Mdthbs

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 *)