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Equal area, different boundary · Prove the square · Inside border paths · Wall along one side · Chapter index
A square and rectangle can enclose the same area while requiring unequal fence length and leaving different usable central areas after border treatment. The exact equations and their assumptions are worked through.
Chapter index
Additional worked checks: gates and scaling
Advanced worked companion: aspect ratio and fencing budgets
Chapters 1–4
1. Equal area is not equal boundary length · 2. Define a consistent rectangular model · 3. Calculate three explicit areas and perimeters · 4. Cost materials under one transparent rate
Chapters 5–8
5. Parameterise every rectangle with area 100 · 6. Prove the square minimises rectangle perimeter · 7. Check the optimum independently with calculus · 8. The square is not the global minimum among all shapes
Chapters 9–12
9. Measure the central area after an interior edge path · 10. Derive the inset area loss from perimeter · 11. Change the feasible dimensions, change the optimisation problem · 12. Work a three-sided fence along an existing wall
Chapters 13–15
13. Make shape compactness explicit · 14. Measurement precision can affect both area and perimeter estimates · 15. Questions and connected Geometry learning
Two fictional grounds can contain identical areas but demand different lengths of fencing. Ground S is a square 10 metres by 10 metres. Ground R is a rectangle 5 metres by 20 metres. Both contain exactly 100 square metres, yet their full outer boundary lengths are 40 and 50 metres.
That difference is material whenever a project has a continuous boundary treatment, such as fencing, garden edging, an inspection route or a perimeter drain. The area says how much surface is enclosed, whereas the perimeter says how long its outer edge is.
The two grounds and any associated costs in this guide are invented mathematical examples, not approved plans or price quotations for any real Singapore site. We will calculate equal-area rectangles, derive which rectangle minimises perimeter, evaluate inner border paths and examine what changes when a wall replaces one fence side.
The question to remember is not simply ‘which shape is best?’ It is ‘best for which constraint and measured quantity?’ A shape that is optimal for perimeter may not be admissible, accessible or operationally useful for a particular site.
A positive-length rectangle with sides x metres and y metres has area A=xy in m² and full perimeter P=2(x+y) in metres. The area formula multiplies two lengths; the perimeter formula sums the lengths of the four sides.
We assume straight sides, right angles and no gaps, gates or shared walls in our first comparison. Each result refers to the complete outside boundary. A real fence contract may count only certain segments and can include extra costs for posts and foundations.
The measurement boundary must be named before the formula is applied. A cadastral lot, a buildable footprint and the central playable area after setbacks need not have identical dimensions, even if they are drawn on one site plan.
The unit check helps: comparing 100 m² with 50 m as though they were two versions of the same quantity is a category mistake. Both numbers can be correct while answering separate physical questions.
Ground S is 10×10, giving A=100 m² and P=2(10+10)=40 m. Ground R is 5×20, also giving A=100 m² but P=2(5+20)=50 m. A third extremely long ground L, 2×50, again gives 100 m² but P=104 m.
R uses 10 m more perimeter than S. Relative to S, that is 10/40=25 percent more. The long narrow L uses 64 m more than S, or 160 percent more perimeter.
This is an exact comparison from stipulated dimensions. It does not rely on a drawing that might misrepresent scale. Each area should be recomputed from its own two dimensions before a boundary comparison is accepted.
These shapes also have different aspect ratios and layout possibilities. Equal area cannot determine whether two rectangles offer enough width for a particular activity; operational suitability needs further constraints beyond the area formula.
Suppose continuous boundary material costs twelve illustrative currency units per metre, independent of location and segment. Ground S would require 40×12=480 units, R 50×12=600 and L 104×12=1,248.
The difference in hypothetical cost follows exactly from the difference in complete perimeter. It is not a market estimate and excludes gates, corners, posts, labour, drainage and waste allowances.
If a project has a real gate opening of fixed length common to all plans, the purchased straight fence length could be full perimeter minus that shared opening. Both the definition of boundary and the pricing rule must be stated.
The lesson is not that square land is invariably cheap to develop. It is that one known surface area does not determine the quantity of a perimeter-based resource. A complete construction decision needs the actual edges and site constraints.
