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Removing a sector from a circle removes an arc from the original circumference, but it can expose two straight radii on the boundary of the remaining region. If a 60° sector is cut out of a circle of radius 6 cm, the remaining boundary contains the 300° major arc and two 6 cm edges. Its perimeter is 10π + 12 cm, not just 10π cm.
For G3 Mathematics, trace the boundary of the region that remains before choosing a formula. Identify every curved and straight piece that a complete trip around that region would follow. Area subtraction and perimeter subtraction behave differently: removing part of a shape can create new exposed edges, so subtracting the removed arc from the circumference does not necessarily give the whole new perimeter.
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Sections 1–5
Sections 6–10
Sections 11–15
Sections 16–20
Sections 21–23
Imagine an ordinary wedge-shaped 60° sector removed from a paper circle of radius 6 cm. The curved part left outside the wedge measures 300/360 × 2π × 6 = 10π cm. To travel completely around the remaining region, you must also follow the two cut edges from the arc endpoints to the centre and back. They contribute 12 cm.
The remaining area is 300/360 × π × 6² = 30π cm². That calculation uses the fraction of area left. The perimeter calculation uses the actual exposed boundary, including straight edges created by the removal.
These statements describe one specified geometry: a sector extending from the centre to the circumference is removed and the remaining major-sector region is the target. A circular segment, a hole away from the edge or an internal construction line gives a different boundary. Read the shape rather than reacting only to the word removed.
This guide concentrates on that boundary decision. Use the broader composite-area and perimeter guide for the foundational distinction between covering a region and following its edge.
The SEAB G3 Mathematics syllabus for 2027 includes arc length, sector area, circular segments and radian measure. Composite questions require the learner to decide which of those quantities belongs to the requested region before calculating it.
This article uses exact expressions in π where convenient. Numerical approximations should follow the accuracy and value-of-π instructions supplied in the actual question. The diagrams described here are mathematical teaching configurations, not real cutting instructions.
The difficult step is often not the arc formula. It is deciding whether the answer needs the minor arc, major arc, two radii, a chord, an inner curve or some combination. A correctly evaluated formula can answer the wrong geometric question if that selection is wrong.
The guide to matching diagram marks supports the first reading step. Use stated centres, radii, angles and boundaries. Do not assume that an internal line is part of the outside edge simply because it is drawn clearly.
For the wider subject labels, see what G1, G2 and G3 Mathematics mean. Choose the degree or radian sections according to what has been taught, while retaining the same boundary-tracing habit throughout.
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Area measures the size of a region in square units. Perimeter measures the total length of its boundary in length units. A smaller area does not automatically imply a shorter perimeter because the boundary can become more complicated as material is removed.
For a simple rectangle, cutting a rectangular notch into one side removes area while exposing the notch’s sides. The new perimeter depends on those exposed edges. The circle-sector example follows the same principle with a curved outer boundary.
Do not treat perimeter as an amount that must decrease whenever the shape loses material. Ask what happens to the edge. Some old edge disappears, and some new edge may appear.
The unit provides a useful check but not a complete solution. An answer in centimetres could still omit two straight edges. An area formula produces square units and is wrong for perimeter, but using a length formula does not guarantee that every required length has been included.
Before computing, name the target: remaining region’s perimeter, removed sector’s perimeter, remaining curved edge, or total cut length. These targets can use some of the same pieces but combine them differently. The wording is part of the mathematics.
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A circle of radius 6 cm has circumference 2π × 6 = 12π cm. A 60° sector occupies one sixth of the full angle, so its outer arc has length 60/360 × 12π = 2π cm.
Subtracting that arc from the circumference leaves 10π cm of original curved edge. This is a correct calculation of the remaining arc. It is not yet a calculation of the entire remaining perimeter.
The distinction is easiest to preserve by labelling the result “remaining arc = 10π cm.” If you label it perimeter immediately, you may stop before noticing the newly exposed straight edges.
The same remaining arc can be calculated directly from its angle: 360° − 60° = 300°, so length = 300/360 × 12π = 10π cm. The direct and subtraction routes agree because they identify the same curved portion.
Use one route as the main calculation and the other as a check when useful. Their agreement verifies the arc selection and arithmetic, but both can still omit the radii if you mistake the arc for the full boundary. A second calculation of the same partial quantity does not check whether the target was selected correctly.
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The removed sector is bounded by its outer arc and two radii from the centre to the arc endpoints. When the wedge is removed from the circle, those radii separate the remaining material from the empty wedge. They are now exposed edges of the target region.
A complete traversal of the remaining region follows the major arc from one endpoint to the other, then follows one radius to the centre and the other radius back to the starting endpoint. The two radii together contribute 2r.
