This isn't really a Mathematica problem. It is a Euclidean geometry problem and can be solve by a little classic geometry reasoning. Like ubpdqn I will work in the 1st quadrant and invoke symmetry.
By construction
$\qquad$OC = OE = R
$\qquad$OB = OD = DA = BA = R/2
By observation
$\qquad$Quadrant perimeter = EA + AO + OA + BA + EC
$\qquad$OBAD is a square
Arcs EA + OA and AO, + BA are one half the circumference of the equal circles centered at B and D, which have diameters R/2, so EA + OA + AO, + BA = circumference of an inner circle = π R. EC is one quarter of the circumference of the outer circle, so EC = (2 π R)/4 = π R/2. The quadrant perimeter is therefore π R + π R/2 = 3 π R/2.
It follows that the full perimeter, 4 x (quadrant perimeter), is 6 π R.
Point A is one of the points where the inner circles intersect and it clearly lies at {R/2, R/2}. By symmetry, the four points of intersection are
$\qquad${{R/2, R/2}, {-R/2, R/2}, {-R/2, -R/2}, {R/2, -R/2}}.
Finding the area is a little more complicated, but not much.
The area, a1, between the two arcs ending at points O and A is clearly twice the difference of the area between the arc OA and the dashed line OA. This in turn is the area of a quadrant of inner circle centered at B less the half the square OBAD. Thus,
$\qquad$a1 = 2 ((π (R/2)^2)/4 - ((R/2)^2)/2)= = 1/8 (π -2 + π 2) R^2
The area, a2, bordered by the arcs EC, EA and AC is the area of the quadrant less the area of 2 quadrants of an inner circle less the area of the square OBAD. This is given by
$\qquad$a2 = (π R^2R^2)/4 - (π (R/2)^2)/2 - (R/2)^2 = 1/8 (π -2 + π 2) R^2
Note that a1 = a2 (which I find an interesting result in itself). Therefore, the full area is
$\qquad$4 (2 a1) = (π -2 + π 2) R^2