Sedrakyan's area formula:
A Unified Distance-Based Area Formula for Convex Polygons

Abstract

This work fills a 2000-year-old gap and is the structural “missing link”: Heron's formula dates back to $\sim\!60$ AD and Bretschneider's formula to 1842, yet no such unified formula existed for polygons with more than 4 sides. This work shows that these two formulas—which looked completely different for centuries—are particular cases of a single general formula. The remarkable thing is not that it uses only side lengths and certain distances, but rather that one single unified formula computes the area of any convex $n$-gon for any number of sides $n \geq 3$ (which is the key novelty), previously thought impossible for $n \geq 5$. This novel formula is overcoming the “impossibility” of rigidity for $n \geq 5$. Applications include distance geometry, knot theory, drone networks, robotics—where coordinates are unavailable, only pairwise distances are known, and area calculations are mission-critical. The geometrically constructed point $B_2$ can be viewed as a representative point associated with the figure’s area distribution—analogous in spirit to the center of mass, which serves as a representative point for the volume of a three-dimensional body.

Introduction

In computer graphics, geographic information systems, and structural engineering, calculating the area of irregular polygons usually requires breaking them down into a mesh of triangles or using Cartesian coordinates. In many real-world scenarios (flexible structures, robotics, distance-only sensors) coordinates are genuinely unavailable, making this formula not just a niche alternative but the only clean option. This newly discovered formula allows for the area calculation of a flexible or moving convex structure using only distance constraints (sides and specific internal diagonal lengths), bypassing the need to calculate coordinate positions or interior angles. The formula is expressed via a beautifully alternating sum of squared side lengths and a geometrically constructed point $B_2$, which elegantly bypasses the structural flexibility of the polygon by introducing a single reference point relative to the diagonal $A_1A_3$. The distance $A_2B_2$ accounts for the degrees of freedom of the polygon without requiring trigonometry, coordinates, or triangulation.

Remark. This formula has been presented at international conferences as Sedrakyan's area formula. The formula was initially realized by Hayk Sedrakyan, experimentally verified with Christophe Pinarbasi, and formally proved by Nairi Sedrakyan and Hayk Sedrakyan. The author thanks Christophe Pinarbasi for stimulating discussions. Having recognized the result but lacking a formal proof at the time, the authors have, since 2022, shared this topic with several people; however, none of them was able to prove the result. Nevertheless, the authors would like to thank them for their efforts.

Particular cases (Heron's and Bretschneider's formulas)

Theorem ($n = 3$, Heron's formula). The area $A_T$ of a triangle $A_1A_2A_3$ with side lengths $a$, $b$, $c$ is (in the section main results and proofs we show that for a triangle point $B_2$ coincides with vertex $A_1$):

In the next section we show that this formula of $A_T$ is actually Heron's formula.

Theorem ($n = 4$, Bretschneider's formula). The area $A_Q$ of a quadrilateral $A_1A_2A_3A_4$ with side lengths $a$, $b$, $c$, $d$ and diagonals $e$, $f$ is (in the section main results and proofs we show that for a quadrilateral point $B_2$ coincides with $A_4$):

In the next section we show that this formula of $A_Q$ is actually Bretschneider's formula.

Remark. In applied sciences, we often encounter situations where we can measure distances, but we have absolutely no way to map out an $(x, y)$ coordinate grid. For example:

Distance-Only Sensor Networks: Imagine a swarm of drones or localized robotic nodes. They can ping each other to measure their exact mutual distances (the side lengths and select diagonals of an $n$-gon), but they do not know their orientation in global space. They have no coordinate origin.

Flexible Structural Mechanics: Imagine an engineering framework or a architectural mesh under stress. The beams (sides) are shifting and flexing. In these scenarios, you cannot use the coordinate-based formula because you have no coordinates. This area formula provides the only direct mathematical pathway to compute or track the changing area purely out of a distance matrix.