If one side is x>0 metres and the area is fixed at 100 m², the other side must be y=100/x. Substitute that constraint into full perimeter to obtain P(x)=2x+200/x.
At x=10 the perimeter is 40. At x=5 it is 50. At x=2 it is 104. The function describes a continuous family of rectangles all having exactly the same area.
Swapping x and 100/x only rotates a rectangle and leaves perimeter unchanged. The examples x=5 and x=20 are therefore the same geometric shape with its side labels exchanged.
As x approaches zero, its companion side tends to infinity and the perimeter grows without bound. Fixed area alone permits wildly elongated shapes in the ideal model, demonstrating why the perimeter cannot be read back from area without additional information.
For positive x and y, the arithmetic-geometric mean inequality gives (x+y)/2≥sqrt(xy). Multiply both sides by four to obtain P=2(x+y)≥4sqrt(A).
For area A=100 the minimum possible rectangular perimeter is at least 4sqrt100=40 m. Equality in this inequality occurs exactly when x=y, giving the 10×10 square.
This is a global proof over every positive rectangle with the stated area. It establishes a precise result rather than merely reporting that the square performed well among three handpicked examples.
The Mathematics LibreTexts discussion of arithmetic and geometric means develops this rectangular fencing inequality.
Differentiate P(x)=2x+200/x. The first derivative is P′(x)=2−200/x². It equals zero at x²=100, so the positive stationary side length is x=10.
The second derivative P″(x)=400/x³ is positive for every positive x, including x=10. The stationary point is a minimum, and the perimeter grows towards infinity at the extreme positive domain boundaries.
Therefore the global minimum is P(10)=40, again at the 10×10 square. Agreement between the derivative method and the AM–GM inequality is a useful independent check.
The Calculus optimisation textbook in Mathematics LibreTexts contains the fixed-area 100-square-unit example and its perimeter minimum.
The fixed-area result applies to rectangles, not every plane curve. The classical isoperimetric theorem states that a circle has the smallest perimeter among suitable simple plane regions enclosing a given area.
For a circle with A=100 m², radius r=sqrt(100/π)≈5.64190 m and perimeter 2πr=2sqrt(100π)≈35.4491 m. This is shorter than the 40 m boundary of the square.
A circle may nonetheless be unsuitable for an orthogonal building layout, equipment spacing or existing road geometry. Comparing mathematically efficient shapes is a starting point, not a complete practical recommendation.
Always state the class of admissible shapes before claiming optimality. The square is perimeter-optimal within rectangles; allowing circular boundaries changes the answer.
Suppose every ground has a continuous 0.5 m wide strip just inside all four edges, with no overlaps counted twice. The central rectangle then has length and width reduced by 1 m each.
Ground S leaves a central area of 9×9=81 m². Ground R leaves 4×19=76 m². Ground L leaves 1×49=49 m², although each original outer area was 100 m².
These are original geometric insets, not claims about any real required footpath width, accessibility standard or safety rule. They demonstrate how a boundary feature can consume usable central space differently in shapes of equal gross area.
The narrow ground loses a much larger share of its area because a long boundary surrounds relatively little width. Site function, entry points and clearance requirements still need assessment beyond this simple inset calculation.
| Metric | Square S | Rectangle R | Strip L |
|---|---|---|---|
| Dimensions | 10×10 m | 5×20 m | 2×50 m |
| Area | 100 m² | 100 m² | 100 m² |
| Perimeter | 40 m | 50 m | 104 m |
| Fence cost at 12 units/m | 480 | 600 | 1,248 |
| Central area after 0.5 m inside border | 81 m² | 76 m² | 49 m² |
For an x-by-y rectangle with inside strip width w, central area is A_c=(x−2w)(y−2w), valid when both dimensions exceed 2w.
Expanding gives A_c=xy−2w(x+y)+4w². Since A=xy and P=2(x+y), the strip area is A−A_c=wP−4w². The dependence on perimeter is now explicit.
