# $\Large{\purple{Squaring}\space \blue{Any}\space \red{Polygon!}}$

$\large{\green{{\underline{The\space Problem}}}}$ $Given\space a \space polygon \space A_1A_2A_3...A_n \space we \space need \space to \space construct \space a \space square\space PQRS \space such \space that$ $\boxed{ar(A_1A_2A_3...A_n)=ar(PQRS)}$ $Using\space Straight \space - \space edge \space and \space compass \space construction$

$\large{\purple{\underline{The\space Solution}}}$

The problem may seem very hard but it isn't really is.

The idea is that we can break any polygon $A_1A_2A_3...A_n$ into $n-2$ triangles, these triangles will be of the form $\triangle A_1A_{i}A_{i+1},i\in\mathbb{N}\cap (1,n)$

We can then find square $P_iQ_iR_iS_i$ such that $ar(\triangle A_1A_iA_{i+1})=ar(P_iQ_iR_iS_i),i\in\mathbb{N}\cap (1,n)$

And then we can find a square $PQRS$ such that $ar(PQRS)=\sum_{i=2}^{n-1}ar(P_iQ_iR_iS_i)$

And if we are able two add at least 2 square then we can find the sum.

So we are first going to break the problem in two parts :

• Squaring A Triangle

$\large{\underline{\blue{Squaring\space A \space Triangle}}}$ $Given \space \triangle ABC \space we \space need \space to \space find \space square \space PQRS \space such \space that$ $ar(\triangle ABC)=ar(PQRS)$ $Steps \space of \space construction$ $1) \space Draw \space \overline{AD}\perp\overline{BC} \space such \space that \space D \space lies \space either \space on \space \red{\overline{BC}} \space or \space \blue{extended \space \overline{BC}}$ $2) \space Draw \space a \space rectangle \space EFGH \space such \space that \space \overline{EF}=\overline{AD},\overline{FG}=\dfrac{\overline{BC}}{2}$ $3) \space Draw \space a \space square \space PQRS \space such \space that \space ar(PQRS)=ar(EFGH)$

$\large{\underline{\red{Adding\space Two\space Squares}}}$ $Given \space square \space ABCD \space and \space EFGH \space we \space need \space to \space find \space square \space PQRS \space such \space that$ $ar(PQRS)=ar(ABCD)+ar(EFGH)$ $Steps \space of \space contruction$ $1) \space Draw \space \triangle \space KLM \space such \space that \space \angle KLM=90\degree,\overline{KL}=\overline{AB},\overline{LM}=\overline{EF}$ $2) \space Draw \space square \space PQRS \space such \space that \space \overline{PQ}=\overline{KM}$

So we are done!

$\large{\underline{NOTE}}$

• $ar(X)$ denotes the area of planar figure $X$

• If anybody need justification of any of the construction he or she may ask. (Although they are much easier to be justified)

• The number of steps for squaring an $n-sided$ polygon shall be increasing linearly if we used these steps, if anybody can do this in less amount of steps then please comment.

• If anybody can prepare an animation of this either in Desmos or in .gif format etc, please share it here. Note by Zakir Husain
3 months, 1 week ago

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Sort by: Talking about a convex $n$-gon, we can start successively reducing the number of sides one at a time, keeping the area unchanged, until we come up with a triangle with the same area as the initial polygon. The rest has already been described by @Zakir Husain.

Here is the course of action (referring to the figure):

Form ${{A}_{n}}$ draw a line parallel to ${{A}_{1}}{{A}_{n-1}}$ wich intersects ${{A}_{n-1}}{{A}_{n-2}}$ at point ${{{A}'}_{n-1}}$. $\triangle {{A}_{1}}{{{A}'}_{n-1}}{{A}_{n-1}}$ and $\triangle {{A}_{1}}{{A}_{n}}{{A}_{n-1}}$ have the same base $A_{1} A_{n-1}$ and the same corresponding height, therefore the have the same area. Thus, polygon ${{A}_{1}}{{A}_{2}}\ldots {{A}_{n-2}}{{A}_{n-1}}{{A}_{n}}$ and ${{A}_{1}}{{A}_{2}}\ldots {{A}_{n-2}}{{{A}'}_{n-1}}$ also have the same area, but the new polygon has $n-1$ sides. By repeating this procedure, we can transform the $n$-gon to a triangle with the same area.

- 1 month, 2 weeks ago

Great Idea

- 1 month, 2 weeks ago

@Mahdi Raza Can you make an animation on this?

- 3 months, 1 week ago

To square a triangle, how do you move from step 2 to step 3?

- 3 months, 1 week ago

What did you mean by move ?

- 3 months, 1 week ago

I meant, how to go from step 2 of construction to step 3. In step 2, we have a rectangle and in step 3 we have a square with the same area. So how do you go from step 2 to step 3?

- 3 months, 1 week ago

See You basically made a rectangle ABCD in step 2. Now there is only one problem, we need to find a square with that area (area of triangle) not a rectangle. Now from my previous note we know that we can just square the rectangle. Then we will find a square PQRS with the same area as the rectangle, but since the area of the rectangle ABCD is equals to the area of the triangle. Therefore the areas of the triangle, the rectangle, the square are the same.

Therefore in summary we are able to find the square PQRS with the same area as the triangle.

Note :

• The construction of the rectangle here is really just a transition period which will be useful only for some time, after we find the square there is no need of the rectangle so we can just remove it from the paper it will not make any changes.

- 3 months, 1 week ago

Oh, I see, I didn't know you had posted a note before this one. I'll try for an animation. Look interesting

- 3 months, 1 week ago

My final question, sir, is how do I square a circle?

- 3 months, 1 week ago

Well it have been proved that we can't square a circle.

In 1837, Pierre Wantzel showed that any number $x$ that can be constructed using straightedge and compass have to be an algebraic number.

In 1882 Ferdinand von Lindemann proved the transcendence of $π$ hence theoretically we can never construct $\pi\space units$, But rather we can construct some of it's approximations

Now your question gave me a new Theorem

$\boxed{"Suppose \space S \space is \space the \space set \space of \space all \space constructible \space approximations \space of \space \pi,f:S\to\mathbb{R},f(x):=|\pi-x|\space \Rightarrow\space \forall x_1\in{S} \space \exists\space x_2\in S : \red{0

- 3 months, 1 week ago

Or how to construct a line that is equal to $\sqrt{\pi}$?

- 3 months, 1 week ago