In a circle with center $C$ and radius $6\sqrt{2}$ cm, $PQ$ and $SR$ are two parallel chords separated by one of the diameters. If $\angle PQC = 45^\circ$, and the ratio of the perpendicular distance of $PQ$ and $SR$ from $C$ is $3:2$, then the area, in sq. cm, of the quadrilateral $PQRS$ is

Draw a circle with center $C$. Since $PQ$ and $SR$ are separated by a diameter, $C$ lies between the two chords, with $PQ$ on one side and $SR$ on the other.

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In triangle $PQC$, both $CP$ and $CQ$ are radii, so:

$CP = CQ = 6\sqrt{2}$ cm

Since the triangle is isosceles, the base angles are equal:

$\angle QPC = \angle PQC = 45^\circ$

$\angle PCQ = 180^\circ - 45^\circ - 45^\circ = 90^\circ$

So triangle $PQC$ is a right-angled isosceles triangle with the right angle at $C$.

$PQ = \sqrt{(6\sqrt{2})^2 + (6\sqrt{2})^2}$

$= \sqrt{72 + 72}$

$= \sqrt{144}$

$= 12$ cm

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Drop a perpendicular from $C$ to $PQ$, meeting it at $M$. The perpendicular from the center always bisects a chord, so $M$ is the midpoint of $PQ$.

In right triangle $CMQ$, with $\angle MQC = 45^\circ$ and hypotenuse $CQ = 6\sqrt{2}$:

$CM = CQ \cdot \sin 45^\circ$

$= 6\sqrt{2} \times \dfrac{1}{\sqrt{2}}$

$= 6$ cm

So the perpendicular distance from $C$ to $PQ$ is $6$ cm.

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Let the perpendicular distance from $C$ to $SR$ be $d_2$. The ratio of perpendicular distances from $C$ to $PQ$ and $SR$ is $3:2$, so:

$\dfrac{6}{d_2} = \dfrac{3}{2}$

$d_2 = 4$ cm

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Using the relation between the perpendicular distance from center and the half-chord length, with radius $r = 6\sqrt{2}$ and distance $d_2 = 4$:

Half of $SR = \sqrt{r^2 - d_2^2}$

$= \sqrt{(6\sqrt{2})^2 - 4^2}$

$= \sqrt{72 - 16}$

$= \sqrt{56}$

$= 2\sqrt{14}$

$SR = 4\sqrt{14}$ cm

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Since $PQ$ and $SR$ are parallel chords on opposite sides of the center, the perpendicular distance between them is:

$h = 6 + 4 = 10$ cm

The quadrilateral $PQRS$ is a trapezium with parallel sides $PQ = 12$ and $SR = 4\sqrt{14}$, and height $h = 10$.

Area $= \dfrac{1}{2} \times (PQ + SR) \times h$

$= \dfrac{1}{2} \times (12 + 4\sqrt{14}) \times 10$

$= 5 \times (12 + 4\sqrt{14})$

$= 60 + 20\sqrt{14}$

The area of quadrilateral $PQRS$ is $60 + 20\sqrt{14}$ sq. cm.