A Geometry Problem.

Since $A_{\triangle DBA} = x^2 - 1$ , $\implies AB = 2(x-1)$ so that $A_{\triangle DBA} = \dfrac 12 AB \cdot DE = \dfrac 12 \cdot 2(x-1)(x+1) = x^2 -1$ . The $BD = \sqrt{AB^2+AD^2} = \sqrt{(2(x-1))^2 + x^2} = \sqrt{5x^2 - 8x + 4}$ . Note that $\triangle BDE$ and $\triangle BCD$ are similar, then we have:

$\begin{aligned} \frac {BD}{DE} & = \frac {CD}{BD} \\ \frac {\sqrt{5x^2-8x+4}}{x+1} & = \frac {2x+1}{\sqrt{5x^2-8x+4}} \\ 5x^2-8x+4 & = 2x^2 + 3x + 1 \\ 3x^2 - 11x + 3 & = 0 \\ \implies x & = \begin{cases} \dfrac {11+\sqrt{85}}6 \approx 3.369924076 \\ \dfrac {11-\sqrt{85}}6 \approx 0.29674259 \end{cases} \end{aligned}$

From $A_{\triangle DBE} = x^2 -1 \implies x > 1$ , therefore $x \approx \boxed{3.37}$ .

Since ${{A}_{\vartriangle ADE}}={{x}^{2}}-1$ , it follows that $x>1$ .

Referring to the figure we have

$\dfrac{y}{x}=\tan \left( 2\alpha +\beta \right)=\dfrac{\tan \alpha +\tan \left( \alpha +\beta \right)}{1-\tan \alpha \cdot \tan \left( \alpha +\beta \right)}=\dfrac{\dfrac{y}{3x+1}+\dfrac{y}{2x+1}}{1-\dfrac{{{y}^{2}}}{\left( 3x+1 \right)\left( 2x+1 \right)}}$

Hence,

$\dfrac{1}{x}=\dfrac{\dfrac{1}{3x+1}+\dfrac{1}{2x+1}}{1-\dfrac{{{y}^{2}}}{\left( 3x+1 \right)\left( 2x+1 \right)}}\Leftrightarrow {{y}^{2}}={{x}^{2}}+3x+1\Rightarrow y=\sqrt{{{x}^{2}}+3x+1}$ Now we have an equation for $x$ :

$\begin{aligned} & {{A}_{\vartriangle ADE}}={{x}^{2}}-1\Leftrightarrow \dfrac{1}{2}y\cdot \left( x+1 \right)={{x}^{2}}-1 \\ & \Leftrightarrow \sqrt{{{x}^{2}}+3x+1}\cdot \left( x+1 \right)=2\left( x-1 \right)\left( x+1 \right) \\ & \Leftrightarrow \sqrt{{{x}^{2}}+3x+1}=2\left( x-1 \right) \\ & \Rightarrow {{x}^{2}}+3x+1=4{{x}^{2}}-8x+4 \\ & \Leftrightarrow 3{{x}^{2}}-11x+3=0 \\ & \overset{x>1}{\mathop{\Leftrightarrow }}\,x=\dfrac{11+\sqrt{85}}{6}\approx \boxed{3.3699} \\ \end{aligned}$

$m = 180 - (\alpha + \beta) = 180 - \lambda \implies \lambda = \alpha + \beta$

$\implies$ $\tan(\lambda) = \dfrac{h}{x} = \tan(\alpha + \beta)$ , where $\tan(\beta) = \dfrac{h}{2x + 1}$ and $\tan(\alpha) = \dfrac{h}{3x + 1}$

$\implies$ $\dfrac{h}{x} = \tan(\alpha + \beta) = \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} =$ $\dfrac{\dfrac{h}{3x + 1} + \dfrac{h}{2x + 1}}{1 - \dfrac{h^2}{(3x + 1)(2x + 1)}} = (\dfrac{5x + 2}{6x^2 + 5x + 1 - h^2})h$

$\implies \dfrac{1}{x} = \dfrac{5x + 2}{6x^2 + 5x + 1 - h^2} \implies 6x^2 + 5x + 1 - h^2 = 5x^2 + 2x$

$\implies h^2 = x^2 + 3x + 1 \implies h = \sqrt{x^2 + 3x + 1}$

$\implies A_{\triangle{DBE}} = \dfrac{1}{2}(x + 1)(\sqrt{x^2 + 3x + 1}) = x^2 - 1 = (x - 1)(x + 1) \implies$

$\sqrt{x^2 + 3x + 1} = 2(x - 1) \implies x^2 + 3x + 1 = 4x^2 - 8x + 4 \implies$

$3x^2 - 11x + 3 = 0 \implies x = \dfrac{11 \pm \sqrt{85}}{6}$ and $x > 1 \implies x = \dfrac{11 + \sqrt{85}}{6}$ $\approx \boxed{3.369924}$ .