With w=0.5, strip area equals 0.5P−1. Ground S loses 19 m², R loses 24 m², and L loses 51 m². Subtracting each from original area 100 reproduces the central areas 81, 76 and 49.
This identity is useful because it checks several independent numerical examples with one general expression. It also highlights the assumptions: equal strip width, straight rectangle edges and sufficient room for a central rectangle.
CHAPTER 11 OF 15
11. Change the feasible dimensions, change the optimisation problem
A theoretical 10×10 square may not fit a real narrow site. Suppose the project fixes one side at exactly 5 m while requiring 100 m². Then the other side must be 20 m, leaving no freedom to select the square.
The perimeter is necessarily 50 m under the full-boundary model. This does not disprove the square minimum; it demonstrates that a tighter feasibility constraint excludes the square from the admissible alternatives.
Similarly, an existing building wall may substitute for one new fence side. In that situation the quantity to minimise is the length of new fence, not necessarily the full mathematical perimeter.
A sound optimisation never hides the geometry of the admissible site. A lower theoretical material requirement is not a legitimate construction recommendation when its required shape cannot be built.
Assume a straight usable wall runs along one side of a hypothetical rectangular enclosure. Let x be the side parallel to the wall and y the perpendicular side. Only one x-side and two y-sides need new fencing, so new-fence length is F=x+2y.
Under fixed area xy=100, substitute y=100/x to get F(x)=x+200/x. Differentiate to obtain F′(x)=1−200/x². The minimum occurs at x=sqrt200≈14.1421 m and y=100/x≈7.0711 m.
The required new fence at that point is about 28.2843 m, less than the full-boundary square’s 40 m—but these values describe different objectives and building assumptions. It is incorrect to compare them as though the same boundary were treated in both cases.
Actual wall compatibility, foundations and permission would have to be checked in a real project. The exercise shows that changing the quantity to be minimised changes the optimal dimensions.
One dimensionless measure for rectangles is P²/(16A). It equals one for a square, because P=4sqrtA, and exceeds one for nonsquare rectangles by AM–GM.
For Ground S it equals 40²/(16×100)=1. For Ground R it equals 50²/1600=1.5625. For Ground L it equals 104²/1600=6.76. These values rank their boundary efficiency under the chosen reference.
A different scale-independent compactness index, such as 4πA/P², uses the circle as an optimum and produces different numbers. The chosen index and its benchmark must be defined.
Neither compactness statistic establishes traffic flow, safe supervision or usable equipment layouts. It describes geometrical shape only. A project could deliberately choose a less compact rectangle for a valid functional reason.
CHAPTER 14 OF 15
14. Measurement precision can affect both area and perimeter estimates
Real measurements are uncertain. If side lengths x and y have small independent standard uncertainties u_x and u_y, first-order uncertainty propagation gives u_A≈sqrt[(y u_x)²+(x u_y)²] for area A=xy.
For perimeter P=2(x+y), the analogous independent-input expression is u_P≈2sqrt(u_x²+u_y²). If measurements share a tape calibration error or the land shape is not truly rectangular, covariance and geometric modelling need additional treatment.
An exact 100 m² in our teaching case is not the same evidential claim as an apparently precise measured 100.00 m² on a real survey. More displayed decimal places cannot cure unsuitable measurement conditions.
Before ordering material, a real project requires appropriate site measurements, gates, posts, corner treatment, installation tolerances and relevant approvals. The mathematics helps structure the questions but does not replace field verification.
Does equal area mean equal perimeter? No. The 10×10 m square and 5×20 m rectangle each have area 100 m², but perimeters 40 and 50 m.
Is a square always the least-perimeter shape? Only among rectangles of fixed area in this result. A circle encloses a given area with less boundary than the square.
Would both shapes leave the same central activity area after a 0.5 m inside border? No. S leaves 81 m² and R leaves 76 m². The loss reflects their different perimeters.
Practice: compare a 12×12 m square and 6×24 m rectangle; then prove the smallest rectangular perimeter at fixed area 144 m². Next redo the wall-fencing calculation for area 144, carefully naming which side is parallel to the wall.
