Michael Rack offers a solution that produces an acurate algebraic value for pi.
This proposed solution produces a square that has the same area as the
circle. For r=1, each side of the square is equal to the square root of π.
The angle theta is derived from asin(sqrt(pi))/4. R*sin(theta)=aqrt(pi)/2 and
dia.*sin(theta)=sqrt(pi) and pow(dia.*sin(theta),2)=pi.
This is achieved by drawing a series of circles along both axes and
joining the points of intersection, so creating a grid. This ensures that every
constructed square of the grid, has sides of equal legnth, regardless of the legnth of the
radius.This enables the construction of a right triangle. The hypotenuse terminate
at the circumferance creating a point with height equal
to square root pi/2.
