Equation for the Radius of a Circle
Posted: Sat Nov 15, 2025 7:34 pm
Now, once you have generalized the Pythagorean right triangle so that you can work any triangle, you arrive at the equation of the radius which circumscribes any triangle,
For brevity sake, sr/ (ab) for square root of ab.
Side 1 is a, side 2 b, side 3 c.
which gives us a + b + c, a + b - c, a - b + c, and b -a + c as working groups
You will find this is the equation.
(a x b x c) /(sr/ (a + b + c) x sr/ (a + b - c) x sr/ (a - b + c) xsr/ (a - b + c) )
And if you draw it in a geometry program that also lets you write equations, you find that no matter what you do, the radius of the figure is equal to the results of the equation.
You have paralleled common grammar, arithmetic, algebra and geometry, all three saying exactly the same thing.
For brevity sake, sr/ (ab) for square root of ab.
Side 1 is a, side 2 b, side 3 c.
which gives us a + b + c, a + b - c, a - b + c, and b -a + c as working groups
You will find this is the equation.
(a x b x c) /(sr/ (a + b + c) x sr/ (a + b - c) x sr/ (a - b + c) xsr/ (a - b + c) )
And if you draw it in a geometry program that also lets you write equations, you find that no matter what you do, the radius of the figure is equal to the results of the equation.
You have paralleled common grammar, arithmetic, algebra and geometry, all three saying exactly the same thing.