There is a lot of tosh talked about brake pads so in an attempt to try to clarify at least one part of this here is some information regarding pad area.
Even if the pad has a smaller area the braking force can be the same as a pad with a larger area.
As an example.
If you apply a 100lb force on the brake pedal with the mechanical advantage/leverage
of the pedal say at x4, the force at the master cylinder will be 400lb
and let's say the master piston has an area of 1sqin so you have a pressure
That pressure is then transferred down the pipes (in an enclosed system).
(For this example and simplicity we will only consider one slave cylinder and piston as pressure in an enclosed system transfers to all surfaces anyway, it doesn't matter apart from if you were calculating flow rates, which is fairly negligible in a car braking system anyway, unless you have a leak.)
So now at the slave cylinder you have a pressure of 400lb/sqin, if the piston has an area of say 4sqin then you have a total force of 1600lb.
That force is then transferred to the back of the pad, and as the back of the pad is not part of the hydraulic system the force is applied simply as a force. If the pad has an area on the disc of say 8sqin then divide the force by the area to get pressure, so the pressure is 200lb/sqin.
8 x 200 = 1600
But even if you reduce the area of the pad on the disc to 4sqin the total force will still be at 1600lb, so the pressure will be 400lb/sqin.
4 x 400 = 1600
So no loss involved.
(Did I mention that I worked at Mintex for 8 years as an engineer on the hydraulic systems, it helps.)