Perimeter Calculator | 2D Shapes

Perimeter is the total length of the boundary of a shape, measured in linear units (m, cm, ft). Area is the amount of 2D surface enclosed within that boundary, measured in square units (m², cm², ft²). The two quantities are completely independent, a shape can have a large perimeter with a small area, or a small perimeter with a large area.

A classic illustration: a 10×10 m square and a 1×100 m rectangle both have an area of 100 m², but the square has a perimeter of 40 m while the rectangle has 202 m. This independence is the basis of the isoperimetric problem: which shape maximises area for a given perimeter? The answer is always the circle.

• ›Perimeter: measured in metres, centimetres, feet, one-dimensional length
• ›Area: measured in m², cm², ft², two-dimensional surface measure
• ›Same area → many possible perimeters (rectangle vs. square vs. triangle)
• ›Same perimeter → maximum area is achieved only by the circle

The isoperimetric inequality states that for any closed curve with perimeter P enclosing area A, the relationship A ≤ P²/(4π) always holds, with equality if and only if the curve is a perfect circle. This is one of the oldest results in mathematics, known informally since antiquity and proved rigorously in the 19th century.

The practical consequence is that the circle is the most area-efficient shape. For the same perimeter, no other shape encloses more area. Equivalently, for a fixed area, the circle requires the least boundary material. This efficiency principle appears throughout nature and engineering:

• ›Soap bubbles are spherical, surface tension minimises area for a fixed enclosed volume
• ›Circular cross-section pipes carry maximum flow for a given pipe circumference
• ›Bees use hexagonal cells, the best tessellating approximation to circles, to minimise wax
• ›Circular storage tanks and silos use less material per unit of stored volume than rectangular ones

Heron's formula computes the area of any triangle purely from its three side lengths a, b, c, without needing a height or any angle. First compute the semi-perimeter s = (a + b + c) / 2, then Area = √(s(s−a)(s−b)(s−c)). The formula is named after Hero of Alexandria (c. 60 AD) but may have been known to Archimedes earlier.

Use Heron's formula when you know all three sides but not the altitude, this is common in:

• ›Land surveying, three measured distances define a triangular plot without needing a perpendicular
• ›Navigation, GPS-based triangle calculations from known distances between waypoints
• ›Coordinate geometry, the distance formula gives side lengths; Heron gives area from those
• ›Construction, verifying triangular roof sections or land boundaries from measured edges

A rhombus has four equal sides, and its diagonals bisect each other at right angles. This means each side of the rhombus is the hypotenuse of a right triangle whose legs are half of each diagonal. If the diagonals are d₁ and d₂, then each side = √((d₁/2)² + (d₂/2)²), and perimeter = 4 × side. The diagonals also give the area directly: Area = ½ × d₁ × d₂.

• ›A square is a special rhombus with d₁ = d₂ (equal perpendicular diagonals)
• ›The diagonal method is useful when it is easier to measure across than along a slanted side
• ›For a given perimeter, the square (d₁ = d₂) maximises the area among all rhombuses

Geometric scaling follows a consistent pattern: perimeter scales linearly with the scale factor, area scales as the square, and volume scales as the cube. Doubling all dimensions doubles the perimeter, quadruples the area, and octuples the volume. This has profound consequences in biology and engineering, larger organisms have a smaller surface-area-to-volume ratio, which is why elephants need large ears to dissipate heat and small mice lose heat far faster relative to body size.

• ›Scale factor k: new perimeter = k × original perimeter
• ›Scale factor k: new area = k² × original area
• ›Scale factor k: new volume = k³ × original volume
• ›Doubling all dimensions (k = 2): perimeter ×2, area ×4, volume ×8
• ›Halving all dimensions (k = ½): perimeter ÷2, area ÷4, volume ÷8

In packaging and manufacturing, a container that is twice as large in every dimension uses 4× the material (surface area) but holds 8× the product, the economy-of-scale effect that makes large containers proportionally cheaper per unit of contents.

The apothem is the perpendicular distance from the centre of a regular polygon to the midpoint of any one of its sides. It equals the inradius, the radius of the largest inscribed circle that fits inside the polygon and touches all sides. For a regular n-gon with side length s, the apothem = s / (2 tan(π/n)). The apothem gives a clean formula for the polygon's area: Area = ½ × perimeter × apothem, because the polygon can be divided into n identical isosceles triangles each with base s and height equal to the apothem.

• ›Equilateral triangle (n = 3): apothem = s / (2√3) ≈ 0.2887s
• ›Square (n = 4): apothem = s/2
• ›Regular hexagon (n = 6): apothem = s√3/2 ≈ 0.8660s
• ›As n → ∞: the polygon approaches a circle, and apothem → circumradius

π (pi) is defined as the ratio of a circle's circumference to its diameter: π = C/d. This ratio is the same for every circle, regardless of size, a circle 1 mm across and one 100 km across both have C/d = π. Rearranging gives C = πd = 2πr. The ancient Egyptians approximated π ≈ 3.16; Archimedes (c. 250 BC) proved 223/71 < π < 22/7 by inscribing and circumscribing 96-sided polygons.

π is irrational (cannot be expressed as any fraction) and transcendental (not the root of any polynomial with integer coefficients). Its decimal expansion is non-repeating and non-terminating: 3.14159265358979…

• ›C = 2πr, circumference from radius
• ›C = πd, circumference from diameter
• ›Area = πr², the companion area formula, derived by integrating 2πr dr from 0 to r
• ›π ≈ 3.14159265, use full precision in engineering to avoid accumulated rounding errors
• ›22/7 ≈ 3.14286, a handy fraction approximation, error is only 0.04%