Let $x \text{ cm}$ be the length of $AD$. Then we have
$\begin{array}{rcl}
\dfrac{1}{2} \times AD \times BD & = & \dfrac{1}{2} \times CD \times BD + 24 \\
\dfrac{1}{2} \times x \times 12 & = & \dfrac{1}{2} \times (14-x) \times 12 + 24 \\
6x & = & 108 – 6x \\
x & = & 9 \\
\end{array}$
Therefore, $AD=9\text{ cm}$ and $DC = 5\text{ cm}$.
Since $\Delta ABD$ is a right-angled triangle, then by Pythagoras Theorem, we have
$\begin{array}{rcl}
AB^2 & = & AD^2 + BD^2 \\
AB^2 & = & (9)^2 + (12)^2 \\
AB^2 & = & 225 \\
AB & = & 15 \text{ cm}
\end{array}$
Since $\Delta BCD$ is a right-angled triangle, then by Pythagoras Theorem, we have
$\begin{array}{rcl}
BC^2 & = & DC^2 + BD^2 \\
BC^2 & = & (5)^2 + (12)^2 \\
BC^2 & = & 169 \\
BC & = & 13 \text{ cm}
\end{array}$
Therefore, the perimeter of $\Delta ABC$
$\begin{array}{cl}
= & AB + BC + AC \\
= & 15 + 13 + 14 \\
= & 42 \text{ cm}
\end{array}$
