答案：(a) $11\text{ cm}$ (b) $624\text{ cm}^2$
- 考慮該柱體的體積，可得
$\begin{array}{rcl}
\dfrac{12(6 + AD)}{2} \times 10 & = & 1020 \\
AD & = & 11 \mbox{ cm}
\end{array}$ - 在 $AD$ 上加點 $Q$ 使得 $CQ\perp AD$。所以 $\Delta CQD$ 為直角三角形，$CQ=AB=12\mbox{ cm}$，且
$\begin{array}{rcl}
QD & = & AD – BC \\
& = & 11 – 6 \\
& = & 5\mbox{ cm}
\end{array}$所以，可得
$\begin{array}{rcl}
CD & = & \sqrt{CQ^2+QD^2} \\
& = & \sqrt{12^2+5^2} \\
& = & 13 \mbox{ cm}
\end{array}$由此，柱體 $ABCDEFGH$ 的總表面面積
$\begin{array}{cl}
= & \dfrac{12(6+11)}{2}\times 2 + (6 + 12 + 11 + 13)\times 10 \\
= & 624 \mbox{ cm}^2
\end{array}$
