Areas and Perimeters!

The diagonals of rectangle $OACD$ are congruent $\implies \overline{AD} \cong \overline{OC} = r \implies$

$P = r + (r - w) + r - (r + 2 - w) + \dfrac{\pi r}{2} = \dfrac{(4 + \pi)r - 4}{2}$

Using point $C$ above we have: $(8 - w)^2 + w^2 = r^2 \implies$

$2w^2 - 16w + 64 - r^2 = 0 \implies$ $w = \dfrac{8 \pm \sqrt{2(r^2 - 32)}}{2}$

Let $A_{R}$ be the red shaded region.

Using either value of $w \implies$

$A_{R} = \dfrac{1}{4}(\pi r^2 - \dfrac{1}{2}(8 - \sqrt{2(r^2 - 32})(8 + \sqrt{2(r^2 - 32)}) =$

$\dfrac{1}{4}(\pi r^2 - (64 - r^2)) = \dfrac{1}{4}((\pi + 1)r^2 - 64) = 9\pi - 7 \implies$

$\implies (\pi + 1)r^2 = 36(\pi + 1) \implies r = 6 \implies P = \dfrac{20 + 6\pi}{2} = 10 + 3\pi =$

$\alpha + \beta\pi \implies \alpha + \beta = \boxed{13}$ .

Let $AO=a$ and $OD=b$ . Then $a+b=8$ and the area of the red region is $\dfrac {\pi r^2}4 - \dfrac {ab}2 = 9\pi - 7$ . Assuming $\dfrac {\pi r^2}4 = 9 \pi$ , $\implies r = 6$ and $\dfrac {ab}2 = 7 \implies ab = 14$ . And the perimeter of the red region:

$\begin{aligned} P & = BA + AD + DE + \overbrace{BCE}^{\rm arc} \\ & = (r-a) + \sqrt{a^2+b^2} + (r-b) + \frac {2\pi r}4 \\ & = 2r - (a+b) + \sqrt{(a+b)^2 -2ab} + \frac {\pi r}2 \\ & = 2 \cdot 6 - 8 + \sqrt{8^2-2\cdot 14} + \frac {6\pi}2 \\ & = 10 + 3\pi \end{aligned}$

Therefore $\alpha + \beta = 10 + 3 = \boxed{13}$ .

Let $x$ be the width of the rectangle , $r$ be the radius of the circle.

The area of the red shaded region :

$9\pi - 7 =\cfrac{\pi r^2}{4} - \cfrac{x (8-x)}{2}$

$36\pi - 28 = \pi r^2 - 2 x (8-x)$

$(36-r^2) \pi = 2x^2 - 16 x + 28 \hspace{15mm} [1]$

Radius:

$r^2 = x^2 + (8-x)^2 = 2x^2 -16x + 64$

$r^2 - 36 = 2x^2 -16x + 28 \hspace{20mm} [2]$

By [1] and [2], $r^2 = 36 \implies r = 6$

Perimeter $P = (2r - 8) + r + \cfrac{2\pi r}{4} = 3r - 8 + \cfrac{\pi r}{2} = 18 - 8 + \cfrac{\pi 6}{2} = 10 + 3\pi$

$\therefore \alpha + \beta = \boxed{13}$