Perimeters !

The diagonals of a rectangle are congruent $\implies$ in rectangle $ABDC$ we have $AD = r = a - 2$

$x + y = a \implies y = a - x \implies x^2 + (a - x)^2 = (a - 2)^2 \implies$

$2x^2 - 2ax + 4(a - 1) = 0 \implies x = \dfrac{a \pm \sqrt{a^2 - 8a + 8}}{2}$

$( x = \dfrac{a - \sqrt{a^2 - 8a + 8}}{2} \implies y = \dfrac{a + \sqrt{a^2 - 8a + 8}}{2})$

and

$(x = \dfrac{a + \sqrt{a^2 - 8a + 8}}{2} \implies y = \dfrac{a - \sqrt{a^2 - 8a + 8}}{2})$

For either pair $(x,y)$ we obtain the same perimeter since the square root part will cancel out

so choosing $( x = \dfrac{a - \sqrt{a^2 - 8a + 8}}{2}, y = \dfrac{a + \sqrt{a^2 - 8a + 8}}{2})$

$\implies$ The Perimeter $P = (a - 2) + [(a - 2) - \dfrac{1}{2}(a - \sqrt{a^2 - 8a + 8})] +$

$[(a - 2) - \dfrac{1}{2}(a + \sqrt{a^2 - 8a + 8})] + \dfrac{\pi}{2}(a - 2) =$

$2a - 6 + \dfrac{\pi}{2}(a - 2) = \dfrac{(4 + \pi)a - 12 - 2\pi}{2} = 10 + 3\pi \implies$

$(4 + \pi)a = 32 + 8\pi = 8(4 + \pi) \implies a = \boxed{8}$ .

We have $\dfrac{π(a-2)}{2}=3π\implies a=3\times 2+2=\boxed 8$