Proof of equivalence for the cases $n = 3$ and $n = 4$

The case $n = 3$ (Heron's formula). Let us prove that the formula of $A_T$ is actually Heron's formula. By difference of squares formula, we have

Thus, it follows that

The case $n = 4$ (Bretschneider's formula). Let us prove that the formula for $A_Q$ is actually Bretschneider's formula. To find the area $A_Q$ of any quadrilateral Bretschneider's formula states:

where $\alpha, \gamma$ are opposite angles. Let $\theta$ be the angle between diagonals $e$ and $f$. Let us prove that

Step 1: Our formula equals $\frac{1}{2}ef\sin\theta$. Place the quadrilateral in the plane with $A_1$ at the origin and let $\vec{e}=\overrightarrow{A_1A_3}$, $\vec{u}=\overrightarrow{A_1A_2}$, $\vec{v}=\overrightarrow{A_1A_4}$. We get

Thus

Step 2: Bretschneider's formula also equals $\frac{1}{2}ef\sin\theta$. Splitting the quadrilateral along diagonal $e$ into two triangles gives:

and by the law of cosines $e^2=a^2+b^2-2ab\cos\alpha=c^2+d^2-2cd\cos\gamma$. Using the identity $2\sin\alpha\sin\gamma=\cos(\alpha-\gamma)-\cos(\alpha+\gamma)$ and expanding $(s-a)(s-b)(s-c)(s-d)$, one verifies by direct algebra that:

and hence Bretschneider's formula gives $A_Q=\frac{1}{2}ef\sin\theta$.

Conclusion. Both formulas equal $\frac{1}{2}ef\sin\theta$, so they are equivalent.

The main result

Important (constructing the point $B_2$). Looking at the form of $y_0$ in the previous equation, it initially appears that the area of $(A_1A_2\dots A_n)$ is required to construct line $\ell_2'$ and, consequently, to locate point $B_2$. This seems to introduce a circular argument, implying that we must already know the area of $(A_1A_2\dots A_n)$ to find it. However, in Remark 1 of the next section, we demonstrate that the value of this area is not actually needed to construct $\ell_2'$ and determine the location of $B_2$.

The meaning of the formula

At first glance, the introduction of the point $B_2$ may seem somewhat unusual. However, consider the analogous concept of the center of mass of a three-dimensional body. The center of mass is a single point that represents the distribution of mass throughout the body, and it is well known that this point may lie outside the body itself. In two dimensions, instead of volume, we consider area. The concept of volume in three dimensions corresponds to the concept of area in two dimensions. In a similar way, the point $B_2$ can be viewed as a representative point associated with the area distribution of the figure.

The proof of the main result

Let us denote the area of a convex $n$-gon $A_1A_2\ldots A_n$ by $(A_1A_2\ldots A_n)$. Let $A_1A_2=a_1$, $A_2A_3=a_2$, ..., $A_{n-1}A_n=a_{n-1}$, $A_nA_1=a_n$.

Problem 1. Find the locus of all points $M$ in the plane such that

Solution of Problem 1. Choose a coordinate system as shown in the figure.

We have

Thus

Let us prove that the locus of all points $M$ in the plane is line $l_2$ which passes through the point

and is perpendicular to line $A_1A_3$.

If $n = 3$, then $a=a_3$ and line $l_2$ passes through the point $A_1(0, 0)$. If $n = 4$, then line $l_2$ is passing through point $A_4$, as

Problem 2. Find the locus of all points $M$ in the plane such that $(MA_1A_2A_3)=(A_1A_2\ldots A_n)$ and that points $M$ and $A_2$ are not on the same side of line $A_1A_3$.

Solution of Problem 2. Choose a coordinate system as shown in the solution to problem 1. Let $M(x,y)$, where $y\leq 0$, we have

Thus

Let us prove that the locus of all points $M$ is line $l_2^{\prime}$ passing through point

and is perpendicular to line $l_2$.

If $n = 3$, then line $l_2^{\prime}$ coincides with line $A_1A_3$, as

If $n = 4$, then line $l_2^{\prime}$ passes through point $A_4$, as

Let us denote by $B_2$ the intersection point of lines $l_2$ and $l_2^{\prime}$, in a similar way denote by $B_i$ the intersection point of lines $l_i$ and $l_i^{\prime}$, where $i = 1,\ldots,n$.

