The final form is $A = {a²√3}/4$ A is area. a is the known side length. 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. 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√...