Transforming irregular n-gon to regular n-gon through pure construction

A polygon is called regular, if all the sides are equal and all the angles are equal. Else, it is called an irregular polygon. In mathematics, a regular polygon which can be constructed with compass and straight edge is called a constructible regular polygon.

Finding the length of the sides of a regular n-gon :

By equating the area of an irregular polygon with a square of same area, we obtain a ratio. Then with the ratio of the side of the square to the side of the required regular n-gon, we equate both the ratios to find the length of sides of the regular n-gon.

For example, the ratio between sides of an equilateral triangle and a square of the same area
ie; 1:\sqrt[\scriptstyle 4]{\frac{3}{2}}

RATIO BETWEEN THE SIDES OF AN EQUILATERAL TRIANGLE AND A SQUARE OF THE SAME AREA

Consider an equilateral triangle of side ‘a’

In ∆ACD, AC = AD = DC = ‘a’ units

Height = {\frac{\sqrt{3}a}{2}}

Area = {\frac{\sqrt{3}a^2}{4}}

Consider a square having the same area.
Area of square = Area of the equilateral triangle

Side² = {\frac{\sqrt{3}a^2}{4}}

Side = {\frac{\sqrt[\scriptstyle 4]{3}a}{2}}

Therefore, there exists a relationship between the sides of the equilateral triangle and a square of
same area i.e. a: {\frac{\sqrt[\scriptstyle 4]{3}a}{2}}

1: {\frac{\sqrt[\scriptstyle 4]{3}}{2}}

RATIO BETWEEN THE SIDES OF A REGULAR PENTAGON AND A SQUARE OF SAME AREA

Consider a regular pentagon with side ‘x’ units.

AC = CE = EF = FG = GA = ‘x’ units.
Internal angle of regular pentagon = ∠𝐺𝐴𝐶 = 108°
Height of triangle ACD ={\frac{x}{2}}tan{54°}

Area of ∆𝐴𝐶𝐷 = {\frac{x}{2}}.{\frac{x}{2}}tan{54°}

                      ={\frac{x^2}{4}}tan{54°}

Area of the regular pentagon = 5 * area of one triangle

= {\frac{5{x^2}}{4}}tan{54°}

= {\frac{5{x^2}}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }

(\because tan 54° = \sqrt{1+ \frac{2 \sqrt{5} }{5} })

Consider a square having the same area
Area of square = Area of the regular pentagon

Side² = {\frac{5{x^2}}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }

Side= x\sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }}

We get, the side of the square having the same area as a regular pentagon is
\sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }}times the side of the regular pentagon.
∴ The ratio between sides of regular pentagon and square of the same area is
x: x\sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }}
1: 1\sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }}

RATIO BETWEEN THE SIDES OF A REGULAR HEXAGON AND A SQUARE OF SAME AREA

Consider a regular hexagon ABDEFG, Let the length of a side of the hexagon be ‘x’ units.

Consider ∆𝐴𝐵𝐶.
Height of ∆𝐴𝐵𝐶 = AH tan60°
                             = \frac{x}{2} \sqrt{3}

Area of ∆𝐴𝐵𝐶 = \frac{\sqrt{3}}{4} {x^2}

Area of the regular hexagon =6. \frac{\sqrt{3}}{4} {x^2}

= \frac{3 \sqrt{3}}{2} {x^2}
Consider a square having the same area as the regular hexagon.
Area of square = Area of the regular pentagon

side² = \frac{3 \sqrt{3}}{2} {x^2}

Side = \sqrt{ \frac{3 \sqrt{3}}{2}} {x}

It means that the regular hexagon of side ‘x’ will always have the same area as a square with one side
\sqrt{ \frac{3 \sqrt{3}}{2}} {x}
∴ The ratio between sides of regular hexagon and square of the same area is
1: \sqrt{ \frac{3 \sqrt{3}}{2}}

RATIO BETWEEN SIDES OF REGULAR OCTAGON AND SQUARE OF THE SAME AREA

Consider the regular octagon with sides of length ‘x’ units.
Area of regular octagon = 2(1+\sqrt{2}){x^2}
Consider a square having the same area
Side² =Area of the regular octagon
Side²= 2(1+\sqrt{2}){x^2}

Side= \sqrt{2(1+\sqrt{2}){x^2}}
So the square of the same area of a regular octagon with one side ‘x’ unit will always have
\sqrt{2(1+\sqrt{2})}{x} units
∴ The ratio between sides of regular octagon and square of the same area is,
1: \sqrt{2(1+\sqrt{2})}

FOR A REGULAR N-GON

Consider an N-gon with side ‘a’ unit.
Internal angle = \frac{(𝑛−2).(180)}{n}

= 180-\frac{360}{n}
Consider a triangle with a base on one side of the polygon and an opposite vertex at the center of the
polygon.
∴ Half of the internal angle of the polygon = one angle of triangle = 90-\frac{180}{n}
∴ Height of triangle = (\frac{1}{2} \hspace{0.1cm} of \hspace{0.1cm} base)tanθ

=\frac{a}{2}tanθ

=\frac{a}{2}tan(90-\frac{180}{n})°

Area of triangle = \frac{1}{2}.a.\frac{a.tanθ}{2}

= \frac{a^2}{4}tanθ

Area of N-gon = Area of ‘n’ triangles

= \frac{n{a^2}}{4} tanθ

= \frac{n}{4}tanθ.{a^2}

= \frac{n}{4}tan(90-\frac{180}{n})°{a^2}

Consider a square of the same area.
Area of square = Area of regular N-gon

side²=\frac{n}{4}tanθ.{a^2}

Side= \sqrt{\frac{n}{4}tan(90-\frac{180}{n})°}{a}

Which implies that there exists a general ratio between sides of an ‘n’ sided regular polygon and a square of equivalent area. A side of the square having the same area of an n-gon with a side ‘a’ unit is
\sqrt{\frac{n}{4}tan(90-\frac{180}{n})°}{a}
∴ The common ratio of sides of n-sided regular polygon and square of equivalent area,
a: \sqrt{\frac{n}{4}tan(90-\frac{180}{n})°}{a}

ie; 1: \sqrt{\frac{n}{4}tan(90-\frac{180}{n})°}

Similarly, we can obtain the ratios between regular polygon with any number of sides and a square having the same area if tan of half of one angle of it is a constructible value.