If $n\geq 5$, then let us start with the following remark.

Remark 1. Let a convex $n$-gon be given. Construct point $M_n$ on line $C_2C_3$ such that

Let us construct point $M_n$ be mathematical induction.

The base case. If $n = 4$, then from point $C_4$ let us draw a line parallel to line $C_1C_3$. Denote by $M_4$ the intersection point of that line and ray $C_2C_3$ (see figure).

Indeed, we have

The inductive hypothesis. Assume that for $n = k$ the point $M_k$ is already constructed, then let us show how point $M_{k+1}$ is constructed. Consider the convex polygon $C_1C_2C_4\ldots C_{k+1}$. According to the inductive hypothesis on ray $C_2C_4$ we can construct point $M_k$ such that

Let us draw a line from point $M_k$ parallel to line $C_1C_4$ (see figure). Denote by $N_k$ the intersection point of that line and line $C_3C_4$.

We have

Thus

Thus, we get that

It is left for quadrilateral $C_1C_2C_3N_k$ on ray $C_2C_3$ to construct point $M_{k+1}$, such that

From (1) and (2), we get

This ends the proof for $n = k+1$. Thus, this ends the proof for any $n\geq 4$.

To construct line $l_2^{\prime}$ it is sufficient to construct point $M_{n-1}$ for the polygon $A_1A_3A_4\ldots A_4$ and to draw a line passing through point $M_{n-1}$ and parallel to line $A_1A_3$.

Let us prove the following theorem.

Theorem. The area of a convex $n$-gon $A_1A_2\ldots A_n$ is:

To prove this theorem, we use the following lemma.

Lemma (the Trapezoid Law [Sedrakyan et al.], see [1], chapter 5). Let sides $AB$ and $CD$ of quadrilateral $ABCD$ be parallel to each other, then

Let us continue the proof of the theorem, consider the the following figure.

We have

Moreover

On the other hand, according to the Pythagorean theorem, we have

According to the lemma, we have

Point $A_2^{\prime}$ is on line $l_2$, thus according to Problem 1, we have

From (1), (2), (3), we get that

Thus

This ends the proof of the theorem.

Conclusion. Triangles ($n=3$) are inherently rigid because fixing their side lengths completely locks their interior angles and area into place. Quadrilaterals, pentagons, and higher-order polygons ($n\geq 4$) are flexible, meaning their sides can deform and flex like a mechanical linkage, constantly changing their angles and overall area. This fundamental fluidness is why a unified, side-length-based area formula was long considered impossible for shapes with five or more sides. Conversely, in geometry, you cannot find the area of an irregular $n$-gon purely from its sides because the shape can bend. By capturing that exact architectural fluidness inside a single point $B_2$ and an alternating sum of squared sides, this area formula accomplishes something structurally revolutionary that did not have an existing. It proves that the chaotic flexibility of any arbitrary polygon can be structurally bound and unified under a single algebraic law, opening the door to solving distance-only constraints that were previously considered mathematically impossible.

Example showing how the formula works

Example. Let us find the area of a regular hexagon $A_1A_2\ldots A_6$, where $A_1A_2=1$.

Solution. Line $l_2$ is determined by all points $M$ satisfying the following equation

So, it will be the perpendicular bisector of line segment $A_1A_3$ (see figure).

Now, let us construct line $l_2^{\prime}$ and find the intersection point $B_2$ of lines $l_2$ and $l_2^{\prime}$. Using the theorem we have that $l_2^{\prime}$ is parrallel to $A_1A_3$ and is passing through the point $(0, y_0)$, where

Notice that $y_0 < 0$, so $-y_0 > 0$. We get

Therefore, point $B_2$ is a point on line $l_2$, such that

When $A_1A_5\parallel A_6B_2$ and $A_3A_5\parallel A_4B_2$, then $(A_1MA_6)=(B_2MA_5)$ and $(A_3NA_4)=(B_2NA_5)$. Hence

as $A_1A_3NA_5M$ is in common for triangle $B_2A_1A_3$ and pentagon $A_1A_3A_4A_5A_6$. We obtain

Applying the area formula (note that $a_1=a_2=a_3=a_4=a_5=a_6$), we get

Verification of the answer. A regular hexagon with side length 1 can be divided into six equilateral triangles of side length 1. The area of one equilateral triangle with side length 1 is

Therefore, the area of the hexagon is

Applications

Application 1. A triangle has side lengths $\sqrt{3}$, $\sqrt{5}$, and $\sqrt{7}$. Find its area.

