<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] Another approach would be using calculus. For instance, consider the line $f(x) = \sqrt{3}x $. Notice the line from $(0,0)$ to $(\frac{s}{2}, f( \frac{s}{2}) ) $ is the hypothenuse of half the equilateral triangle. The other sides are $\frac{s}{3}$ and $\sqrt{3} \frac{s}{2} $. So, the area of half of the equilateral triangle is
$$ \int\limits_0^{\frac{s}{2}} \sqrt{3} x = \frac{ \sqrt{3} s^2}{8}$$
And hence the area of the equilateral triangle is twice this area which is 
$$ \frac{ \sqrt{3} s^2}{4}$$
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