Math: Area of Equilateral Triangle (With Given Side Length)

The final form is $A = {a²√3}/4$

A is area.

a is the known side length.

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Heron's Formula

The formulation for arbitrary triangle, Heron's formula, is:

$A = √{s(s-a)(s-b)(s-c)}$

With $a ≥ b ≥ c$

Where:

$s = (a + b + c)/2$

a, b, c are the lengths of each side.

The $1/2$ in $s$ is related to the center point (centroid) of the triangle.

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Derivation

The equilateral triangle area formulation (final form above) is derived from Heron's formula.

As in ➡️ a = b = c

Plug that into Heron's formula, from s then to A, we will get same final (simplified) form.

So then, $a = b = c$. We just use one letter, $a$.

Substitute $b$ and $c$ as $a$ in $s$. Therefore, $s = {(a + a + a)/2} = {3/2}a$.

Plug the $s$ to $A$, also substitute $b$ and $c$ as $a$ in $A$.

So then ➡️ $A = √{{3/2}a({3/2}a-a)({3/2}a-a)({3/2}a-a)}$

Continue:

$A = √{{3/2}a({1/2}a)({1/2}a)({1/2}a)}$

$A = √{{3/2}a{1/8}a³}$

$A = √{{3/16}a⁴}$

$A = √3√{1/16}√a⁴$

$A = √3{1/4}a²$

$A = {a²√3}/4$

Ἥρων ὁ Ἀλεξανδρεύς ἐν τῇ στοᾷ καθήμενος, λίθους μέγιστους ἀναφέρει, μυῶνας αὐξάνων ὥσπερ ἄτλας.

Translation: "Heron of Alexandria, sitting in the stoa, lifts massive stones, growing muscles like Atlas."

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Basic: Area of a Triangle With Base and Height Lengths

Area of a triangle with known base and height is:

$A = {b×h}/2$

b is base

h is height

The $1/2$ (divided by two or multiplied by half) is based on a diagonally and evenly cut rectangle or square, from one corner to the the other diagonally opposite corner into two pieces. Each piece is called a (right) triangle. You see, the area of rectangle is length times width.

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Why is the area of a rectangle calculated as length times width?

Let's visualize this using identical sticks or foot-length markers.

Imagine placing them side by side, covering the entire rectangular area without overlapping. The total number of sticks (or foot-lengths) that fit within the rectangle represents its area. Since we arrange them along both the length and width, multiplying length by width gives us the total count of these units, fully covering the shape in two dimensions.

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Heron's Formulation

Heron's can be proven via law of cosines, or Pythagorean, or law of cotangents, or other mathematical methods.

How did Heron get to that?

Well, actually, prior Heron there were also mathematicians who developed similar ideas related to triangle area calculation.

Who was the first to discover it?

I don't know.

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Heron's Formula Forms

Base form is $A = √{s(s-a)(s-b)(s-c)}$

If we do not want to abbreviate the ${a + b + c}/2$ as $s$, we will find other equivalent forms, such as:

$A = {√{(a + b + c)(-a + b + c)(a - b + c)(a + b - c)}}/4$

$A = {√{2(a²b² + a²c² + b²c²) - (a⁴ + b⁴ + c⁴)}}/4$

$A = {√{(a² + b² + c²)² - 2(a⁴ + b⁴ + c⁴)}}/4$

$A = {√{4(a²b² + a²c² + b²c²) - (a² + b² + c²)²}}/4$

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