# Irregular Polygons

Polygons are two dimensional geometric objects composed of points and straight lines connected together to close and form a single shape. **Irregular polygon** are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles.

As the concept of irregular polygons is extremely general, knowledge about this concept can be very valuable, both for problem solving and for simple problems in the real world. For example, this rug can be described as a single irregular polygon.

Using the information in this wiki, the area of this rug could be easily calculated with a few measurements.

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## Simple and Self-Intersecting Polygon

Simple polygon is a polygon such that no side crosses itself.

Here are some examples of simple polygons.

As opposed to simple polygon, self-intersecting polygon is a polygon that has at least one side crosses itself.

Here are some examples of self-intersecting polygons.

## Convex and Concave Polygon

A convex polygon is a simple polygon that has all its interior angle less than \(180^\circ\)

As opposed to convex polygon, a concave polygon is a simple polygon that has at least one interior angle greater than \(180^\circ\).

Classify these polygons as convex, concave, or neither.

We begin with polygon A. All of it's angles are less than \(180^\circ\), so it is a

convexpolygon. Polygon B has two sides which intersect, so it is a self intersecting polygon, not a simple polygon. Therefore, it cannot be convex or concave, so it isneither. Polygon C has one angle which is greater than \(180^\circ\), so it isconcave.

## Cyclic and Tangential Polygon

A cyclic polygon is a polygon with vertices upon which a circle can be circumscribed. Every triangle is cyclic polygon.

A tangential polygon is a simple polygon formed by the lines tangent to the circumcircle of a polygon. Similarly, every triangle is a tangential polygon.

## Angle Sum Property

Main Article: General Polygons - Angles

For a simple \(n\)-gon, the sum of all **interior angles** is

\[S = (n-2)\times 180^\circ \quad \text{or} \quad S=(n-2)\times \pi \text{ rad}\]

One of the proof can be found in Polygon Triangulation section.

Sameer was given homework in geometry. He was stuck with a question. The question said that the sum of the interior angles of a polygon did not exceed 2014 degrees. It asked him to find the Max. number of sides the polygon would contain. Enter your answer as the maximum number of sides it could contain.

For any simple polygon, the sum of all **exterior angles** is \(360^\circ\). The proof of this is given below.

## Perimeter of Irregular Polygon

Main Article: Perimeter

The perimeter of a two-dimensional figure is the length of the boundary of the figure. If the figure is a polygon, then the perimeter is the sum of all side lengths of the polygon.

## What is the perimeter of a rectangle with width \(4\) and height \(7\)?

Since a rectangle has two sides with equal width and two sides with equal height, and the perimeter of a rectangle is the sum of its side lengths, the perimeter of the rectangle is

\[ 4 + 4 + 7 + 7 = 2(4) + 2(7) = 8 + 14 = 22 . \ _\square \]

## Area - Grid

A grid is useful to measure the area of an irregular polygon. The technique is to split the polygon into several basic shapes such as triangles and rectangles.

Below is a grid formed by \(1\times 1\) squares. What is the area of the polygon?

We split the polygon into several pieces.

Independently compute the area of each pieces, we have

\[\begin{aligned} A &= \frac{1}{2}\times 3\times 5 \\ &= 7\frac{1}{2} \\ B &= \frac{1}{2}\times 2\times 3 \\ &= 3 \\ C &= 3\times 3 \\ &= 9 \\ D &= 2\times 3 \\ &= 6 \end{aligned}\]

Finally, we add up all the small areas to get the overall area,

\[A=7\frac{1}{2} +3+9+6=25\frac{1}{2}\]

## Area - Pick's Theorem

Main article: Pick's Theorem

Pick's Theorem provides a method to determine the area of any polygon whose vertices are points on a lattice.

Let \(P\) be a lattice polygon, \(B(P)\) and \(I(P)\) be the number of points on the boundary and in the interior of the polygon respectively. Then the area \[A= I(P) + \frac{1}{2} B(P) -1.\]

This is illustrated in the following example:

Consider a quadrilateral with verices \((1,1) ; (1,-1) ; (-1,-1); (-1,1)\). Find it's area using Pick's Theorem.

Let's calculate \(B(P)\) first.

On drawing the figure and counting the number of points on the boundary (with integer coordinates), we see that \(B(P)=8\).

Now, we see that the only interior point with integer coordinates is \((0,0)\). Hence \(I(P)=1\).

Therefore the area (\(A\)) is given by: \[ A=(1) + \frac{1}{2} (8) - 1 = 4 \]

## Area - Coordinate Geometry

For an \(n\)-sided polygon, if we sort the points in a counterclockwise manner and let the coordinate of point \(i\) be \((x_i,y_i)\) and the area of polygon be \(A\), we have

\[A = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ \vdots & \vdots & \vdots \\ x_n & y_n & 1 \end{vmatrix}\]

In other words,

\[A = \frac{1}{2} \sum_{i=1}^n \bigg ( x_iy_{i+1} - x_{i+1}y \bigg )\]

Note that this is a generalization of the shoelace formula.

What is the area of the polygon with vertices at point \((1,1),(2,3),(0,5),(-1,0),(0,-3)\)?

By the shoelace formula,

\[A = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 1 \\ 0 & 5 & 1 \\ -1 & 0 & 1 \\ 0 & -3 & 1 \end{vmatrix}\]

## Centroid

Main Article: Centroid

A centroid is the center of mass of a two dimensional object with evenly distributed mass. The centroid of a simple polygon defined by \(n\) vertices \((x_0,y_0),(x_1,y_1),\dots,(x_n,y_n)\) is the point \((C_x,C_y)\) where

\[\begin{aligned} C_x &= \frac{1}{6A} \sum_{i=0}^{n-1} (x_i+x_{i+1})(x_iy_{i+1}-x_{i+1}y_i) \\ C_y &= \frac{1}{6A} \sum_{i=0}^{n-1} (y_i+y_{i+1})(x_iy_{i+1}-x_{i+1}y_i) \end{aligned}\]

Note that because the polygon has \(n\) vertices, \(x_0 = x_n\) and \(y_0=y_n\).

## Polygon Triangulation

Main Article: Polygon Triangulation

Polygon Triangulation is the process of decomposing a polygon into triangles. Every simple \(n\)-gon can be decomposed to \(n-2\) triangles. This immediately follows that the sum of interior angles in a \(n\)-gon is \((n-2)\times 180^\circ\).

## Art Gallery Problem

Main article: Art Gallery Problem

The art gallery problem studies the minimum number of guards needed to guard every point in the museum, which layout can be seen as a polygon. It turns out that any museum (polygon) with \(n\) walls (sides) can be guarded by *at most* \(\lfloor \frac{n}{3} \rfloor\) guards.

Details: A point \(P\) in the museum is visible to a guard if the line segment from the guard to \(P\) lies within the triangle (or along the boundary).

See Guarding a Museum for more details on guarding a museum.

## Problem Solving

In this section we will go through some examples that is related irregular polygon followed by problem to try it yourself.

The interior angles of a polygon are in arithmetic progression. The smallest angle is \(120^\circ\) and the common difference is \(5^\circ\). Find the number of sides of the polygon.

A hexagon is inscribed in a circle of radius \(r\). Two of its sides have length 1, two of them have length 2, two of them have length 3. If \(r\) is a root of \(ax^{3}+bx^{2}+cx+d=0\), where \(|a|, |c|\) and \(|d|\) are prime numbers, then find the value of \(|a|+|b|+|c|+|d|\).

###### This is a problem found in most math books for IIT JEE.

## See Also

**Cite as:**Irregular Polygons.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/irregular-polygons/