Find the area of the yellow region...

We note the areas $[CDB]=[CFD]=7$ . Since $\triangle CDB$ and $\triangle CFD$ are sharing the same height $h_1$ , this means that $BD=DF$ . Similarly, $\triangle BED$ and $\triangle BDC$ are sharing the same height $h_2$ . This implies that $\dfrac {ED}{DC} = \dfrac {[BED]}{[BDC]} = \dfrac 37$ .

Now divide the yellow region into $x=[AED]$ and $y=[ADF]$ as shown. Note that $\triangle ABD$ and $\triangle ADF$ are sharing the same height $h_3$ . Therefore,

$\begin{aligned} \frac {[ABD]}{[ADF]} & = \frac {BD}{DF} & \small \color{#3D99F6} \text{Note that }BD=DF \\ \frac {3+x}y & = 1 \\ \implies y & = x+3 \end{aligned}$

Similarly, $\triangle AED$ and $\triangle ADC$ are sharing the same height $h_4$ .

$\begin{aligned} \frac {[AED]}{[ADC]} & = \frac {ED}{DC} \\ \frac x{7+y} & = \frac 37 \\ 7x & = 21+3\color{#3D99F6}y & \small \color{#3D99F6} \text{Note that }y = x + 3 \\ 7x & = 21 + \color{#3D99F6}3x + 9 \\ \implies x & = 7.5 \\ \implies y & = 7.5+3 = 10.5 \end{aligned}$

Therefore the area of the yellow region $x+y = 7.5+10.5 = \boxed{18}$ .

This can be solved by proving the Ceva's Theorem

$\dfrac{3+7}{3}=\dfrac{a+b+7}{a}$ $\implies$ $\dfrac{10}{3}=\dfrac{a+b+7}{a}$ $\implies$ $7a=3b+21$ $\color{#D61F06}(1)$

$\dfrac{a+b+3}{b}=\dfrac{7+7}{7}$ $\implies$ $\dfrac{a+b+3}{b}=2$ $\implies$ $a=b-3$ $\color{#D61F06}(2)$

Solving $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ , we get

$a=10.5$ and $b=7.5$ .

The desired area is $7.5+10.5=\boxed{18}$ .