Shoelace formula

Basic idea: Any polygon edge determines the signed area of a trapezoid. All these areas sum up to the polygon area.

Given: A planar simple polygon with a positively oriented (counter clock wise) sequence of points ${\displaystyle P_{i}=(x_{i},y_{i}),i=1,\dots ,n}$ in a Cartesian coordinate system.
For the simplicity of the formulas below it is convenient to set ${\displaystyle P_{0}=P_{n},P_{n+1}=P_{1}}$.

The formulas:
The area of the given polygon can be expressed by a variety of formulas, which are connected by simple operations (see below):
If the polygon is negatively oriented, then the result ${\displaystyle A}$ of the formulas is negative. In any case ${\displaystyle |A|}$ is the sought area of the polygon.[8]

Trapezoid formula

The trapezoid formula sums up a sequence of oriented areas ${\displaystyle A_{i}={\tfrac {1}{2}}(y_{i}+y_{i+1})(x_{i}-x_{i+1})}$ of trapezoids with ${\displaystyle P_{i}P_{i+1}}$ as one of its four edges (see below):

$${\displaystyle {\begin{aligned}A&={\frac {1}{2}}\sum _{i=1}^{n}(y_{i}+y_{i+1})(x_{i}-x_{i+1})\\&={\frac {1}{2}}{\Big (}(y_{1}+y_{2})(x_{1}-x_{2})+\cdots +(y_{n}+y_{1})(x_{n}-x_{1}){\Big )}\end{aligned}}}$$

Triangle formula

The triangle formula sums up the oriented areas ${\displaystyle A_{i}}$ of triangles ${\displaystyle OP_{i}P_{i+1}}$:[9]

$${\displaystyle {\begin{aligned}A&={\frac {1}{2}}\sum _{i=1}^{n}(x_{i}y_{i+1}-x_{i+1}y_{i})={\frac {1}{2}}\sum _{i=1}^{n}{\begin{vmatrix}x_{i}&x_{i+1}\\y_{i}&y_{i+1}\end{vmatrix}}={\frac {1}{2}}\sum _{i=1}^{n}{\begin{vmatrix}x_{i}&y_{i}\\x_{i+1}&y_{i+1}\end{vmatrix}}\\&={\frac {1}{2}}{\Big (}x_{1}y_{2}-x_{2}y_{1}+x_{2}y_{3}-x_{3}y_{2}+\cdots +x_{n}y_{1}-x_{1}y_{n}{\Big )}\end{aligned}}}$$

Shoelace formula

Shoelace scheme, vertical form: With all the slashes drawn, the matrix loosely resembles a shoe with the laces done up, giving rise to the algorithm's name.

The triangle formula is the base of the popular shoelace formula, which is a scheme that optimizes the calculation of the sum of the 2×2-Determinants by hand:

$${\displaystyle {\begin{aligned}2A&=\underbrace {{\begin{vmatrix}x_{1}&x_{2}\\y_{1}&y_{2}\end{vmatrix}}+{\begin{vmatrix}x_{2}&x_{3}\\y_{2}&y_{3}\end{vmatrix}}+\cdots +{\begin{vmatrix}x_{n}&x_{1}\\y_{n}&y_{1}\end{vmatrix}}} \\&={\begin{vmatrix}x_{1}&x_{2}&x_{3}\cdots &x_{n}&x_{1}\\y_{1}&y_{2}&y_{3}\cdots &y_{n}&y_{1}\end{vmatrix}}&{\text{(horizontal shoelace scheme)}}\\&={\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\\\vdots &\vdots \\x_{n}&y_{n}\\x_{1}&y_{1}\end{vmatrix}}&{\text{(vertical form)}}\end{aligned}}}$$

Other formulas

$${\displaystyle {\begin{aligned}A&={\frac {1}{2}}\sum _{i=1}^{n}y_{i}(x_{i-1}-x_{i+1})\\&={\frac {1}{2}}{\Big (}y_{1}(x_{n}-x_{2})+y_{2}(x_{1}-x_{3})+\cdots +y_{n}(x_{n-1}-x_{1}){\Big )}\end{aligned}}}$$

$${\displaystyle A={\frac {1}{2}}\sum _{i=1}^{n}x_{i}(y_{i+1}-y_{i-1})}$$

A particularly concise statement of the formula can be given in terms of the exterior algebra. If ${\displaystyle v_{1},\dots ,v_{n}}$ are the consecutive vertices of the polygon (regarded as vectors in the Cartesian plane) then

$${\displaystyle A={\frac {1}{2}}\cdot \left|\sum _{i=1}^{n}v_{i}\wedge v_{i+1}\right|.}$$

Example

Horizontal shoelace form for the example.

For the area of the pentagon with

$${\displaystyle {\begin{aligned}&P_{1}=(1,6),P_{2}=(3,1),P_{3}=(7,2),\\&P_{4}=(4,4),P_{5}=(8,5)\end{aligned}}}$$

one gets

$${\displaystyle {\begin{aligned}2A&={\begin{vmatrix}1&3\\6&1\end{vmatrix}}+{\begin{vmatrix}3&7\\1&2\end{vmatrix}}+{\begin{vmatrix}7&4\\2&4\end{vmatrix}}+{\begin{vmatrix}4&8\\4&5\end{vmatrix}}+{\begin{vmatrix}8&1\\5&6\end{vmatrix}}\\&=1-18\;+6-7\;+28-8\;+20-32\;+48-5=33\\A&=16.5\end{aligned}}}$$

The advantage of the shoelace form: Only 6 columns have to be written for calculating the 5 determinants with 10 columns.

