如图所示，POR 是一个半径为 6 cm 的圆的四分之一扇形。OR 与 PQ 平行。QR 是以 P 为圆心的一段圆弧。

1. Identify the angle using the right-angled triangle and parallel lines

Since $OP=OR=6$ and $\angle POR=90^\circ$, triangle $OPR$ is an isosceles right-angled triangle.

Hence
$\angle OPR=45^\circ.$

Since $PQ \parallel OR$, the angle between $PQ$ and $PR$ is equal to the angle between $OR$ and $RP$.

Therefore
$\angle QPR=45^\circ.$

2. Convert the angle from degrees to radians

$45^\circ=45\times\frac{\pi}{180}=\frac{\pi}{4}\text{ radians}.$

1. Find the radius and angle of sector $PQR$

Since $QR$ is an arc with centre $P$, the radius of sector $PQR$ is
$PQ=PR.$
In right-angled triangle $OPR$,
$PR=\sqrt{6^2+6^2}=6\sqrt2.$
From part (a),
$\angle QPR=\frac{\pi}{4}.$

2. Calculate the area of sector $PQR$

Area of sector $PQR$:
$\frac12 r^2\theta=\frac12(6\sqrt2)^2\left(\frac{\pi}{4}\right)$
$=\frac12(72)\left(\frac{\pi}{4}\right)=9\pi.$

3. Calculate the area of the circular segment between chord $PR$ and the quadrant arc

Area of quadrant $OPR$:
$\frac14\pi(6)^2=9\pi.$
Area of triangle $OPR$:
$\frac12(6)(6)=18.$
So the area of the circular segment between chord $PR$ and arc $PR$ is
$9\pi-18.$

4. Add the two required parts of the shaded region

Shaded area
$=\text{area of sector }PQR+\text{area of segment between }PR\text{ and the quadrant arc}$
$=9\pi+(9\pi-18)$
$=18\pi-18$
$=38.5\text{ cm}^2\quad\text{(to 3 s.f.)}.$