For r = 6 cm, this is 12 cm. Adding the remaining arc gives perimeter 10π + 12 cm. The centre is a corner on this boundary even though it was not a point on the original circle’s circumference.
This explains why a boundary can acquire pieces that were internal before the change. The original whole circle had no straight perimeter edges. The cut changes the region whose boundary is being measured.
Do not add the radii simply because every sector formula you remember contains them somewhere. Add them because they are on the boundary of the particular remaining region. That reason will continue to work when the target changes to a different shape and the required straight edges are no longer two full radii.
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Let a circle have radius r and let an ordinary sector of angle α, with 0° < α < 360°, be removed from centre to circumference. The remaining connected major- or minor-sector region has angle 360° − α.
Its remaining arc length is [(360 − α)/360] × 2πr. Its two radial edges contribute 2r. Therefore its perimeter is [(360 − α)/360] × 2πr + 2r for this specified configuration.
The angle condition matters. At a zero-width cut or a complete removal, the ordinary region and boundary model changes. Do not substitute endpoint angles into a formula without inspecting what shape remains.
The formula is a compact record of the tracing argument. It should not replace the tracing argument when the question has a different cut, an internal hole or a composite region with shared edges.
A useful written solution lists the pieces first: “remaining arc plus two radii.” Then substitute the numerical angle and radius. This makes the final expression reviewable. A reader can see which physical edges each term represents rather than guessing why an extra 2r appeared after the arc calculation.
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For radius 6 cm and removed angle 60°, the removed sector’s perimeter is 2π + 12 cm: its 60° arc plus two radii. The remaining region’s perimeter is 10π + 12 cm: its 300° arc plus two radii.
Adding these two perimeters gives 12π + 24 cm. This exceeds the original circumference because the two cut edges are counted once on each separate piece. When pieces are joined into the original circle, those shared edges are internal and do not contribute to the outside perimeter.
That is why perimeter does not follow the simple subtraction rule “whole perimeter minus removed perimeter.” Using 12π − (2π + 12) would give 10π − 12, which wrongly subtracts the cut edges instead of including them on the remaining piece.
Area behaves differently for non-overlapping pieces that exactly partition the circle. The areas of the removed sector and remaining region add to the original circle’s area. Shared boundary lines have no area in the ideal plane model.
This comparison provides a strong explanation for the common mistake. The learner is not merely forgetting a formula; they may be transferring an area decomposition rule to boundary length. Repair the distinction between region and edge before giving more numerical arc exercises.
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Use a circle of radius 6 cm and remove a narrow 30° sector. The remaining arc is 330/360 × 12π = 11π cm. Adding two radii gives 11π + 12 cm.
The original circumference is 12π cm. The difference is 12 − π cm, which is positive. The remaining region has less area but a longer perimeter than the original circle.
The reason is geometric: a relatively short arc of length π cm has been removed, while two straight edges of total length 12 cm have been exposed. The added boundary exceeds the lost boundary.
This is a useful reasonableness check. A learner may reject the correct result because “removing something must make the perimeter smaller.” The example shows why that intuition does not apply to every shape change.
Do not reverse the intuition into “removing a sector always increases perimeter.” A larger removed arc can exceed the two added radii, so the new perimeter can also be shorter. Compare the lost arc length with the total new straight-edge length. The sign of that difference determines the result for this model.
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If a 180° sector is removed, the remaining region is a semicircle. Its curved edge is πr and its straight boundary is a diameter of length 2r. The perimeter is πr + 2r.
The two radii in the sector description lie along one straight line in this special case. Their total is still 2r, which is the diameter. The perimeter agrees with the familiar semicircle result.
For r = 5 cm, the perimeter is 5π + 10 cm. The area is half the circle’s area, 12.5π cm². The numerical forms and units remind you that the two targets use different relationships.
If a question asks only for the semicircular arc, report 5π cm. If it asks for the whole perimeter, include the 10 cm straight edge. A familiar shape does not remove the need to read the requested target.
The semicircle and quarter-circle guide provides foundational practice with this distinction. The general sector-removal problem extends the same boundary reasoning to other angles and to less familiar remaining shapes.
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A circular segment is the region between a chord and its corresponding arc. It is different from a sector, which is bounded by two radii and an arc. Removing a segment from a circle can leave a boundary containing a chord, not two full radii.
For a chord joining points A and B on a circle, inspect whether the removed region reaches the centre. A wedge from the centre to the circumference is a sector. A cap cut off by straight chord AB is a segment.
