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Consider one fourth part of the circle.

$\tan~\phi=\dfrac{3}{6}=\dfrac{1}{2}=0.5$ $\implies$ $\phi=\tan^{-1}0.5$

$\theta=45-\phi=45-\tan^{-1}0.5$

By pythagorean theorem, $NL=6\sqrt{2}$

$\tan~\theta=\dfrac{x}{\frac{NL}{2}}=\dfrac{x}{3\sqrt{2}}$ $\implies$ $x=3\sqrt{2}\tan(45-\tan^{-1}0.5)=3\sqrt{2}\left(\dfrac{1}{3}\right)=\sqrt{2}$

Let $A_G$ be the area of the green region, then $A_G=\dfrac{1}{2}\sqrt{2}(6\sqrt{2})=6$

Let $A_Y$ be the area of yellow region, then $A_Y=\dfrac{1}{4} \pi(6^2)-\dfrac{1}{2}(6)(6)=9 \pi-18$

Let $A_R$ be the area of the red region, then $A_R=\dfrac{1}{4} \pi(6^2)-(9 \pi-18)-6=12$

Finally, the area of the whole red region is $4(12)=$ $\color{#D61F06}\large \boxed{48}$

Since the symmetry, $[KOM]=[MOL]=[LON]=[NOK]$ So let's consider the $\triangle KOM$ . The $MA, KB$ are medians. If we draw the third median, then we divided the triangle to six equal parts. From these six parts, exactly four is red, i.e. part of the red area. We know that $[KOM]=\dfrac{6*6}{2}=18$ If the intersection of $MA$ and $KB$ is $Q$ , then $[KQMO]=\dfrac{4}{6}[KOM]=12=[MOL]=[LON]=[NOK]$ Therefore $\text{red area}=4*12=\boxed{48}$

Let $AN$ meet $KD$ at $P$ , $PQ$ be the perpendicular from $P$ to $KO$ , and $PQ=h$ . We note that $\triangle KDO$ , $\triangle KPQ$ and $\triangle PAQ$ are similar. Therefore,

$\begin{aligned} \frac {KQ}{\color{#3D99F6}PQ} & = \frac {KO}{DO} = \frac 63 = 2 & \small \color{#3D99F6} \text{Note that }PQ = h \\ \frac {KQ}{\color{#3D99F6}h} & = 2 \\ \implies KQ & = 2h \end{aligned}$

Also that

$\begin{aligned} \frac {AQ}{\color{#3D99F6}PQ} & = \frac {DO}{KO} = \frac 36 = \frac 12 & \small \color{#3D99F6} \text{Note that }PQ = h \\ \frac {AQ}{\color{#3D99F6}h} & = \frac 12 \\ \implies AQ & = \frac h2 \end{aligned}$

Note that

$\begin{aligned} KQ & = {\color{#3D99F6}KA} + AQ & \small \color{#3D99F6} \text{Note that }KA = 3 \\ 2h & = {\color{#3D99F6}3} + \frac h2 \\ \implies h & = 2 \end{aligned}$

We note that the area of the red polygon $A_{\color{#D61F06}red} = 8 [KPO] = 8 \times \dfrac 12 \times 6 \times h = 8 \times \dfrac 12 \times 6 \times 2 = \boxed{48}$ .

[APK]=[APO]=[OPD] = 1/3 (3 x 6/2) = 3. So the red area = 6(12)/2 + 4(3) = 48