Application 2. Consider a trapezoid with bases of lengths 5 and 9. If the sum of the squares of its diagonals is 196, find the sum of the squares of its legs.

Application 3. The area of triangle with sides $a$, $b$, $c$ is equal to the area of triangle with sides $a$, $b$, $d$, where $d\neq c$, and that $a^{2}+b^{2}=16$, find $c^{2}+d^{2}$.

Application 4. Each of two non-congruent triangles has two sides of lengths 5 and 7, and the triangles have equal circumradii. Find the product of the lengths of their third sides.

Application 5. Let $a_1$, $a_2$, $a_3$ be the side lengths of a triangle, prove that

Application 6. Let $ABCD$ be a convex quadrilateral, such that $AB=a$, $BC=b$, $CD=c$, $AD=d$, $AC=e$, $BD=f$. Prove that

Application 7. Let $A_1A_2\ldots A_{2k}$ be a convex polygon such that $A_1A_3=A_2A_4=\cdots=A_{2k-2}A_{2k}=A_{2k-1}A_1=A_{2k}A_2$. Prove that

Solutions of applications

Application 1. A triangle has side lengths $\sqrt{3}$, $\sqrt{5}$, and $\sqrt{7}$. Find its area.

Solution. According to this area formula, we have

Application 2. Consider a trapezoid with bases of lengths 5 and 9. If the sum of the squares of its diagonals is 196, find the sum of the squares of its legs.

Solution. Let the diagonals be $e$ and $f$ (see the figure). Given that $e^{2}+f^{2}=196$. It is a known result that if $e^{2}+f^{2}=(a+b)^{2}$, where $a$ and $b$ are the bases then this condition is equivalent to the condition that the diagonals of the trapezoid must be perpendicular to each other. In this case the bases are 5 and 9, given that $e^{2}+f^{2}=196=(5+9)^{2}$. This means that the diagonals of this trapezoid must be perpendicular to each other. Then, we have

On the other hand (by the considered area formula), we have

Thus, we get

Squaring both side of this equation, we get

This means that

So, we get

Application 3. The area of triangle with sides $a$, $b$, $c$ is equal to the area of triangle with sides $a$, $b$, $d$, where $d\neq c$, and that $a^{2}+b^{2}=16$, find $c^{2}+d^{2}$.

Solution. As the areas of both triangles are equal, then we have

Squaring both sides and canceling the expression $4a^{2}b^{2}$ on both side, we get

Therefore,

Note that this can be factored as follows:

Simplifying, we get

Given that $a^{2}+b^{2}=16$, hence

Given also that $c\neq d$, thus $c^{2}-d^{2}\neq 0$. Therefore,

We obtain

Application 4. Each of two non-congruent triangles has two sides of lengths 5 and 7, and the triangles have equal circumradii. Find the product of the lengths of their third sides.

Solution of Application 3. Denote their areas by $A_1$ and $A_2$, denote the length of their third sides by $a$ and $b$, then by this area formula, we have

Denote by $R_1$ and $R_2$ their circumradii, then by the law of sines, we have

As $R_1=R_2$, we get

Thus, it follows that

Squaring both sides, we get

Simplifying and using difference of squares formula, we get

We obtain

As triangles are different, then $a\neq b$. Canceling $a^{2}-b^{2}$ and $(a-b)(a+b)$, we get

Hence

Therefore

Application 5. Let $a_1$, $a_2$, $a_3$ be the side lengths of a triangle, prove that