The perimeter of the remaining region after removing a cap consists of the remaining major arc plus chord AB. Adding 2r would generally be wrong because the boundary travels directly along the chord rather than through the centre.
You may need a taught method such as trigonometry to calculate the chord length, depending on the data supplied. Select that method only after the boundary has identified which straight length is needed.
The article’s title uses can deliberately. Removing a sector can leave the two radii on the boundary in the specified wedge configuration. It does not claim that every circular cut produces the same edge pattern. The name and geometry of the removed region determine which formula applies.
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Many circle diagrams show radii to mark angles or connect useful triangles. A drawn radius inside a shaded region is not part of its perimeter merely because it is visible. It may be a construction line that helps with calculation.
Before adding a segment, ask whether the target region lies on one side of it with empty space or another region outside the target on the other. If the same target region continues across the segment, it is internal to that region rather than an exposed boundary.
For the cut-out sector, the two radial sides separate remaining material from the removed wedge, so they belong. For an intact circle with two radii drawn inside, the region continues across them, so the circumference alone is the perimeter.
The diagram-marking guide distinguishes givens from appearance. A radius label can establish a length without establishing that the whole segment is part of the requested boundary.
This is a particularly useful reading check for shaded diagrams. Colour identifies the target region, while lines may serve several roles. Trace the edge of the colour or the explicitly named region rather than adding every labelled segment you can find.
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An annular sector is a sector-shaped part of a ring between concentric circles. Suppose its outer radius is R, inner radius r and central angle is θ. Its boundary contains an outer arc, an inner arc and two straight radial pieces of length R − r each.
If θ is in degrees, the curved total is θ/360 × 2π(R + r). The straight total is 2(R − r). The two straight pieces are not two full outer radii because the central portion inside the inner circle is absent.
For R = 8 cm, r = 5 cm and θ = 90°, the outer arc is 4π cm and the inner arc is 2.5π cm. The straight pieces total 6 cm, giving perimeter 6.5π + 6 cm.
This extension makes the boundary method more useful than a memorised extra 2r. The diagram determines the straight lengths and the number of arcs. You need not use a new rule blindly; list the pieces and calculate each.
The area uses the difference of two sector areas: θ/360 × π(R² − r²). That is a separate target. The presence of both radii in the area formula does not tell you to subtract one arc from the other for perimeter, because both arcs are exposed boundaries of the ring-shaped region.
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If an angle θ is measured in radians, an arc of radius r has length rθ. For a full circle the angle is 2π radians. Removing a sector of angle θ leaves an arc of length r(2π − θ).
For the ordinary cut-out sector shape, add the two radii to obtain remaining perimeter r(2π − θ) + 2r. This is the radian version of the degree formula, with the same boundary interpretation.
For r = 6 cm and removed angle π/3 radians, the remaining arc is 6(2π − π/3) = 10π cm. Adding 12 cm gives 10π + 12 cm, matching the 60° example because π/3 radians equals 60°.
Do not enter 60 directly into rθ while intending degrees. That would treat 60 as a radian measure and produce a different arc. Check the angle unit before choosing the numerical expression.
The coefficient of π can help with an exact check, but it does not replace reading the angle unit. If the question supplies degrees, use the degree fraction or convert correctly. If it supplies radians, use the radian relationship with the full angle 2π. The boundary pieces remain the same under either notation.
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For the radius-6, removed-60° example, the exact perimeter is 10π + 12 cm. If the question requests a decimal approximation, evaluate this expression using the instructed value or calculator convention for π and round the final answer accordingly.
Rounding the arc early can change the final value, especially if the result is later multiplied by a cost per unit length or combined with other boundaries. Retain the expression or calculator value through the later steps.
The reverse-percentage rounding guide illustrates the same principle in another topic: a shortened intermediate value can become a different final answer. Here an exact π expression is often a convenient way to preserve the quantity.
Also keep units consistent. If the radius is in centimetres and a later quantity is per metre, convert the boundary length before using that rate. An exact expression in the wrong unit is still an unsuitable input to the next calculation.
Write a final sentence that identifies perimeter rather than leaving an unexplained decimal. A value can look plausible while representing remaining arc length, total cut length or another nearby target. The label makes the calculation’s purpose visible.
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A circle of radius 9 cm has a 120° sector removed. Find the remaining area and perimeter in terms of π. The remaining angle is 240°, or two thirds of a full turn.
The remaining area is 240/360 × π × 9² = 54π cm². The remaining arc is 240/360 × 2π × 9 = 12π cm. The two exposed radii contribute 18 cm, so the remaining perimeter is 12π + 18 cm.
The removed sector has area 27π cm², and 27π + 54π = 81π, the whole circle’s area. This is a useful area partition check.
