10. Let $A B C D$ be a trapezium with $A D$ parallel to $B C$ and $\angle A D C=90^{\circ}$, as shown in the figure below. Given that $M$ is the midpoint of $A B$ with $C M=\frac{13}{2} \mathrm{~cm}$ and $B C+C D+D A=17 \mathrm{~cm}$, find the area of the trapezium $A B C D$ in $\mathrm{cm}^{2}$.
(A) 26
(B) 28
(C) 30
(D) 33
(E) 35
10. Answer: (C)

Extend $D A$ and $C M$ to meet at $E$ as shown in the figure above. Since $A M=M B$, $\angle A E M=\angle B C M$ and $\angle A M E=\angle B M C$, we conclude that $\triangle A E M$ is congruent to $\triangle B C M$. Therefore $A E=B C$ and $C M=E M$. Thus $C E=2 C M=13$.
Let the area of trapezium $A B C D$ be $S \mathrm{~cm}^{2}$. Then $S=\frac{1}{2}(D E)(D C)$, and we have
$$
(D E+D C)^{2}=D E^{2}+D C^{2}+2(D E)(D C)=C E^{2}+4 S=13^{2}+4 S=169+4 S \text {. }
$$

Now $D E+D C=D A+A E+D C=D A+B C+D C=17$. Hence $17^{2}=169+4 S$, and it follows that $S=30$.