For the subject foundation use The Gold Standard of Geometry, the calculus optimisation example and the arithmetic–geometric mean proof. Our ground designs and costs are invented. Continue via the Gold Standard hub.
Advanced worked companion: aspect ratio, fencing budget and measurement checks
Rewrite the shape using a dimensionless aspect ratio
Suppose a rectangle has side lengths x and y, area A=xy and aspect ratio k=y/x≥1. Solving gives x=sqrt(A/k) and y=sqrt(Ak). Substituting into perimeter yields P=2sqrt(A)(sqrt(k)+1/sqrt(k)). This makes the amount of elongation a visible parameter rather than hiding it inside an unspecified pair of side lengths.
At A=100 m² and k=1, the perimeter is 2×10×(1+1)=40 m. At k=4, it becomes 2×10×(2+1/2)=50 m, corresponding to 5×20. At k=25, it becomes 2×10×(5+1/5)=104 m, corresponding to 2×50. The formula independently verifies our three constructed grounds.
The expression is symmetrical under replacing k by 1/k, representing a ninety-degree rotation rather than a fundamentally new shape. Because the arithmetic–geometric mean of sqrt(k) and 1/sqrt(k) is at least one, the expression is smallest at k=1. Elongation has a measurable boundary penalty for rectangles of fixed area.
If a site designer specifies an acceptable maximum aspect ratio, that constraint limits the family of possible perimeters. A statement such as ‘100 m² of available land’ remains incomplete until the admissible geometry is known. Both shape and area belong in a material estimate.
Reverse the question: maximise area for a fixed amount of fencing
Suppose there is exactly 50 m of full four-sided boundary material and we seek the largest rectangle it can enclose. Write P=2(x+y)=50, hence x+y=25. For a rectangle, area A=xy is maximal when x=y=12.5, giving A=12.5²=156.25 m².
The original 5×20 m rectangle already uses that same 50 m of perimeter, yet encloses only 100 m². Choosing the 12.5×12.5 square would increase the theoretical enclosed area by 56.25 m² under the ideal unconstrained shape comparison. It is not a claim that the redesigned square would fit a specific school site.
The fixed-area and fixed-perimeter optimisation problems are reciprocal forms of the same rectangle inequality. Under fixed area, the square minimises boundary length. Under fixed full perimeter, the square maximises enclosed rectangle area. A student should be able to move between the two questions without changing the underlying geometry.
A practical plan may instead minimise cost, satisfy a frontage rule or reserve a corridor. In such cases the variables being held fixed are different, and the square result may no longer be feasible or applicable.
Turn a fencing budget into an allowable range of dimensions
With fixed area 100 m², suppose the full-perimeter material budget permits at most 50 m of boundary. Our model requires 2(x+100/x)≤50. Divide by two and multiply by positive x: x²−25x+100≤0.
The quadratic factors as (x−5)(x−20)≤0, giving 5≤x≤20. Every x within that interval produces a rectangle of area 100 m² whose perimeter is no greater than 50 m. The endpoints correspond to the 5×20 shape under reversed orientations; the midpoint of the feasible shape family includes the more compact square.
This is a genuine feasibility result rather than an arbitrary shortlist. The area restriction alone allowed all positive x, including impractically narrow extreme shapes. Adding the perimeter budget removes the configurations that consume too much material.
If a design must also have each side at least six metres, its feasible interval contracts further: 6≤x≤100/6≈16.667. A mathematical model can combine multiple restrictions explicitly rather than claiming that one area number contains all these conditions.
When fence sides cost different amounts
The simple material comparison assumed the same price per metre around every edge. Suppose instead the pair of sides parallel to x costs ten hypothetical units per metre and the pair parallel to y costs fourteen units per metre. New-fence cost is C=20x+28y.
For a fixed 100 m², substitute y=100/x to obtain C(x)=20x+2800/x. Its derivative is C′=20−2800/x², so a positive optimum satisfies x²=140. Thus x≈11.8322 m and y≈8.4515 m, rather than the 10×10 square.
At this point the cost is approximately 473.29 hypothetical units. This example changes the pricing model and therefore changes the objective. It does not disprove the result that the square minimises unweighted full perimeter.
