Question about the derivation of RC Filters frequency cutoff
"Basically, this low pass filter passes low frequencies but blocks high frequencies."

At first, no lowpass filter can "block high frequencies" - and, in particular, not the most simple first order RC lowpass. Each lowpass only can - increasingly with rising frequencies - attenuate the amplitudes of frequencies above the corner frequency. Please note that the attenuation at this corner frequency is 3db (attenuation factor 1.414).

Secondly, regarding the squareroot expression: Please note that the capacitive impedance is an imaginary expression (Xc=1/jwc) - and, hence, application of the classical voltage divider rule (between R and Xc) leads to a COMPLEX expression for the transfer ratio T(jw) consisting of a real (Re) and an imaginary part (Im). 

Then, the magnitude of the transfer function T(jw) (output-to-input ratio) is |T(jw)|=SQRT(Re²+Im²) and the phase is Phi=arctan(Re/Im).

Setting |T(jw)|=1/1.414=0.7071 and solving for w gives the corner frequency fc=1/2Pi*RC.

Regarding C-R highpass:
The above described calculation (based on SQRT(2)=1.414) reveals that at the corner frequency w=wc the magnitude of the capacitive resistance (1/wC) is equal to the real resistance (R). Therefore, the first-order C-R highpass has the same corner frequency as the R-C lowpass.