图示为一个棱柱，其横截面为边长为 $(5 + \sqrt{3})$ cm 的正六边形。

1. Express the area of the regular hexagon in terms of its side length

A regular hexagon can be divided into 6 equilateral triangles.
Area $= 6 \times \left( \frac{1}{2} s^2 \sin 60^\circ \right)$
$= 3 s^2 \left( \frac{\sqrt{3}}{2} \right)$
$= \frac{3\sqrt{3}}{2} (5 + \sqrt{3})^2$

2. Expand the squared term

$(5 + \sqrt{3})^2 = 5^2 + 2(5)(\sqrt{3}) + (\sqrt{3})^2$
$= 25 + 10\sqrt{3} + 3$
$= 28 + 10\sqrt{3}$

3. Multiply and simplify to the required form

Area $= \frac{3\sqrt{3}}{2} (28 + 10\sqrt{3})$
$= 3\sqrt{3} (14 + 5\sqrt{3})$
$= 42\sqrt{3} + 15(3)$
$= 42\sqrt{3} + 45$

1. Calculate the cross-sectional area of the regular hexagon from part (a)

Area $= 6 \times \frac{1}{2} \times (5 + \sqrt{3})^2 \times \sin(60^\circ)$
$= 3 \times (25 + 10\sqrt{3} + 3) \times \frac{\sqrt{3}}{2}$
$= \frac{3\sqrt{3}}{2} (28 + 10\sqrt{3})$
$= 3\sqrt{3} (14 + 5\sqrt{3})$
$= 42\sqrt{3} + 45$

2. Set up the equation for the height of the prism

Volume $= \text{Cross-sectional area} \times \text{height}$
$3(32\sqrt{3} + 138) = (45 + 42\sqrt{3}) \times h$
$h = \frac{414 + 96\sqrt{3}}{45 + 42\sqrt{3}}$
$h = \frac{3(138 + 32\sqrt{3})}{3(15 + 14\sqrt{3})}$
$h = \frac{138 + 32\sqrt{3}}{15 + 14\sqrt{3}}$

3. Rationalise the denominator to find the height in the required form

$h = \frac{138 + 32\sqrt{3}}{15 + 14\sqrt{3}} \times \frac{15 - 14\sqrt{3}}{15 - 14\sqrt{3}}$
$= \frac{138(15) - 138(14\sqrt{3}) + 32\sqrt{3}(15) - 32(14)(3)}{15^2 - (14\sqrt{3})^2}$
$= \frac{2070 - 1932\sqrt{3} + 480\sqrt{3} - 1344}{225 - 588}$
$= \frac{726 - 1452\sqrt{3}}{-363}$
$= -2 + 4\sqrt{3}$
$= 4\sqrt{3} - 2$