In real work, per-metre prices might differ because one edge needs stronger foundations or faces different weather exposure, and corners and gates can introduce fixed charges. Every such addition belongs in the actual cost model if it matters. Geometry reveals lengths; pricing converts them into a monetary decision.
Measurement uncertainty depends on shape as well as the reported area
Imagine each independently measured side has standard uncertainty 0.05 m. For the 10×10 square, the first-order propagated area standard uncertainty is sqrt[(10×0.05)²+(10×0.05)²]≈0.7071 m². For a 5×20 rectangle with the same side uncertainties, it is sqrt[(20×0.05)²+(5×0.05)²]≈1.0308 m².
The area values are both 100 m², but the uncertainty of that estimated area can differ because the derivative of xy with respect to each side depends on the other side’s length. This introduces another important qualifier: even identical central values do not ensure identical measurement evidence.
For perimeter P=2(x+y), those same independent side uncertainties give u_P≈2sqrt(0.05²+0.05²)≈0.1414 m in either rectangular case. This equality is a consequence of the chosen independent error model, not a universal real-survey property.
If measurements share a calibration error, the independence assumption is inappropriate and covariance must be considered. A real property boundary may not form a perfect rectangle. Qualified measurement and local compliance review are needed before ordering building material.
Practical design checks before making a material claim
A useful plan states the gross footprint, the exact boundary that will receive treatment, the side lengths, minimum clearance requirements and any site obstacles. It should separate a mathematically elegant closed rectangle from a real gate, wall connection or irregular boundary.
For boundary strips, include the position of the strip relative to the perimeter and ensure the remaining central dimensions are positive. A simple inset of width w removes wP−4w² of area only for the ideal straight-edged rectangular model; curved or highly irregular sites require their own geometry.
Distinguish the aims of minimising fence length, maximising usable interior area and accommodating a particular activity. An allocation that saves fence material may have a less suitable internal layout. The two goals need explicit weights, constraints or separate evaluation rather than being conflated.
These teaching calculations connect directly to algebra, measurement, units, ratios and optimisation. The Secondary 1 Mathematics learning guide provides a broader route through mathematical structure and disciplined working. The worked grounds and costs here are fictional, not actual tuition-centre facility plans.
Return to The Gold Standard of Geometry and the Gold Standard hub for related mathematics. Keep the area units, boundary definition and admissible shapes visible before calling a plan efficient.
Additional worked checks: gate allowances and scale changes
Suppose both rectangular grounds require exactly one metre of gate opening along their external edge and all other outer segments use the same fencing material. Then the purchased straight fence length would be P−1 rather than P. The square requires 39 m and the 5×20 m rectangle requires 49 m. The ten-metre difference remains even though both plans received the same deduction.
At the illustrative constant price of twelve units per metre, those lengths would imply 468 and 588 currency units of straight fence. This excludes the gate itself, posts, foundation, labour and material waste. A gate allowance does not erase the distinction between the two full perimeter values; it simply adds another explicit term to the costing model.
Now scale both dimensions of a rectangle by a factor of two. Area multiplies by 2²=4, while its perimeter multiplies only by two. The 10×10 m square becomes 20×20 m, with area 400 m² and perimeter 80 m. The original 5×20 becomes 10×40, with area 400 m² and perimeter 100 m.
For a family of geometrically similar shapes, perimeter is proportional to the square root of area, not directly to area. That scaling rule offers another quick check against unrealistic material estimates that multiply fence length by four merely because area increased fourfold.
For an inside border strip, changing w from 0.5 m to 0.25 m also changes the useful centre: the square retains 9.5×9.5=90.25 m² and the 5×20 rectangle retains 4.5×19.5=87.75 m². At w=0.5 m those values were 81 and 76 m². Both gross areas remain 100 m², but the net-area difference depends on the chosen border width.
These scale and allowance examples reinforce the main lesson: a meaningful material or usable-space estimate requires more than an area total. Identify the length, shape, boundary treatment, allowed dimensions and uncertainty before turning a geometrical result into a practical plan.