Quadrilateral area quest

Since $\angle A = 60^\circ$ , $\angle B = 90^\circ$ , and $\angle C = 120^\circ$ , $\implies \angle D = 360^\circ - 60^\circ - 90^\circ - 120^\circ = 90^\circ$ . Note that opposite angles $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$ . This means that $\implies ABCD$ is a cyclic quadrilateral . Let the center of the circumcircle be $O$ . Then $\angle DOB = 2 \angle A = 120^\circ$ . Let the circumradius be $r$ . Then $r = \dfrac {BD}2 \csc \dfrac {\angle DOB}2 = \dfrac 32 \times \dfrac 2{\sqrt 3} = \sqrt 3$ . By sine rule ,

$\begin{cases} \dfrac {\sin \angle BOM}{MB} = \dfrac {\sin \angle BMC}{OB} \\ \dfrac {\sin \angle DOM}{MD} = \dfrac {\sin \angle DMC}{OD} = \dfrac {\sin \angle BMC}{OB} \end{cases} \implies \dfrac {\sin \angle BOM}{MB} = \dfrac {\sin \angle DOM}{MD}$

Let $\angle BOM = \theta$ . Then

$\begin{aligned} \frac {\sin \theta}1 & = \frac {\sin (120^\circ - \theta)}2 \\ 2 \sin \theta & = \frac {\sqrt 3}2 \cos \theta + \frac 12 \sin \theta \\ \tan \theta & = \frac 1{\sqrt 3} \\ \implies \theta & = 30^\circ \end{aligned}$

Then the area of quadrilateral $ABCD$ :

$[ABCD] = \frac 12 AC \cdot BN + \frac 12 AC \cdot OD = \frac {AC}2 (BN + OD) = \frac {2\sqrt 3}2(r \sin 30^\circ + r) = \frac 92 = \boxed{4.5}$