Find the area...

$\text{The area of the rectangle is 6*12=72.}\\ \text{Areas of triangles FDA and EAB each is }\dfrac{72} 4. \\ \text{So the area of }FAECF=72-\dfrac{72} 4-\dfrac{72} 4=\dfrac{72} 2=36.\\ So ~ area ~ of~\color{#D61F06}{\triangle ~FAE}=~ area ~ of ~ FAECF~-~area ~ of ~\triangle~ CFE.\\ =36~- ~\dfrac 1 2 *3*6=\color{#D61F06}{27}.\\ \text{Property of a rectangle Q and T trisects the diagonal DB.}\\ \therefore QT=\dfrac 1 3 *DB.\\ \therefore ~in~\triangle~CDB,~~~FE=\dfrac 1 2 *DB.\\ \therefore ~\color{#EC7300}{\dfrac{QT}{FE}=\dfrac 2 3 .}\\ In ~ \triangle~TDA, ~Q~is~midpoint, ~ QR ||DA.\\ \therefore~ R ~is~ midpoint~of~ AT.\\ Areas~\triangle~ QRT~=\dfrac 1 2 *\triangle~QAT \\$ $=\dfrac 1 2 * \left ( \color{#EC7300}{ \dfrac{QT}{FE} }\right )^2* \color{#D61F06}{\triangle ~FAE}\\ =\dfrac 1 2 *\left(\dfrac2 3 \right)^2*27\\ = \Large \color{#3D99F6}{6}$ $Proof~~of~~Trisection$ $\text{Draw the other diagonal CA to bisect diagonal DB at at O. }\\ \text{Let G be the midpoint of AB. Draw GC}\\ \implies~in~\triangle ~CAB,~BO,~CG,~AE, ~are ~medians ~co-incident ~at~T.\\ So~BT=\dfrac 2 3 * BO=\dfrac 1 3 * BD.\\ \implies~T trisects~ the ~diagonal.$

Use coordinates geometry, with A(0,0).The equation of lines AF : y=x, BD:y=-0.5x+6,AE:y=0.25x, Then we will find the coordinates of intersection points Q(4,4),R(4,1), and T(8,2). The area of the triangle QRT = (1/2)(4)(3) = 6.

$\triangle ETB$ and $\triangle ATD$ are similar trianlges. So, $\displaystyle\frac{ET}{AT}=\frac{BT}{DT}=\frac{EB}{AD}=\frac{1}{2}$

$\displaystyle \implies a=ET=\frac{AE}{3}=\frac{\sqrt{12^2+3^2}}{3}=\sqrt{17}$ ,

$\displaystyle b=BT=\frac{BD}{3}=\frac{\sqrt{12^2+6^2}}{3}=2\sqrt{5}$

and $c=EB=3$

$\begin{aligned}\triangle ETB&=\sqrt{s(s-a)(s-b)(s-c)}\quad\quad \color{#3D99F6} \Big[\text{where } s=\frac{a+b+c}{2}\Big]\\ &=\frac{1}{4}\sqrt{2a^2 b^2+2b^2 c^2+2c^2 a^2 -a^4-b^4-c^4}\\ &=\frac{1}{4}\sqrt{576}=6\end{aligned}$

Now, $\triangle AQB$ and $\triangle FQD$ are also similar triangles.

$\implies \displaystyle\frac{DQ}{BQ}=\frac{DF}{AB}=\frac{1}{2} \quad \implies TQ=BT=DQ=\frac{BD}{3}$

But $PS$ and $CB$ are parallel straight lines. So, $\angle TQR=\angle TBE$ and $\angle TRQ=\angle TEB$ .

Then, $\triangle ETB \cong \triangle QRT \quad \implies \boxed{\triangle QRT=6}$