Ans: A
By applying the Pythagoras Theorem to $\Delta ACD$, we have
By applying the Pythagoras Theorem to $\Delta ACD$, we have
$\begin{array}{rcl}
CD^2 & = & 17^2 – 15^2 \\
CD & = & 8 \text{ cm}
\end{array}$
Therefore, the area of the trapezium $ABCD$
$\begin{array}{cl}
= & \dfrac{1}{2} (8 + 38) \times 15 \\
= & 345 \text{ cm}^2
\end{array}$
