Ans: C
By applying the Pythagorus Theorem to $\Delta CDN$, we have
By applying the Pythagorus Theorem to $\Delta CDN$, we have
$\begin{array}{rcl}
CD^2 & = & CN^2 + DN^2 \\
10^2 & = & CN^2 + 6^2 \\
CN^2 & = & 64 \\
CN & = & 8 \text{ cm}
\end{array}$
By applying the Pythagorus Theorem to $\Delta AEN$, we have
$\begin{array}{rcl}
AE^2 & = & EN^2 + AN^2 \\
13^2 & = & 5^2 + AN^2 \\
AN^2 & = & 144 \\
AN & = & 12 \text{ cm}
\end{array}$
By applying the Pythagorus Theorem to $\Delta ABC$, we have
$\begin{array}{rcl}
AC^2 & = & AB^2 + BC^2 \\
(12+ 8)^2 & = & AB^2 + 16^2 \\
AB^2 & = & 144 \\
AB & = & 12 \text{ cm}
\end{array}$
Hence, the area of $\Delta ABC$
$\begin{array}{cl}
= & \dfrac{1}{2} \times BC \times AB \\
= & \dfrac{1}{2} \times (16) \times (12) \\
= & 96 \text{ cm}^2
\end{array}$
