<s>[INST] Proving the area of an equilateral triangle How do you prove that How do you prove that for any equilateral triangle with side length s, area is $\frac{s^2 √3}{4}$ ? I tried using an equilateral triangle in a square, but I keep coming up with a $2x^2√3$ , as shown below. What am I doing wrong?
I started with the following:

The area of the full square is:
$ 2x * 2x = 4x^2$
To find the area of the triangle, I will subtract the non-triangle parts from the square.
The part shaded green is:
$ (2x - x√3) * 2x = 4 x^2-2x^2√3$
The parts shaded blue are:
$ \frac{x   *   ( x√3)}{2} + \frac{x   *   ( x√3)}{2} = x^2√3$
Adding blue and green:
$(x^2√3) + (4 x^2-2x^2√3) = 4 x^2-x^2√3 $
Subtract blue and green from whole square:
$(4x^2) -(4 x^2-x^2√3) = x^2√3$
Multiply by 2 because I am referring to the $2x$ side, not half of it ($x$):
Final answer: $2 * (x^2√3) = 2x^2√3$
And of course, $ 2x^2√3 \neq \frac{x^2 √3}{4}$
 [/INST] Let's consider a rectangle wit height $x\sqrt{3}$ and length $x$ such that the right angled triangle with legs as the sides forms half of the equilateral triangle (as shown in the picture in the question). Then, the area of the right angled triangle is $\dfrac{1}{2}x^2\sqrt{3}$. Two times this is the area of the equilateral triangle. This gives $x^2\sqrt{3}$. Now, $x$ is half of the side $s$ of the equilateral triangle, i.e., $x=s/2$, implying that the area is $\dfrac{s^2}{4}\sqrt{3}$.
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