Ans: A
By applying the Pythagoras Theorem to $\Delta ABD$, we have
By applying the Pythagoras Theorem to $\Delta ABD$, we have
$\begin{array}{rcl}
BD^2 & = & AD^2 – AB^2 \\
BD^2 & = & 30^2 – 18^2 \\
BD & = & 24 \text{ cm}
\end{array}$
Note that $\angle BDC = \angle ABD = 90^\circ$ (alt. $\angle$s, $AB//DC$).
By applying the Pythagoras Theorem to $\Delta BCD$, we have
$\begin{array}{rcl}
CD^2 & = & BC^2 – BD^2 \\
CD^2 & = & 26^2 – 24^2 \\
CD & = & 10 \text{ cm}
\end{array}$
Therefore, the area of the trapezium $ABCD$
$\begin{array}{cl}
= & \dfrac{1}{2} (AB + CD) \times BD \\
= & \dfrac{1}{2} (18 + 10) \times 24 \\
= & 336\text{ cm}^2
\end{array}$
