Xjyuk

They would be kind enough to give me some indications of using the disk command in any of these cases (calculation of area and perimeter of the scratched)

I have looked for something that tells me how to do it, but I can't find anything acceptable, or maybe I look bad.
<<<< any help is welcome

Here is an alternative of
the previous answer
that might give you the plots in your question (after a sufficient number of experiments.)

Clear[RandomDisk]
RandomDisk[] := Opacity[RandomChoice[Range[0, 1, 0.25]]], 
 FaceForm[RandomChoice[None, Pink, Gray, LightBlue]], 
 EdgeForm[Black], 
 Disk[RandomChoice[
 Append[Flatten[Outer[List, 0, 1, 0, 1], 1], 0.5, 0.5]], 
 RandomChoice[0.5, 1]];

Clear[RandomRectangle]
RandomRectangle[] := EdgeForm[
 RandomChoice[None, Black, Blue, Red, Gray, Orange, LightBlue]], 
 FaceForm[None], Rectangle[]

Multicolumn[
 Table[Graphics[Flatten[Table[RandomDisk[], RandomChoice[Range[4]]], 
 1], RandomRectangle[], Frame -> True, PlotRangeClipping -> True,
 PlotRange -> 0, 1, 0, 1], 16], 4]

You can specify the quarter disks using the three-argument form of Disk.

For the first picture:

a = 1;
d1 = Disk[0, 0, a, 0, Pi/2];
d2 = Disk[a, 0, a, Pi/2, Pi];
d3 = Disk[0, a, a, -Pi/2, 0];

Graphics[EdgeForm[Gray], Opacity[.25], Red, d1, Blue, d2 , Green, d3]

ri = RegionIntersection[d1, d2, d3];

Perimeter[ri]

2.61799

N @ Area[ri]

0.442972

A simpler alternative is to take the intersections of full disks with Rectangle[0,0,a,a]:

d1b=RegionIntersection[Rectangle[0,0, a,a], Disk[0,0, a]];
d2b=RegionIntersection[Rectangle[0,0, a,a], Disk[a,0, a]];
d3b=RegionIntersection[Rectangle[0,0, a,a], Disk[0,a, a]];

Graphics[EdgeForm[Gray],Opacity[.25], Red, d1b, Blue, d2b, Green, d3b]

same picture

Similarly, for the third picture:

Graphics[Opacity[.25], Blue, d2b, Red, Disk[a, a/2, a/2]]

rd = RegionDifference[ Disk[a, a/2, a/2], d2b];
Through[Perimeter, N@*Area@rd]

2.18282, 0.146381

Clear["Global`*"]

For the first image

reg[1, a_] = Disk[0, 0, a, 0, Pi/2];
reg[2, a_] = Disk[a, 0, a, Pi/2, Pi];
reg[3, a_] = Disk[0, a, a, -Pi/2, 0];
reg[4, a_] = RegionIntersection[reg[1, a], reg[2, a], reg[3, a]];

Show[
 Graphics[
 EdgeForm[Black],
 Lighter[Blue, 0.6],
 Opacity[0.75],
 reg[1, 1], reg[2, 1], reg[3, 1]],
 Region[reg[4, 1],
 BaseStyle -> Opacity[0.5, Blue]]]

EDIT: The gap at the lower-left corner can be filled by using DiscretizeRegion

Graphics[
 EdgeForm[Black],
 Lighter[Blue, 0.6],
 Opacity[0.75],
 reg[1, 1], reg[2, 1], reg[3, 1],
 DiscretizeRegion[reg[4, 1],
 MeshCellStyle -> Opacity[0.5, Blue],
 MaxCellMeasure -> 1]]

The area is proportional to a^2

And @@ Table[
 Area[reg[4, a]] == a^2*Area[reg[4, 1]],
 a, 1, 10]

(* True *)

area1 = a^2*Area[reg[4, 1]]

(* 1/12 a^2 (-6 Sqrt[3] + 5 π) *)

area1 // N

(* 0.442972 a^2 *)

Perimeter[reg[4, 1]]

(* 2.61799 *)

For the second image

reg[5, a_] = Disk[a, a, a, Pi, 3 Pi/2]; reg[6, a_] = 
 RegionUnion[
 BooleanRegion[#1 && #2 && ! #3 && ! #4 &, #] & /@
 reg[1, a], reg[2, a],
 reg[3, a], reg[5, a],
 reg[2, a], reg[5, a], reg[1, a], reg[3, a],
 reg[1, a], reg[3, a], reg[2, a], reg[5, a],
 reg[3, a], reg[5, a], reg[1, a], reg[2, a]];

Show[
 Graphics[
 EdgeForm[Black],
 White, Opacity[0.25],
 reg[1, 1], reg[2, 1], reg[3, 1], reg[5, 1]],
 Region[reg[6, 1], BaseStyle -> LightGray],
 Frame -> True]

The area is proportional to a^2

And @@ Table[
 Area[reg[6, a]] == a^2*Area[reg[6, 1]],
 a, 1, 10]

(* True *)

area2 = a^2*Area[reg[6, 1]] // Simplify

(* -(1/3) a^2 (3 (-4 + Sqrt[3]) + 2 π) *)

area2 // N

(* 0.173554 a^2 *)

Perimeter[reg[6, 1]]

(* 7.11792 *)

This number for the perimeter is too low since each of the four subregions has a perimeter that must exceed 2. Looking at 4 times the perimeter of a subregion

reg[6 sr, a_] = BooleanRegion[#1 && #2 && ! #3 && ! #4 &,
 reg[1, a], reg[2, a], reg[3, a], reg[5, a]];

4*Perimeter[reg[6 sr, 1]]

(* 8.18879 *)

For the last image

reg[7, a_] = Disk[a/2, a/2, a/2];

reg[8, a_] = BooleanRegion[#1 && ! #2 &, reg[7, a], reg[2, a]];

Show[
 Graphics[
 EdgeForm[Black],
 White, Opacity[0.25],
 Rectangle[0, 0],
 reg[2, 1], reg[7, 1]],
 Region[reg[8, 1], BaseStyle -> Red]]

The area is proportional to a^2

And @@ Table[
 Area[reg[8, a]] == a^2*Area[reg[8, 1]] // Simplify,
 a, 1, 10]

(* True *)

area3 = a^2*Area[reg[8, 1]] //
 TrigToExp // FullSimplify

(* 1/8 a^2 (Sqrt[7] + π - ArcCot[3/Sqrt[7]] - 4 ArcTan[(5 Sqrt[7])/9]) *)

area3 // N

(* 0.146381 a^2 *)

Perimeter[reg[8, 1]]

(* 2.18282 *)