You can notionally build as many stages as you want with a single amplifier, and AFAIR I have seen a 5 stage design implemented just to make the point BUT it becomes increasingly hard to "realise" (= construct) as you add stages around a single amplifier. To obtain the correct ratios of components requires increasingly precise component values and increasingly stable components. Capacitors are hard to get with extremely high precision and resistors are only slightly better. For a two stage or 3 stage design you can in most cases manage with 1% parts. Beyond that, the fun begins.
Note: "Pole" used generally here rather than saying "pole or zero as is applicable ..." in each case.
While you will notionally get the same result from a bandpass filter by cascading stages in any order, you will find that in limiting cases aspects such as stage Q and signal magnitude will have some effect. The same applies to stage order in a multiple stage low or high pass.
Your circuits are unusual in separately providing gain for the amplifier. This is acceptable, but the norm is to use a unity gain buffer in this application - amplifier Vout connected to amplifier inverting input. The addition of gain will also affect filter Q and you will end up not realising a classic filter polynomial if you alter the gain - assuming the designer implemented a 'proper' filter in the first place. In the case of the multipole design, varying the gain arbitrarily as shown will influence the "shape" of the resultant response rather than just its amplitude.
For one and two pole designs that need a unity gain buffer, you can use a 1 transistor emitter follower with usually acceptable results. As shown below, the results with a transistor with relatively low gain are inferior to results usually available from an opamp, but can still be very useful..
The above diagram is from this extremely good page -
Elliott Sound products: Active filters - Characteristics, Topologies, Examples
Lots more on the above, and related, here - Gargoyle search.
There's many ways to answer that, here are some to throw out ideas:
1) since it is a second order filter with 2X slope per octave then you can argue that it should be sqrt(10) X.
2) typically though that measurement might also take into account the ripple and constrain it so that a fixed factor may not be apropos. In your case of critically damped this doesn't hold, but in a more general sense perhaps a simple factor isn't the right criteria.
3) Since I work with S&H system and switched cap we don't use a 10X but how many time periods until settled to within the error budget of the system (so it's variable depending upon what is being designed).
Best Answer
Attenuating to 10% means the dB reduction is 20 dB because 20 log 10 = 20 dB.
A 2nd order low-pass filter will normally reduce the high frequencies at 40 dB per decade therefore, you need to have the filter's cut-off frequency half a decade below 100 Hz at 31.62 Hz (\$\sqrt{10\times 100}\$).
Here's an approximation: -
This was modelled with an RLC low pass filter interactive tool having a butterworth response (\$\zeta\$ = 0.7071). Make sure that if you are using a sallen key filter the op-amp has a suitable GBWP or it might not meet your expectations.
Note that I didn't go to great lengths to choose LC to make exactly a cut-off 31.62 Hz but it's near enough to demonstrate that the marker at 100.66 Hz is 20 dB down.