For example, a 10-GON

1: \sqrt{\frac{10}{4}tan(90-\frac{180}{10})°}

1: \sqrt{\frac{5}{2}tan(72°)}

tan72°= \sqrt{5+2√5}

∴ The ratio between sides 1: \sqrt{\frac{5}{2}\sqrt{5+2√5}}

CONVERTING A REGULAR POLYGON OR IRREGULAR SHAPE TO SQUARE OF EQUAL AREA / FINDING SIDE OF SQUARE

1: TRIANGLE

Here the irregular triangle is ABC.
CD⊥AB
CE = ED = (1/2)CD
ED = BF
H = Midpoint of AF
AGF is the semicircle of AHF
BG⊥AF
BG is one side of the square having the same area of Δ𝐴𝐵C

2: IRREGULAR PENTAGON

Here ABCDE is an irregular pentagon.
By reducing each corner without changing the area, at last we get a triangle. Ie, Δ𝐹𝐼𝐶.
Steps:
• AC ∥ GF
• By joining FC
• Area of Δ𝐹𝐴𝐶 = Δ𝐴𝐵𝐶
• CE ∥ HI
• By joining CI
• Area of ΔCEI = ΔEDC
∴ Area of irregular pentagon = area of Δ𝐹𝐼C
CJ⊥FI
CK = KJ = (1/2)CJ
KJ = IL
FM = ML (Midpoint of FL is M)
FNL is the semicircle of FML
IN ⊥ FL
IN is one side of the square having the same area of 𝐼𝑟𝑟𝑒𝑔𝑢𝑙𝑎𝑟 𝑝𝑒𝑛𝑡𝑎𝑔𝑜𝑛.

Similarly, we can convert an irregular polygon into a triangle of the same area by reducing sides. And when reaching the triangle we can draw a square of the same area by using the intersecting chord theorem. By applying the ratio we can get the sides of the regular polygon.

FINDING THE SIDE OF A REGULAR POLYGON GIVEN A SIDE OF THE SQUARE OF SAME AREA AND CONSTRUCTING THE REGULAR POLYGON

In the previous section, we have drawn one side of the square which has an area equal to that of the given irregular polygon. We apply the ratio to obtain a side of the required regular polygon.

1: EQUILATERAL TRIANGLE

First, we can look at how the sides of an equilateral triangle is obtained from a square’s side that’s drawn. (The area of square corresponding to the side considered equals to the area of any irregular shape or
triangle.)

For applying the ratio we extent a line from B to D.
BC = {\dfrac{\sqrt[\scriptstyle 4]{3}}{2}} units & CD = 1 unit
(Both units are proportional to each other only. It is not affected by any other units taken)
Firstly we join C to A.
DE ∥ AC
Now we need to find out the length of one side of the equilateral triangle.

\dfrac{Side \hspace{0.1cm}of \hspace{0.1cm} square}{Side \hspace{0.1cm} of \hspace{0.1cm} equilateral \hspace{0.1cm} triangle}  =\dfrac{AB}{AE} ={\dfrac{\dfrac{\sqrt[\scriptstyle 4]{3}}{2}}{1}} 

\therefore AE= \dfrac{AB}{\dfrac{\sqrt[\scriptstyle 4]{3}}{2}}= \dfrac{2}{\sqrt[\scriptstyle 4]{3}}.AB
∴ AE is the required side of an equilateral triangle, its area equal to the area of a square with side AB. Now we can easily draw an equilateral triangle with a side AE.

2: REGULAR PENTAGON

The ratio between a side of a regular pentagon and a side of the square with same area is
1: 1\sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }}
The proportional value of \sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }} to 1 unit we taking can be drawn through several steps.

Here AB is one side of the square and we have to find out the side of regular pentagon with same
area of the square with a side AB.
BC= \sqrt{{\frac{5}{4}}\sqrt{1+ \frac{2 \sqrt{5} }{5} }} times the length of CD, CD = 1unit
Join A & C.
AC ∥ DE

\therefore\dfrac{𝐵C}{CD}= \dfrac{Square's \hspace{0.1cm}side(AB)}{Regular\hspace{0.1cm} pentagon's \hspace{0.1cm} side(AE)}

\therefore AE=\dfrac{AB}{\sqrt{{\dfrac{5}{4}}\sqrt{1+ \dfrac{2 \sqrt{5} }{5} }}}
So AE is a side of a regular pentagon.
For drawing the pentagon from one side,

Using, side(AE) and height (BF) of the pentagon, we can complete the regular pentagon with an area same as the square.

Similarly, we can construct any regular polygon (constructible) from any irregular polygon through the
above-explained method.

Eg: Converting a triangle to an equilateral triangle, Converting irregular octagon to regular octagon.

Here is how to convert an irregular pentagon to a regular polygon such as square, pentagon, octagon, decagon, dodecagon, hexadecagon, icositetragon and so on.

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Read more: “The Intriguing divine ratio of Nature- Golden Ratio“

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