Is this a valid derivation that the value of $\pi$ is $4\sqrt{\sqrt{5/4}-\sqrt{1/4}}$?
This is a completely different from my last try at a proof of the same result. It is in no way a duplicate of my last question.
When the circumference (c) of a circle with diameter (d), and the perimeter (4h) of a square are equal, we will call it a squared circle.
Our first squared circle. (Fig. 1)
Our second squared circle, which is π/4 times smaller. (Fig. 2)
Next we stack our two squared circles. (Fig. 3)
Then we can add in our triangle. (Fig. 4)
Our full triangle where the hypotenuse equals the diameter (d) of the previous circle. (Fig. 5)
Then we could use Pythagoras to calculate π.
But first, lets dig a little deeper.
Any generic squared circle. (Fig. 6)
From this we get our red and blue squares. (Fig. 7)
The blue squares link the previous squared circle with the next consecutive squared circle, which is always π/4 times smaller. (Fig. 8)
This can be seen more clearly in the table below, the arrows show this relationship between any two consecutive squared circles.
This repeating pattern goes on ad infinitum, where each consecutive squared circle is always π/4 times smaller. (Fig. 9)
This gives us our angles and our generic triangle. The base (x) of the triangle is equal to the height (h) of the triangle minus the side-lengths of the two blue squares. (Fig. 10)
The side-length (h) divided by the diameter (d) is always π/4.
Then we can replace π/4 in our equation.
And then we use this value in the following triangle.
This gives us the generic triangle. (Fig. 11)
A circle with diameter (d) equal to 1. (Fig. 12)
Using Pythagoras we get the following equation.
From this equation we can calculate π.
No, this is not correct. In Figure 3, if we assume that you have placed a right triangle with a side of length $\pi/4$ centered on the bottom edge of the square, then the diagonal of the triangle is not actually a diameter of the circle. It also does not appear that you ever make any argument to why this diagonal should be a diameter, but that's a critical omission since it is exactly where your argument fails.
More or less, you need to convince yourself that this argument fails, since any other counterarguments one can offer rely on being able to calculate $\pi$ by some other means. However, if you're willing to accept that the usual value of $\pi$ is actually true, consider the following:
By the Pythagorean theorem, if we impose Cartesian coordinates on this diagram, labelling the axes to be parallel the sides of the square and putting the origin at the center of the circle, we can find that the intersections of the square with the circle are at points of the following two forms: $$(\pm1,\pm \sqrt{(2/\pi)^2-(1/2)^2})$$ $$(\pm \sqrt{(2/\pi)^2-(1/2)^2},\pm 1).$$ Thus, the length of the side of your triangle would have to be $2\sqrt{(2/\pi)^2-(1/2)^2}$ rather than $\pi/4$. These two values differ by about $.003$, so it would be hard to see the difference on a diagram, but there is a difference.
More generally, $\pi$ is known to be a transcendental number. This means it is not the root of any polynomial with rational coefficients. This implies that it cannot be written using addition, subtraction, division, multiplication, and radicals (i.e. square roots, cube roots, ...), so your value cannot be correct.