For perimeter, the removed sector has arc 6π cm and two radii totalling 18 cm. Subtracting its entire perimeter from the original 18π cm would give 12π − 18, which is not the boundary of the remaining region. The cut edges belong to both separate pieces.
The correct perimeter check is a piece inventory: one 240° arc plus two radius segments. That inventory explains every term. It also shows why an answer of 12π cm is incomplete even though its arc calculation is correct.
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Question A: A 90° sector is removed from a circle of radius 4 cm, leaving the other sector-shaped region. Find the remaining arc and whole perimeter separately in terms of π.
Question B: A 180° sector is removed from a circle of radius 7 cm. Find the perimeter of the remaining region and explain the straight part of its boundary.
Question C: A 60° sector is removed from a circle of radius 10 cm. Find the remaining area and perimeter in terms of π. Keep the length and area units distinct.
Question D: A learner subtracts the removed sector’s whole perimeter from the circle’s circumference. Explain why this treats the cut edges incorrectly.
Question E: An annular sector has outer radius 10 cm, inner radius 6 cm and angle 90°. Find its perimeter by listing both arcs and both straight pieces.
Question F: A circle of radius 5 cm loses a sector of angle π/2 radians. Find the remaining perimeter. Explain which full-turn angle belongs in the radian calculation.
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For A, the remaining angle is 270°. The remaining arc is 270/360 × 8π = 6π cm. The whole perimeter is 6π + 8 cm because the two radii each measure 4 cm.
For B, the remaining region is a semicircle. The perimeter is 7π + 14 cm. The two radial edges lie on a straight diameter and together measure 14 cm.
For C, the remaining angle is 300°. Area is 300/360 × 100π = 250π/3 cm². The remaining arc is 300/360 × 20π = 50π/3 cm, so perimeter is 50π/3 + 20 cm.
For D, the removed sector’s straight sides are new exposed edges on the remaining piece as well. They should be added to the remaining arc, not subtracted from the original circumference.
For E, the outer arc is 5π cm and the inner arc is 3π cm. The two straight pieces are 4 cm each. Perimeter is 8π + 8 cm. For F, the remaining angle is 2π − π/2 = 3π/2 radians. The arc is 15π/2 cm and the whole perimeter is 15π/2 + 10 cm.
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A student gives only the major arc as the perimeter. The repair is to trace a complete closed route around the remaining region and name the missing straight pieces. More practice substituting into the arc formula will not address the omitted edges.
Another student adds every radius drawn in the figure. The repair is to distinguish construction lines from exposed boundary. Ask whether the target region continues across each line.
A third uses a chord where the cut is actually a sector from the centre. The repair is to identify the removed shape. A fourth adds two full outer radii to an annular sector. The repair is to measure only the exposed radial gaps.
The marked-paper guide helps convert such errors into specific next steps. Perimeter problems can fail through target selection, geometry reading, formula choice, unit conversion or arithmetic; those are different repairs.
Ask the learner to point to the piece represented by each term in the expression. If a term has no boundary piece, or a boundary piece has no term, the model needs attention before any numerical calculation is repeated.
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For an ordinary sector cut of radian angle θ from a circle of radius r, the removed arc has length rθ. The cut exposes two radii of total length 2r. The change from the original circumference to the remaining perimeter is therefore 2r − rθ = r(2 − θ).
Because r is positive, the sign depends on whether θ is smaller than, equal to or larger than 2 radians. A cut angle below 2 radians produces a longer remaining perimeter. At exactly 2 radians, the lost arc and new straight edges have equal total length. Above 2 radians, the remaining perimeter is shorter than the original circumference.
This comparison concerns the specified wedge removal and its exposed boundary. It is not a general rule for every way of removing material from a circle. A segment cut or an off-centre hole has a different boundary inventory.
For example, with radius 5 cm and removed angle 1 radian, the arc lost is 5 cm while the straight edges added total 10 cm. The remaining perimeter is 5 cm longer than the original circumference. With the same radius and a removed angle of 3 radians, the arc lost is 15 cm and the straight edges added still total 10 cm, so the perimeter is 5 cm shorter.
The equality case is useful because it shows that area can decrease while perimeter stays unchanged. At θ = 2 radians, a positive sector area is removed, but the boundary lost and boundary added have equal lengths. Equal perimeter does not mean equal area or identical shape.
This extension turns a vague size check into a precise comparison. Instead of deciding whether an answer feels too large, compare the exact pieces that changed. The same habit applies in many composite figures: list what disappears, list what becomes exposed, and compare those lengths under the actual geometry.
