Here you go, Piston area and force.

dusterbing the peace said
The 36% increase from @71340 is correct:
(19.63-12.56)/19.63 = 0.36

12.56/(1-.36) = 19.63

Let’s say that the tunes were matched somehow in the two different bores and everything else remained unchanged:
1,000 lbs/si * 19.63 si = 19,630 lbs force
1,000 lbs/si * 12.56 si = 12,560 lbs force
This is still a 36% increase in force on the piston:
(19,630-12,560)/19,630 = 0.36

The trick is to get the same, if not greater, explosion force out of the larger bore. This is somewhat easier to do when boring an engine for a rebuild, however the net change is far less…4.00 to 4.03 is only 1% increase (when rounding up).

If the premise is increase from the initial piston, then the reference area is 12.57 in sq.
The increase in area is 7.07 in sq.

So when phrased that way, 5" piston has 7.07" square inches more than the 4" piston.
Same as saying 56 % increase in area.
How much is a 56% increase in the area of a 4" piston?
0.56 x 12.57 in sq. = 7.04 in sq

Close enough considering I was using 22/7 for PI.

So it the percentage depends on what the comparison.
If I bought eggs for a 1.00 last week and this week I said the price went up 50%, how much did eggs cost this week?