A multi-concept problem

The convex quadrilateral is a trapezium. $\triangle AOB$ and $\triangle COD$ are similar.
So, $\dfrac{AB}{CD}=\dfrac{2}{3}\Rightarrow AB=\dfrac{2CD}{3}.$
Let the height of $\triangle COD=y.$ So, $9=\dfrac{1}{2}\times CD\times\Rightarrow y=\dfrac{18}{CD}.$
Let the height of $\triangle AOB=x.$ So, $4=\dfrac{1}{2}\times AB\times\Rightarrow x=\dfrac{8}{AB}\Rightarrow x=\dfrac{12}{CD}.$
Therefore the smallest area of the quadrilateral $=\dfrac{1}{2}\times (AB+CD)\times (x+y)\Rightarrow \dfrac{1}{2}\times\dfrac{5CD}{3}\times\dfrac{30}{CD}=\boxed{25}.$

Triangular Areas in Quadrilateral

Since $\triangle ABO$ and $\triangle CBO$ share a co-linear base AOC with $\triangle ADO$ and $\triangle CDO$ , we get ratio of areas (marked in blue) as $\frac{4}{y} = \frac{x}{9}$ . That is xy = 36

For a fixed product (36) and minimum sum, both x and y must be equal i.e. x = y = 6.

Thus the minimum area ABCD is $4 + 6 + 9 + 6 = \color{#D61F06} {25}$