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Suppose a circular hole of radius 2 cm is removed from the interior of a circle of radius 7 cm, with the hole lying entirely inside and not touching the outer edge. If the task asks for the total boundary length of the remaining region, both the outer circumference and the hole’s circumference contribute.
The total boundary length is 2π × 7 + 2π × 2 = 18π cm. The remaining area is π × 7² − π × 2² = 45π cm². One target adds the exposed boundary lengths, while the other subtracts the removed area.
If the question asks only for the outer boundary, the answer is 14π cm. Read whether the task includes the edge of the hole. Terms such as total boundary length, outer perimeter and length of edging should be interpreted from the stated diagram and wording rather than assumed interchangeable.
This case differs from the wedge removal. The original outside circumference is still complete, and the new edge forms a separate closed inner curve. There are no radial cut edges linking the hole to the outer circumference.
If the inner circle touches the outer edge or a channel connects the hole to the outside, inspect the new geometry again. The simple two-disjoint-circumferences inventory may no longer describe the intended region. A formula should follow the boundary structure rather than survive every change to the picture by habit.
The comparison with a sector is valuable practice because the word removed appears in both problems while the boundary changes differently. The learner must identify the actual cut, not select an operation from that single word.
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An invented paper model asks how much straight cutting is needed to remove a sector from a circle, with the two cuts running from the circumference to the centre. If the radius is 8 cm, the two straight cuts have total length 16 cm.
That is not the perimeter of the remaining region. It excludes the remaining circular arc, because the arc was already an outer edge rather than a newly made straight cut. The target is the operation’s cut length, not a full traversal of the resulting shape.
For a removed angle of 90°, the remaining arc is three quarters of the original circumference: 3/4 × 16π = 12π cm. The remaining perimeter is therefore 12π + 16 cm. The cut length and perimeter share the two radial pieces but differ by the entire remaining arc.
Now suppose the question asks for the perimeter of the removed sector instead. Its arc is 4π cm, so its perimeter is 4π + 16 cm. Three legitimate answers, 16, 12π + 16 and 4π + 16, arise from three different targets in the same geometry.
This is why the final sentence should name the quantity. A number with the right unit can still answer the wrong question. Length units alone cannot distinguish cutting distance, arc length and perimeter.
A useful review exercise is to supply the three expressions without labels and ask the learner to attach each to its target. Then change the removed angle while keeping the radius fixed. The straight cut length stays 16 cm, while both arc contributions change. The contrast reveals which part of each expression depends on angle and which depends only on radius.
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A final boundary check concerns the length label. Suppose the original circle has diameter 12 cm and a 60° sector is removed. The radius is 6 cm, so the remaining perimeter is 10π + 12 cm, as in the opening example.
Using 12 as the radius would produce 20π + 24 cm, exactly twice the correct perimeter. The arc fraction could be correct and the two-edge inventory could be correct while the selected radius is still wrong. Different checks protect different decisions.
Trace the labelled segment’s endpoints. A radius runs from the centre to the circumference. A diameter runs from one circumference point to another through the centre. A chord need not pass through the centre and cannot automatically be halved to obtain a radius.
If the centre is labelled O, a segment OA to a circumference point A may supply the radius directly. A segment AB across the circle needs the appropriate geometric information before it is treated as a diameter. A long horizontal line in a sketch is not enough by itself.
Record the radius once before computing arcs or areas. This avoids using one interpretation in the arc calculation and another in the added straight edges. It also makes a factor-of-two disagreement easier to diagnose: inspect the radius-versus-diameter reading before recalculating every formula.
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Does removing material always reduce perimeter?
No. It removes some old boundary and may expose new boundary. The resulting perimeter can increase or decrease depending on the lengths involved. Area reduction alone does not determine the change in perimeter.
Are the two radii always included in a sector question?
Include them when they are exposed edges of the requested region. Do not include them when the question asks only for arc length or when they are internal construction lines. Read the target and trace its boundary.
What if the removed piece is a circular segment?
A segment is bounded by a chord and an arc. The remaining boundary must be analysed using that chord, not automatically two radii. Identify the geometry before choosing the straight-edge calculation.
Can I subtract areas even when I cannot subtract perimeters?
For non-overlapping regions that exactly partition the original shape, area subtraction is valid. Perimeter requires a separate boundary inventory because shared edges become internal when pieces are joined and exposed when they are separated.
Where should I go next?
Use the Mathematics explanation directory for another focused problem. The Clementi Secondary 1 Mathematics small-group guide gives the local teaching context. For this topic, the reliable first move is to follow the boundary of what remains, list each exposed piece, and only then turn that list into a formula.