribed approximations of π produce different decimals · Recover the circle area when the corner material is known · The four corners are equal because of the configuration · Different geometric relationships require different area operations
A square boundary and circular boundary measure different things
The fictional diagram shows a circle exactly touching the four sides of a twenty-eight-centimetre square. Its inscribed radius is fourteen centimetres, half the square’s side.
The square’s outside perimeter is 4 × 28 = 112 cm. The circular boundary length is 2 × 22/7 × 14 = 88 cm using the specified π approximation.
Those numbers are both lengths, but they belong to different curves. A question about the square frame needs 112; a circular string around the disc needs eighty-eight.
The four corner regions lie within the square and outside the circular disc. Their combined area is 168 cm², which cannot be added directly to either perimeter because its unit is squared.
Ask the learner to trace each boundary with a different colour and then shade the corner regions. That separates the geometrical targets before any formula is selected.
Worked example: perimeter around one shaded corner region
Take the upper-left corner of the twenty-eight-centimetre square. The inscribed circle touches the midpoint of the upper and left sides, fourteen centimetres from that corner along each edge.
The single corner region has two straight boundary segments, each fourteen centimetres long, meeting at the ninety-degree square corner.
Its curved inner edge is a quarter of the radius-fourteen circle’s circumference. The complete circular circumference is eighty-eight, so one quarter is 22 cm.
The full boundary of this corner region is 14 + 14 + 22 = 50 cm. This is a length, not the corner’s forty-two-square-centimetre area.
The region’s perimeter needs both the curved and straight parts. A child who reports twenty-two has traced only the circular arc; one who reports twenty-eight has traced only the two straight edges.
Why all four separate corner perimeters total 200 cm
Each of the four symmetric corners has a boundary length of fifty centimetres. If the four corner regions are treated separately, their combined perimeters total 4 × 50 = 200 cm.
Check through the geometry instead of repeating the multiplication. The four corner regions together touch the entire outside square frame of 112 cm and the entire inscribed circle boundary of eighty-eight centimetres.
Adding those two distinct boundaries gives 112 + 88 = 200 cm, agreeing with the sum of the four individual perimeters.
The corner regions meet at their endpoints only and do not share whole boundary segments that would need to be subtracted from this total.
This two-route check is useful for a confident learner, but the question must specify whether it wants one corner, all four boundaries or only the outer square edging.
What percentage of the square lies outside the circle?
The original square has area 784 cm², while the inscribed circle has area 616 cm² using π = 22/7. The unshaded disc plus the four shaded corners fill the complete square.
The combined corners cover 784 − 616 = 168 cm². Their fraction of the square is 168/784.
Divide numerator and denominator by fifty-six to simplify this fraction to 3/14. As a percentage, three fourteenths is approximately 21.43%.
The circle occupies the complementary fraction 616/784 = 11/14, approximately 78.57%. These fractions sum to one whole.
Ask the pupil which shape is the denominator. The forty-two-square-centimetre area of one corner is one quarter of the combined corner area, but it is not one quarter of the entire square.
A corner area as a fraction of the complete square
One corner region has area 42 cm². Compared with the complete 784-cm² square, the fraction is 42/784.
Simplifying by fourteen gives 3/56 of the square, about 5.36%.
Four such corners occupy four times that share, 12/56 or 3/14, consistent with the combined corner calculation.
This is a different fraction from one quarter of the total corner material. The phrase ‘of the shaded region’ uses 168 as the reference whole, whereas ‘of the square’ uses 784.
A tutor can ask the learner to state the relevant whole before writing a denominator. A correct numerator paired with the wrong whole describes a different mathematical relationship.
Verify one corner with a quarter-square minus a quarter-disc
Divide the twenty-eight-centimetre square using its horizontal and vertical centre lines. Each quarter of the square is fourteen by fourteen and has area 196 cm².
A quarter of the inscribed radius-fourteen circle has area 616 ÷ 4 = 154 cm².
The remaining shaded region in that quadrant is 196 − 154 = 42 cm², equal to the one-corner area found by dividing all 168 corner units by four.
This is a second geometric decomposition. It independently confirms that each corner consists of a quarter-square region with the quarter-disc removed.
An error that adds the quarter-disc instead of subtracting it would exceed the available 196-cm² quadrant, which contradicts the diagram.
Changed example: a circle inside a 56 cm square
A second original figure uses a larger square of side fifty-six centimetres with a circle touching all four sides. The inscribed circle has radius twenty-eight centimetres.
The square’s area is 56 × 56 = 3,136 cm². Using π = 22/7, the circle’s area is 22/7 × 28 × 28 = 2,464 cm².
