Is there any reason why $4-\pi$ is quite close to $\frac{\sqrt{3}}{2}$?

Question

In this question obviously the error of our "approximation" is $4-\pi=0.858...$ . I tried to reconstruct the false argument with $\tau=2\pi$, and the error in that case would be $8-\tau=1.716...$, which is obviously quite close to $\sqrt{3}$. Is there any particular reason why $8-\tau$ is quite close to $\sqrt{3}$? Or, equivalently, is there any particular reason why $4-\pi$ is quite close to $\frac{\sqrt{3}}{2}$? Is this just a coincidence?

Answer 1

Demonstration without words (except for these):

Answer 2

Here is one explanation, though I'm not convinced it's the best one. Note that $4\int_0^1(1-\sqrt{1-x^2})\,dx=4-\pi$ (geometrically, this corresponds to $4-\pi$ being the area left over when the unit disk is removed from a square of side-length $2$). We can expand the integrand as $\dfrac{1}{2}x^2+\dfrac{1}{8}x^4+O(x^4)$, giving the integral approximation

$$4\int_0^1(1-\sqrt{1-x^2})\,dx\approx 4\left[\frac{1}{4}+\frac{1}{32}\right]=1+\frac{1}{8}$$

At the same time, we can do a binomial approximation of $\sqrt{3}/2$ as

$$\frac{\sqrt{3}}{2}=\sqrt{1-\frac{1}{4}}\approx 1-\frac{1}{8}$$ If the second term were positive, then the approximation of the OP would be justified; as it is, it only approximates the approximation.