How to draw a certain region and calculate its area and perimeter?

Show it:

RegionPlot[
 region = RegionUnion[
   Sequence @@ 
    RegionIntersection @@@ 
     Subsets[{Disk[{-1, 0}], Disk[{0, -1}], Disk[{1, 0}], 
       Disk[{0, 1}]}, {2}], 
   Fold[RegionDifference, {Disk[{0, 0}, 2], Disk[{-1, 0}], 
     Disk[{0, -1}], Disk[{1, 0}], Disk[{0, 1}]}]], Frame -> False]

Area

Area[region]

4 (-2 + π)

Perimeter

ArcLength@RegionBoundary[region]

12 π

Your question wants a relationship for r which I assume is the radius of the larger circle. You can get it like this:

c1 = ImplicitRegion[(x - r)^2 + y^2 <= r^2, {x, y}];
c2 = ImplicitRegion[x^2 + (y - r)^2 <= r^2, {x, y}];
Assuming[r > 0, Area[RegionIntersection[c1, c2]]]

yields

$\frac{1}{2} (\pi -2) r^2$

All the shaded areas in terms of r:

FullSimplify[\[Pi]*r^2 - 4 \[Pi] (r/2)^2 + 8 1/8 (-2 + \[Pi]) r^2]

$\left ( \pi -2\right )r^2$

Testing for the particular answer given by yode:

(-2 + \[Pi]) r^2 /. r -> 2

yields

$4\pi - 8$

Just for fun. By symmetry we need only consider the first quadrant.

Graphics[{EdgeForm[Black], FaceForm[None], Disk[{0, 0}, 2, {0, Pi/2}],
   Disk[{1, 0}, 1, {0, Pi}], Disk[{0, 1}, 1, {-Pi/2, Pi/2}], 
  Line[{{1, 0}, {1, 1}}], Line[{{0, 0}, {1, 1}}], 
  Line[{{0, 1}, {1, 1}}],
  Text[Style["A", 20], {0, 0}, {1, 1}],
  Text[Style["B", 20], {1, 0}, {1, 1}],
  Text[Style["C", 20], {1, 1}, {-2, -1}],
  Text[Style["D", 20], {2, 0}, {1, 1}],
  Text[Style["arc 1", 20, Background -> White], {0.7, 0.3}, {0, 0}],
  Text[Style["arc 2", 20, Background -> White], {0.3, 0.7}, {0, 0}],
  Text[Style["arc 3", 20, Background -> White], {1.4, 1.4}, {0, 0}],
  Text[Style["arc 4", 20, Background -> White], {0.8, 1.5}, {0, 0}],
  Text[Style["arc 5", 20, Background -> White], {1.5, 0.8}, {0, 0}]
  }]

Let the radius of the small circle be 1 (hence the radius of the large circle is 2).

So the perimeter can be seen to be length of arc1+ arc2+arc3+arc4+arc5:

Let $p_i$ represent arc i length. Now $p_1=p_2=p_4=p_5= \pi/2$ and arc length $p3= \pi/2 \times 2$. Hence total perimeter:

perimeter = 4 (4 Pi/2 + Pi/2 2)

i.e. $12\pi$

For the area: area bounded by arc1 and arc 2 is 2 x (area of sector-area of triangle ABC):

area1 = 2 (Pi/4 - 1/2)

The area bounded by arcs 3,4 and 5= area of quarter circle -area of 2 semicircles+ area of overlap:

area2 = (Pi/2) 2^2/2 - Pi + area1

Note area1=area2=$\pi/2-1$, so the total area is

total = Simplify[4 (area1 + area2)]

yielding:

4 (-2 + \[Pi])