Area Bonanza !

$m_{OA} = \dfrac{1}{\sqrt{3}} \implies m_{\perp} = -\sqrt{3}$

$m_{OB} = -\dfrac{1}{\sqrt{3}} \implies m^{*}_{\perp} = \sqrt{3}$

Let $y = ax^2 + bx + c \implies \dfrac{dy}{dx} = 2ax + b$

$\implies \dfrac{dy}{dx}|_{(x = \dfrac{\sqrt{3}}{2})} = \sqrt{3}a + b = -\sqrt{3}$

and

$\dfrac{dy}{dx}|_{(x = \dfrac{-\sqrt{3}}{2})} = -\sqrt{3}a + b = \sqrt{3}$

Solving the above system we obtain $a = -1$ and $b = 0$ and using $A(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}) \implies$

$\dfrac{1}{2} = -1(\dfrac{\sqrt{3}}{2})^2 + c \implies c = \dfrac{5}{4} \implies$ $y = \dfrac{5}{4} - x^2$

and for the portion of the circle we have $y = \sqrt{1 - x^2}$

$\implies A = 2\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} ((\dfrac{5}{4} - x^2) - \sqrt{1 - x^2}) dx$

$= 2(\dfrac{\sqrt{3}}{2} - \displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} \sqrt{1 - x^2} dx)$

Letting $x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies$

$\displaystyle\int_{0}^{\dfrac{\sqrt{3}}{2}} \sqrt{1 - x^2} dx = \displaystyle\int_{0}^{\dfrac{\pi}{3}} (1 + \cos(2\theta)) d\theta =$ $\dfrac{1}{2}(\theta + \dfrac{1}{2}\sin(2\theta))|_{0}^{\dfrac{\pi}{3}} =$

$\dfrac{1}{2}(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4})$

$\implies A_{1} = \sqrt{3} - (\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4}) = \boxed{\dfrac{3\sqrt{3}}{4} - \dfrac{\pi}{3}}$

$m_{OD} = 1 \implies m_\perp = -1$

$m_{OC} = -1 \implies m^*_\perp = 1$

Let $y = a^{*} x^2 + b^{*} x + c \implies \dfrac{dy}{dx} = 2a^{*} x + b^{*}$

$\dfrac{dy}{dx}|_{(x = -\dfrac{1}{\sqrt{2}}} = -\sqrt{2}a^{*} + b^{*} = -1$

and

$\dfrac{dy}{dx}|_{(x = \dfrac{1}{\sqrt{2}}} = \sqrt{2}a^{*} + b^{*} = 1$

Solving the above system we obtain:

$a^{*} = \dfrac{1}{\sqrt{2}}$ and $b^{*} = 0$ and using $(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}})$ we obtain $c^{*} = \dfrac{-3}{2\sqrt{2}}$

$\implies y = \dfrac{1}{\sqrt{2}}x^2 - \dfrac{3}{2\sqrt{2}}$ and for the portion of the circle we have $y = -\sqrt{1 - x^2}$ .

$\implies A_{2} = 2\displaystyle\int_{0}^{\dfrac{1}{\sqrt{2}}} (\dfrac{3}{2\sqrt{2}} - \dfrac{1}{\sqrt{2}}x^2 - \sqrt{1 - x^2}) dx$

Letting $x = \sin(\theta) \implies dx = \cos(\theta) d\theta \implies$ $A_{2} = 2(\dfrac{2}{3} - \dfrac{1}{2}\displaystyle\int_{0}^{\dfrac{\pi}{4}}$ $(1 + \cos(2\theta)) d\theta)$

$= 2(\dfrac{2}{3} - \dfrac{1}{2}(\dfrac{\pi}{4} + \dfrac{1}{2})) = \boxed{\dfrac{5}{6} - \dfrac{\pi}{4}}$

$\implies A = A_{1} + A_{2} = \dfrac{1}{12}(9\sqrt{3} + 10 - 7\pi) = \dfrac{1}{2^2 * 3}(3^2\sqrt{3} + 2 * 5 - 7\pi)$

$= \dfrac{1}{a^a * b}(b^2\sqrt{b} + a * c - d\pi) \implies a + b + c + d = \boxed{17}$