It's All Areas!

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

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

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

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

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

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

I made it a littble bit different than @Rocco Dalto .

1. calculate the functions of the parabolas

You can make your work easy, because you see, that the functions must be axially symmetrical. So both parabolas must be of the form $ax^2+c$ . Then use the fact that the upper parabola is tangent to A (or C for the lower parabola), to get the second information. Now you can calculate the functions.

2. calculate the area under the parabolas

You can calculate the area between the parabola and the x-axis with the fundamental theorem of analysis . But pay attention! You are dealing with the area, not the signed area (which is the result of an integral).

3. calculate the red and blue area

Because of the axially symmetry, the right and left areas are equal. (So I just calculated the right side and doubled the result).

In order to get the red area, you have to subtract the area of the sircle (in the given intervall) from the result of the integrals in step 2. Now I make an example for the upper right circle area: The area between the line OA and the x-axis is a triangle (easy to calculate; just $\frac{1}{2}$ times base times hight). And you can see, that the angle at O must be 30 degree. Thus the remaining area of this quater circle (in this intervall) must be $\frac{1}{6}$ of the total circle area (because of the 60 degree).

finish

Now just add the areas and bring it into the given form ;)