We lost the center of this Circle
Center another circle on the perimeter of the first
We found the center and made some petals
Connect the intersecting points with lines
These lines are parallel to each other
The sides of this inscribed hexagon are equal length. It is a regular hexagon.
Circle bisectors reveal the six equilateral triangles
A semicircle
Finding its center
Bisect the diameter
It's a right angle
Bisect the quadrant
Bisect the octant
Two more right-angled triangles
Bisect the octants
Complete the chords
Radius of 2
Area of the quadrant is π
Perimeter of the quadrant arc is π
Nothing clever here: this is the definition of π
Calculating the chord lengths is just applying pythagoras's theorem
Now we know we can compute the lengths of successive chords from the previous ones because the triangles are right angled and the radius of the circle is fixed. We can write this algebraically.
The arithmetically simplest recurrence relation to track the sizes of these triangles involves the heights of the triangles - half the width of the corresponding regular pentagon. Just an addtion of 2 and a square root.
We derive the chord length from the height squared to calculate the polygon perimeters.
If we do this long enough with perfect precision the polygon sides will add up to π.
My way is better: add up all the  right-angle triangle areas.
A short C program to explore the method. There is the recurrence followed by the summation.
Here we see the area summation is better. Good results until we run out of floating point precision for the sum.
The perimeter method on the other hand lost accuracy much earlier.
This is why I was inspired to develop this method: the computer I was using, an HP9810 only had fast basic arithmetic and square rooting - the gray keys of the middle keypad.
Affordable calculators had square roots too!
Even this cheap Canon one with a fluorescent display.