The four corner regions therefore occupy 3,136 − 2,464 = 672 cm². Each equal corner occupies 672 ÷ 4 = 168 cm².
Check against the original side-twenty-eight figure: doubling the square side and circle radius multiplies the area of every corresponding region by four. Original combined corners 168 become 672.
An answer of 336 for the new combined corners would apply a one-dimensional doubling rule to an area, overlooking that both width and height doubled.
Why the total corner area scales quadratically
Write the side length of an inscribed-circle square as s. Its area is s², while the inscribed circle has radius s/2 and area π(s/2)².
The combined corner area can be represented as s² − πs²/4 = s²(1 − π/4). The factor in parentheses depends only on the chosen π approximation.
Doubling s multiplies s² by four. Therefore the square, circle and their difference in shaded corner material all gain area by a factor of four.
That symbolic relationship is an optional extension; a P6 learner can see the same result numerically by comparing the 28-cm and 56-cm squares.
Use the extension when the learner already interprets area correctly, rather than replacing the basic shaded-region subtraction with an algebraic expression too early.
Different prescribed approximations of π produce different decimals
The exact mathematical constant π cannot be represented perfectly by the fraction 22/7 or the decimal 3.14. A school problem usually specifies the numerical approximation it expects.
For the twenty-eight-centimetre square, using π = 22/7 gives the inscribed circle area 616 cm² and corner total 168 cm².
Using π = 3.14 instead gives circle area 3.14 × 14 × 14 = 615.44 cm² and corner total 784 − 615.44 = 168.56 cm².
These differences arise from the instructed approximation, not from the learner changing the underlying square or radius.
Do not mix 22/7 for the whole disc with 3.14 for its quarter in the same calculation. Follow one stated approximation and the required final precision consistently.
Recover the circle area when the corner material is known
Suppose a square measures twenty-eight centimetres per side, and the combined corner material outside an inscribed circle is given as 168 cm². Find the circle’s area without calculating π immediately.
The square area is 28 × 28 = 784 cm². Subtract the stated corner material: 784 − 168 = 616 cm² for the disc.
Check that the inscribed circle’s radius must be half the square side, or fourteen centimetres. Using π = 22/7 gives 22/7 × 14² = 616, matching the recovered area.
This inverse route works because the circle and corner regions partition the square exactly without overlap or gaps.
A child who divides 168 by four in response to ‘circle area’ has found one corner’s area instead. The target quantity must be identified before selecting an operation.
The four corners are equal because of the configuration
An inscribed circle is centred in the square and touches each side at its midpoint. The horizontal and vertical lines through that common centre divide both square and circle into four congruent quadrants.
Every quadrant contains one 14-by-14 square region of area 196 cm² and one matching quarter-disc area of 154 cm². The remainder is forty-two in each.
This argument justifies splitting the combined 168-cm² corner area equally among four. Equality is not inferred merely because the corners appear similarly shaded in an imprecise sketch.
If a circle were moved off centre or did not touch all four sides, these four regions would not necessarily be equal, and the one-quarter shortcut could fail.
Ask the pupil to name the symmetry or construction condition that makes the four regions congruent before performing the division.
Different geometric relationships require different area operations
When a circle lies entirely inside a square, the region in the square outside the circle is found by subtracting the circle’s area from the square’s area.
If a semicircle were attached outside a rectangle along a shared edge, its non-overlapping area would instead be added to the rectangle. The shape name alone does not determine the sign.
Likewise a circular hole cut from a rectangular sheet reduces the available material even if the circle’s curved boundary introduces additional edges.
Ask the child to shade the target region and distinguish what is present from what is absent before multiplying any dimensions.
An effective tutorial connects the operation to the region’s physical construction, not to a memorised instruction to always subtract whenever a circle is drawn.
Common errors, weekday/weekend lessons and parent guidance
Six distinct errors in the same figure · When weekday Primary 6 Mathematics tuition may help · When weekend tuition can make the geometry clearer · Three pupils may need three different representations · A flexible four-checkpoint geometry plan · Questions from families around Serangoon North Avenue 4 · Official syllabus, earlier learning and consultation
Six distinct errors in the same figure
A pupil may read the circle’s diameter twenty-eight as though it were a radius of twenty-eight. The calculated disc area then becomes much larger than the enclosing square and can be rejected by a simple area bound.
Another may correctly find the 616-cm² circle but add it to the 784-cm² square rather than subtract it when asked about corners outside the circle.
A third can calculate 168 correctly but divide by two when the diagram contains four equal corner regions. A fourth may report 168 centimetres rather than square centimetres.
