Ans: D
In $\Delta ABD$,
In $\Delta ABD$,
$\begin{array}{rcl}
AB^2 + BD^2 & = & (24)^2 + (32)^2 \\
& = & 1600 \text{ cm}^2
\end{array}$
$\begin{array}{rcl}
AD^2 & = & 40^2 \\
& = & 1600 \text{ cm}^2
\end{array}$
$\because AB^2 + BD^2 = AD^2 = 1600\text{ cm}^2$,
$\therefore$ by the converse of Pythagoras Theorem, $\Delta ABD$ is a right-angled triangle with $\angle ABD=90^\circ$.
Since $ABC$ is a straight line, $\angle CBD = 90^\circ$. Hence $\Delta BCD$ is a right-angled triangle. By Pythagoras Theorem, we have
$\begin{array}{rcl}
BC^2 & = & CD^2 – BD^2 \\
BC^2 & = & 68^2 – 32^2 \\
BC^2 & = & 3600 \\
BC & = & 60 \text{ cm}
\end{array}$
