Green area

$\tan~w=\dfrac{3}{4}$ $\implies$ $w=36.86989765^\circ$

$x=90-w=53.13010235^\circ$

$\tan~z=\dfrac{4}{9}$ $\implies$ $z=23.96248897^\circ$

$y=180-x-z=102.9074087^\circ$

By sine law ,

$\dfrac{9}{\sin~y}=\dfrac{QR}{\sin~x}$ $\implies$ $QR=7.386643352$

Let $A_Y$ be the area of the yellow region, then $A_Y=2\left(\dfrac{1}{2}\right)(9)(7.386643352)(\sin~z)=27$

Let $A_B$ be the area of the blue region, then $A_B=4(3)=12$

Let $A_G$ be the area of the green region, then $A_G=48-27-12=$ $\color{#20A900}\boxed{\large 9}$

Note: In my computation I used the approximate values of the computed angles since they are not whole numbers. But It will arrive in the correct answer since it is accurate to $8$ decimal places.

Since $AP=CQ$ and $AP\mid\mid CQ$ , the $APCQ$ quadrilateral is a parallelogram, $[APCQ]=AP*BC=3*\dfrac{48}{12}=12$ It is clear that $\triangle PBF\sim\triangle FCQ$ , and the similarity ratio is $PB:QC=3:1$ . From that $FY:FX=3:1$ (where $FY$ and $FX$ are altitudes). Since $XY=BC=4$ , $FX=1$ , $[QCF]=\dfrac{3*1}{2}=1.5$ Since $\triangle APE\cong \triangle CQF$ , $[PFQE]=[APCQ]-[APE]-[QCF]=[APCQ]-2*[QCF]=12-2*1.5=\boxed{9}$

We know that $CB$ is equal to $4$ and that $\triangle MAP$ is semelhant to $\triangle PNB$ the ratio of similiarity is $\frac{1}{3}$ also we know that their heights sum up to $4$ now we can conclude $\triangle PNB$ height is 3, so we have $ABCD - (2 \times \triangle PNB + 2 \times \triangle DAP)$ that is equal to $9$

Use coordinate geometry.Let point A(0,0), and the intersection of line AQ and DP is R.Point D(0,4), P(3,0), and Q(9,4). So line DP has equation 4x+3y=12, and AQ: y=(4/9) x.The intersection point R(9/4, 1). The shaded area =2( [DPQ] – [DRQ]) = 2(18 – 9(3/2)) =9.