Finding the cut-off frequency of a filter
For a unity-gain high-pass Sallen Key filter like this: -

The frequency \$\frac{1}{2\pi\sqrt{R_1R_2C_1C_2}}\$ isn't the 3 dB cut-off frequency but the formula for the natural resonant frequency of the filter. However, the 3 dB cut of frequency will be equal to that frequency when Q = \$\frac{1}{\sqrt2}\$. This is because, for a 2nd order low pass (or high pass) filter, the gain at the natural resonant frequency is numerically (and formulaically) equal to Q.
I use the term &quot;natural resonant frequency&quot; but some people (such as @LvW) call it the pole frequency. Anyway, for a 2nd order high-pass filter with unity gain, the \$j\omega\$ transfer function is: -
$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{1}{1-\frac{\omega^2}{\omega_n^2}-j2\zeta\frac{\omega}{\omega_n}}$$
And if you let \$\omega = \omega_n\$ you get: -
$$\dfrac{V_{OUT}}{V_{IN}} = \dfrac{j}{2\zeta}$$
And this implies that at the natural resonant frequency (\$\omega_n\$) there is a phase shift of 90° and the gain is Q.