A fifth may find a curved quarter-circle arc of twenty-two centimetres but omit the straight edges when the question asks for one corner region’s complete boundary.
A sixth may add both the square perimeter and circular circumference when only the outer square frame was requested. Each first unsupported decision calls for a different teaching prompt.
When weekday Primary 6 Mathematics tuition may help
A weekday tutorial can reconnect with a school circle-and-square diagram while the learner still remembers why they selected a particular radius or shaded region.
Bring the original uncorrected attempt. The tutor should ask whether the student distinguished square side from circle diameter, named the target area or perimeter, and applied the specified π approximation.
Consider school dismissal, a meal, co-curricular commitments, travel to the actual teaching venue, homework and sleep. A useful class must leave enough attention for spatial explanation.
An illustrative session may cover the 28-cm square, compare the quarter-region method and end with the unseen 56-cm square variation.
A delayed question a few days later should change a dimension or request a boundary instead of repeating the same answer 168.
When weekend tuition can make the geometry clearer
A weekend session can provide time for a paper circle cut to fit inside a square, leaving four corner shapes. The child can shade the corners and trace the outside square frame separately from the circle boundary.
That model is useful when the arithmetic is correct but the pupil cannot reliably identify which region has been removed from the square.
Weekends may nevertheless contain sports, family visits, other tuition and preparation for school. A longer open period is advantageous only when the learner is rested and able to explain.
A focused lesson should begin with a private attempt, then offer the concrete model for the particular misunderstanding and finish with an independently calculated different-sized figure.
Use a short weekday retrieval question after the weekend so the child must identify radius and corner region without the physical cut-out beside them.
Three pupils may need three different representations
In a three-pupil eduKateSG group, one child may confuse diameter and radius, another may know both but subtract the wrong regions, and the third may solve area correctly while counting only the curve in a perimeter question.
The tutor can provide a centre-to-boundary marker, a shaded square-minus-disc model or a physical trace of the straight and curved corner edges.
An illustrative ninety-minute lesson might include individual diagnoses, the 784-minus-616 worked figure, one quarter-region check and a private exit problem for each pupil.
Group discussion can improve mathematical language but cannot replace each learner’s individual explanation of what is shaded or traced.
Parents can ask which first hint was needed and whether the learner required it again several days later on a different square size.
A flexible four-checkpoint geometry plan
Checkpoint one is dimension identification. State the square side, circular diameter and radius, and verify which pairs are linked by the inscribed-circle condition.
Checkpoint two is area: calculate the square region and full disc, subtract to find the combined corners and divide by four only after justifying symmetry.
Checkpoint three is boundary interpretation: distinguish the square’s perimeter, circle circumference and perimeter of an individual shaded corner, including its straight sides.
Checkpoint four is transfer: change the square side, request a fraction of the whole area or give a shaded corner value and ask for a missing complete region.
These checkpoints are diagnostic stages rather than a guaranteed number of weeks or a prediction of a PSLE result. Timed practice is useful after the child can interpret a new drawing independently.
Questions from families around Serangoon North Avenue 4
What is the area of the 28-cm square? It is 784 cm².
What is the inscribed circle area? With radius fourteen and π = 22/7, it is 616 cm².
What is the area of all four square corners? 784 − 616 = 168 cm², or forty-two for each identical corner.
How long is the outer square frame? One hundred twelve centimetres. The circular boundary separately measures eighty-eight.
What is one corner’s full boundary length? Two straight fourteen-centimetre pieces and a twenty-two-centimetre curved arc, totalling fifty centimetres.
Is weekday or weekend tuition inherently better? Neither. Focus on alertness, targeted feedback, independent retrieval and rest.
Does the location title announce a classroom in the neighbourhood? No. Confirm the actual teaching address and class availability directly.
Official syllabus, earlier learning and consultation
The MOE Primary Mathematics syllabus updated October 2025 provides current course coverage for circle area, circumference and related geometry.
The eduKate Mathematics Learning Hub connects this to earlier area and perimeter learning, including the Serangoon North Avenue 4 Primary 4 line graph guide for separate level-specific Mathematics practice.
All squares and circles on this page are original teaching figures rather than actual property measurements. For a real eduKate Singapore parent–student consultation, confirm school level, actual three-pupil teaching venue, current fees and feasible weekday or weekend travel before arranging a class.
Continue through Serangoon North Avenue 4 Mathematics: Primary 1: Comparing Quantities · Primary 2: Length Conversion · Primary 3: Money and Change · Primary 5: Triangle Area · PSLE: Excess and Shortage · Exam Preparation: Gross and Net Mass. Each article develops a distinct Mathematics learning target.