Answers for "approximate_pi"

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approximate_pi

def approximate_pi(n, d=2):
    r = d/2
    a = 360/n
    x = (2 * r**2 * (1 - math.cos(math.radians(a))))**0.5
    perimeter = x * n
    pi = perimeter / d
    return pi
for n in [3,4,5,10,100,1000,10000,100000]:
    print(n, "-->", approximate_pi(n))

Posted by: Guest on May-07-2022

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