⚡ AZUCATION SHORTCUT:
In $\triangle C M Q$, $CM = CQ \sin(45^\circ) = 6\sqrt{2} \times \frac{1}{\sqrt{2}} = 6$.
Ratio $3:2 \implies$ distances are $6$ and $4$.
Lengths of parallel sides: $PQ = 2(6) = 12$; $SR = 2\sqrt{(6\sqrt{2})^2 - 4^2} = 2\sqrt{56} = 4\sqrt{14}$.
Area = $\frac{1}{2}(12 + 4\sqrt{14}) \times (6 + 4) = 5(12 + 4\sqrt{14}) = \mathbf{20(3 + \sqrt{14})}$.

Step 1: Distance of PQ from center
In $\triangle CMQ$ (where $M$ is the midpoint of $PQ$), $CQ$ is the radius $= 6\sqrt{2}$.
Since $\angle PQC = 45^\circ$ and $\triangle CMQ$ is right-angled at $M$, then $\angle MCQ = 45^\circ$.
$CM = CQ \cos(45^\circ) = 6\sqrt{2} \times \frac{1}{\sqrt{2}} = 6$ cm.
Also, $MQ = 6$ cm, so length $PQ = 2 \times 6 = 12$ cm.

Step 2: Distance of SR and its length
Ratio of distances is $3:2$. Since distance of $PQ = 6$, then distance of $SR = \frac{2}{3} \times 6 = 4$ cm.
Let $N$ be the midpoint of $SR$. In right $\triangle CNS$:
$NS^2 = CS^2 - CN^2 = (6\sqrt{2})^2 - 4^2 = 72 - 16 = 56$.
$NS = \sqrt{56} = 2\sqrt{14}$. So, $SR = 2 \times 2\sqrt{14} = 4\sqrt{14}$ cm.

Step 3: Calculate Area of Trapezium PQRS
The height $h$ is the sum of distances (as they are on opposite sides of a diameter):
$h = 6 + 4 = 10$ cm.
Area $= \frac{1}{2} \times (PQ + SR) \times h = \frac{1}{2} \times (12 + 4\sqrt{14}) \times 10$
Area $= 5 \times 4(3 + \sqrt{14}) = \mathbf{20(3 + \sqrt{14})}$.

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