一个正四棱锥的底面积为 $\left(2\sqrt{5} + 7\right)\text{ cm}^2$，体积为 $\left(34\sqrt{5} + 32\right)\text{ cm}^3$。 求该四棱锥的高，并表示为 $p + q\sqrt{5}$ 的形式，其中 $p$ 和 $q$ 是整数。

1. Set up the equation for the volume of the pyramid and make $h$ the subject

Let the height of the pyramid be $h$.
$V = \frac{1}{3} \times \text{base area} \times h$
$34\sqrt{5} + 32 = \frac{1}{3}(2\sqrt{5} + 7)h$
$h = \frac{3(34\sqrt{5} + 32)}{7 + 2\sqrt{5}}$
$h = \frac{102\sqrt{5} + 96}{7 + 2\sqrt{5}}$

2. Rationalise the denominator by multiplying by the conjugate

$h = \frac{96 + 102\sqrt{5}}{7 + 2\sqrt{5}} \times \frac{7 - 2\sqrt{5}}{7 - 2\sqrt{5}}$
$h = \frac{(96 + 102\sqrt{5})(7 - 2\sqrt{5})}{7^2 - (2\sqrt{5})^2}$

3. Expand the numerator and simplify the expression

$h = \frac{96(7) - 96(2\sqrt{5}) + 102\sqrt{5}(7) - 102\sqrt{5}(2\sqrt{5})}{49 - 20}$
$h = \frac{672 - 192\sqrt{5} + 714\sqrt{5} - 1020}{29}$
$h = \frac{-348 + 522\sqrt{5}}{29}$
$h = -12 + 18\sqrt{5}$