2019-II-19

Ans: A
By applying the Pythagoras Theorem to $\mathrm{\Delta }ABD$, we have

$\begin{array}{rcl}B{D}^2& =& A{D}^2–A{B}^2\\ B{D}^2& =& {30}^2–{18}^2\\ BD& =& 24\text{ cm}\end{array}$

Note that $\mathrm{\angle }BDC=\mathrm{\angle }ABD={90}^{\circ }$ (alt. $\mathrm{\angle }$s, $AB//DC$).

By applying the Pythagoras Theorem to $\mathrm{\Delta }BCD$, we have

$\begin{array}{rcl}C{D}^2& =& B{C}^2–B{D}^2\\ C{D}^2& =& {26}^2–{24}^2\\ CD& =& 10\text{ cm}\end{array}$

Therefore, the area of the trapezium $ABCD$

$\begin{array}{cl}=& {\displaystyle \frac{1}{2}}(AB+CD)\times BD\\ =& {\displaystyle \frac{1}{2}}(18+10)\times 24\\ =& 336{\text{ cm}}^2\end{array}$