Find the area of $\triangle ABC$ if $a = 18 \; cm$, $b = 24 \; cm$ and $c = 30 \; cm$
Area of $\triangle ABC = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ $\;$ where $s$ is the semi-perimeter of $\triangle ABC$
$s = \dfrac{a + b + c}{2} = \dfrac{18 + 24 + 30}{2} = 36 \; cm$
$\begin{aligned}
\therefore \; \text{Area of } \triangle ABC & = \sqrt{36 \left(36 - 18\right) \left(36 - 24\right) \left(36 - 30\right)} \\\\
& = \sqrt{36 \times 18 \times 12 \times 6} \\\\
& = 6 \times 3 \times 2 \times 6 \\\\
& = 216 \; cm^2
\end{aligned}$