${\displaystyle A(P_{1},\dots ,P_{n})}$ indicates the oriented area of the simple polygon ${\displaystyle P_{1},\dots ,P_{n}}$ with ${\displaystyle n\geq 4}$ (see above). ${\displaystyle A}$ is positive/negative if the orientation of the polygon is positive/negative. From the triangle form of the area formula or the diagram below one observes for ${\displaystyle 1<k<n}$:

$${\displaystyle A(P_{1},\dots ,P_{n})=A(P_{1},\dots ,P_{k-1},P_{k+1},\dots ,P_{n})+A(P_{k-1},P_{k},P_{k+1})}$$

In case of ${\displaystyle k=1\;{\text{or}}\;n}$ one should first shift the indices.

Hence:

1. Moving ${\displaystyle P_{k}}$ affects only ${\displaystyle A(P_{k-1},P_{k},P_{k+1})}$ and leaves ${\displaystyle A(P_{1},...,P_{k-1},P_{k+1},...,P_{n})}$ unchanged. There is no change of the area if ${\displaystyle P_{k}}$ is moved parallel to ${\displaystyle {\overline {P_{k-1}P_{k+1}}}}$.
2. Purging ${\displaystyle P_{k}}$ changes the total area by ${\displaystyle A(P_{k-1},P_{k},P_{k+1})}$, which can be positive or negative.
3. Inserting point ${\displaystyle Q}$ between ${\displaystyle P_{k},P_{k+1}}$ changes the total area by ${\displaystyle A(P_{k},Q,P_{k+1})}$, which can be positive or negative.

Example:

${\displaystyle P_{1}=(3,1),P_{2}=(7,2),P_{3}=(4,4),}$

${\displaystyle P_{4}=(8,6),P_{5}=(1,7),\ Q=(4,3)}$

Manipulations of a polygon

With the above notation of the shoelace scheme one gets for the oriented area of the

• blue polygon:
  $${\displaystyle A(P_{1},P_{2},P_{3},P_{4},P_{5})={\tfrac {1}{2}}{\begin{vmatrix}3&7&4&8&1&3\\1&2&4&6&7&1\end{vmatrix}}=15.5}$$

  
• green triangle:
  $${\displaystyle A(P_{2},P_{3},P_{4})={\tfrac {1}{2}}{\begin{vmatrix}7&4&8&7\\2&4&6&2\end{vmatrix}}=-7}$$

  
• red triangle:
  $${\displaystyle A(P_{1},Q,P_{2})={\tfrac {1}{2}}{\begin{vmatrix}3&4&7&3\\1&3&2&1\end{vmatrix}}=-3.5}$$

  
• blue polygon minus point ${\displaystyle P_{3}}$:
  $${\displaystyle A(P_{1},P_{2},P_{4},P_{5})={\tfrac {1}{2}}{\begin{vmatrix}3&7&8&1&3\\1&2&6&7&1\end{vmatrix}}=27.5}$$

  
• blue polygon plus point ${\displaystyle Q}$ between ${\displaystyle P_{1},P_{2}}$:
  $${\displaystyle A(P_{1},Q,P_{2},P_{3},P_{4},P_{5})={\tfrac {1}{2}}{\begin{vmatrix}3&4&7&4&8&1&3\\1&3&2&4&6&7&1\end{vmatrix}}=17}$$

  

One checks, that the following equations hold:

$${\displaystyle A(P_{1},P_{2},P_{3},P_{4},P_{5})=A(P_{1},P_{2},P_{4},P_{5})+A(P_{2},P_{3},P_{4})=20.5}$$

$${\displaystyle A(P_{1},P_{2},P_{3},P_{4},P_{5})+A(P_{1},Q,P_{2})=A(P_{1},Q,P_{2},P_{3},P_{4},P_{5})=17}$$

In higher dimensions the area of a polygon can be calculated from its vertices using the exterior algebra form of the Shoelace formula (e.g. in 3d, the sum of successive cross products):

$${\displaystyle A={\frac {1}{2}}\left\|\sum _{i=1}^{n}v_{i}\wedge v_{i+1}\right\|}$$

(when the vertices are not coplanar this computes the vector area enclosed by the loop, i.e. the projected area or "shadow" in the plane in which it is greatest). This formulation can also be generalized to calculate the volume of an n-dimensional polytope from the coordinates of its vertices, or more accurately, from its hypersurface mesh.[10] For example, the volume of a 3-dimensional polyhedron can be found by triangulating its surface mesh and summing the signed volumes of the tetrahedra formed by each surface triangle and the origin:

$${\displaystyle V={\frac {1}{6}}\left\|\sum _{F}v_{a}\wedge v_{b}\wedge v_{c}\right\|}$$

where the sum is over the faces and care has to be taken to order the vertices consistently (all clockwise or anticlockwise viewed from outside the polyhedron). Alternatively, an expression in terms of the face areas and surface normals may be derived using the divergence theorem (see Polyhedron § Volume).

• Planimeter
• Polygon area
• Pick's theorem
• Heron's formula

• Mathologer video about Gauss' shoelace formula