三角形 ABC 的面积为 $\dfrac{16 + 7\sqrt{10}}{2}\ \text{cm}^2$，AB 的长度为 $(3\sqrt{2} + \sqrt{5})\ \text{cm}$，且 AB 垂直于 BC。

1. Set up the equation for the area of the triangle and isolate $BC$.

$\text{Area} = \frac{1}{2} \times AB \times BC$

$\frac{16 + 7\sqrt{10}}{2} = \frac{1}{2} (3\sqrt{2} + \sqrt{5}) \times BC$

$16 + 7\sqrt{10} = (3\sqrt{2} + \sqrt{5}) \times BC$

$BC = \frac{16 + 7\sqrt{10}}{3\sqrt{2} + \sqrt{5}}$

2. Rationalise the denominator by multiplying by its conjugate.

$BC = \frac{(16 + 7\sqrt{10})(3\sqrt{2} - \sqrt{5})}{(3\sqrt{2} + \sqrt{5})(3\sqrt{2} - \sqrt{5})}$

Denominator $= (3\sqrt{2})^2 - (\sqrt{5})^2 = 18 - 5 = 13$

3. Expand and simplify the numerator.

Numerator $= 16(3\sqrt{2}) - 16\sqrt{5} + 7\sqrt{10}(3\sqrt{2}) - 7\sqrt{10}(\sqrt{5})$

$= 48\sqrt{2} - 16\sqrt{5} + 21\sqrt{20} - 7\sqrt{50}$

$= 48\sqrt{2} - 16\sqrt{5} + 21(2\sqrt{5}) - 7(5\sqrt{2})$

$= 48\sqrt{2} - 16\sqrt{5} + 42\sqrt{5} - 35\sqrt{2}$

$= 13\sqrt{2} + 26\sqrt{5}$

4. Divide by the denominator to obtain the final form.

$BC = \frac{13\sqrt{2} + 26\sqrt{5}}{13}$

$BC = \sqrt{2} + 2\sqrt{5}$

1. Find the length of BC using the area of the triangle

$\frac{1}{2} \times AB \times BC = \text{Area}$
$\frac{1}{2}(3\sqrt{2} + \sqrt{5})BC = \frac{16 + 7\sqrt{10}}{2}$
$BC = \frac{16 + 7\sqrt{10}}{3\sqrt{2} + \sqrt{5}}$
$BC = \frac{(16 + 7\sqrt{10})(3\sqrt{2} - \sqrt{5})}{(3\sqrt{2})^2 - (\sqrt{5})^2}$
$BC = \frac{48\sqrt{2} - 16\sqrt{5} + 21\sqrt{20} - 7\sqrt{50}}{18 - 5}$
$BC = \frac{48\sqrt{2} - 16\sqrt{5} + 42\sqrt{5} - 35\sqrt{2}}{13}$
$BC = \frac{13\sqrt{2} + 26\sqrt{5}}{13} = \sqrt{2} + 2\sqrt{5}$

2. Apply Pythagoras' theorem to find $AC^2$

Since $AB \perp BC$, $\triangle ABC$ is a right-angled triangle.
$AC^2 = AB^2 + BC^2$
$AC^2 = (3\sqrt{2} + \sqrt{5})^2 + (\sqrt{2} + 2\sqrt{5})^2$

3. Expand the squares and simplify to find the final expression

$AC^2 = [18 + 2(3\sqrt{2})(\sqrt{5}) + 5] + [2 + 2(\sqrt{2})(2\sqrt{5}) + 20]$
$AC^2 = (23 + 6\sqrt{10}) + (22 + 4\sqrt{10})$
$AC^2 = 45 + 10\sqrt{10}$