图中所示为一个面积为 18 cm$^2$ 的三角形 $PQR$。底边 $QR$ 的长度为 $(4\sqrt{2} + 8)$ cm。将线段 $RQ$ 延长至点 $S$，使得角 $PSR = 45^\circ$。

1. Set up the equation for the area of the triangle to find the perpendicular height $h$.

$\frac{1}{2} \times \text{base} \times \text{height} = \text{Area}$
$\frac{1}{2}(4\sqrt{2} + 8)h = 18$

2. Isolate $h$ and simplify the fraction.

$(2\sqrt{2} + 4)h = 18$
$h = \frac{18}{4 + 2\sqrt{2}} = \frac{9}{2 + \sqrt{2}}$

3. Rationalise the denominator by multiplying by the conjugate.

$h = \frac{9}{2 + \sqrt{2}} \times \frac{2 - \sqrt{2}}{2 - \sqrt{2}}$
$h = \frac{9(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2}$
$h = \frac{18 - 9\sqrt{2}}{4 - 2}$
$h = \frac{18 - 9\sqrt{2}}{2}$
$h = 9 - \frac{9}{2}\sqrt{2}$

1. Find the perpendicular distance from $P$ to $QR$ (from part a)

Let $h$ be the perpendicular distance from $P$ to $QR$.
$\frac{1}{2} \times (4\sqrt{2} + 8) \times h = 18$
$h = \frac{36}{8 + 4\sqrt{2}} = \frac{9}{2 + \sqrt{2}}$
$h = \frac{9(2 - \sqrt{2})}{(2 + \sqrt{2})(2 - \sqrt{2})} = \frac{18 - 9\sqrt{2}}{4 - 2} = \frac{18 - 9\sqrt{2}}{2}$

2. Use trigonometry to find the length of $PS$

Let $T$ be the foot of the perpendicular from $P$ to $SR$. In the right-angled triangle $PTS$, $\angle PST = 45^\circ$.
$\sin 45^\circ = \frac{h}{PS}$
$PS = \frac{h}{\sin 45^\circ} = \frac{h}{\frac{1}{\sqrt{2}}} = h\sqrt{2}$
$PS = \left( \frac{18 - 9\sqrt{2}}{2} \right) \sqrt{2} = \frac{18\sqrt{2} - 18}{2} = -9 + 9\sqrt{2}$