A Common tangent problem

Using the symmetry about the $y$ axis let ${ f(x) = -x^2 }$ and ${ g(x) = \sqrt{x} \implies }$

$\dfrac{d}{dx}(f(x))|_{x = a} = -2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = \dfrac{1}{2 \sqrt{b}} \implies$

$-2a = \dfrac{1}{2 \sqrt{b}} \implies a = -\dfrac{1}{4 \sqrt{b}}$

$B: (a,-a^2) = (-\dfrac{1}{4 \sqrt{b}} , -\dfrac{1}{16 b})$ and $A: (b, \sqrt{b} ) \implies$

slope $m = \dfrac{1}{2 \sqrt{b}} = \dfrac{1}{4 \sqrt{b}} (\dfrac{16 b^\frac{3}{2} + 1}{4 b^\frac{3}{2} + 1}) \implies$

$16 b^\frac{3}{2} + 1 = 8 b^\frac{3}{2} + 2 \implies 8 b^\frac{3}{2} = 1 \implies b = \dfrac{1}{4}$

$\implies$ slope $m = 1$ and $a = -\dfrac{1}{2}$

$$

Using $B: (-\dfrac{1}{2},-\frac{1}{4}) \implies y = x + \dfrac{1}{4}$

$y = x + \dfrac{1}{4}$ is tangent to $g(x)$ at $A: (\dfrac{1}{4} , \dfrac{1}{2})$ and $f(x)$ at $B: (-\dfrac{1}{2},-\dfrac{1}{4})$

Using the symmetry about the $y$ axis $B:(\dfrac{-1}{2},\dfrac{-1}{4}) \rightarrow E:(\dfrac{1}{2},\dfrac{-1}{4})$ and $A:(\dfrac{1}{4},\dfrac{1}{2}) \rightarrow D:(\dfrac{-1}{4},\dfrac{1}{2})$ .

Using points $E:(\dfrac{1}{2},\dfrac{-1}{4})$ and $D:(\dfrac{-1}{4},\dfrac{1}{2}) \implies y = \dfrac{1}{4} - x$ and $BE = 1, AD = \dfrac{1}{2}$ and the height of the trapezoid $DF = \dfrac{3}{4} \implies$ the area of the trapezoid $A_{BDAE} = \dfrac{1}{2}(\dfrac{3}{4})(\dfrac{3}{2}) = \dfrac{9}{16} = \boxed{0.5625}$ .