18.40 In parallelogram $ABCD$, the area is $10$, $AB=3$, and $BC=5$. Points $E$, $F$, and $G$ are taken on $AB$, $BC$, and $AD$ respectively, such that $AE=BF=AG=2$. A line through $G$ parallel to $EF$ intersects $CD$ at $H$. Then the area of quadrilateral $EFHG$ is 
(A) 4.
(B) 4.5.
(C) 5.
(D) 5.5.
(E) 6.
(43rd American High School Mathematics Examination, 1992)
[Solution] Given $G H / / E F$, then $\triangle B E F \backsim$ $\triangle D H G$.
So $D H=\frac{3}{2}$, and $C H=\frac{3}{2}$.
Also, $S_{\square A B C D}=10=3 \cdot 5 \cdot \sin A$, then
$$
\sin A=\frac{2}{3} .
$$

Therefore, $S_{\triangle A G E}=\frac{1}{2} A G \cdot A E \cdot \sin A=\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{2}{3}=\frac{4}{3}$.
Similarly, $S_{\triangle B E F}=\frac{2}{3}, S_{\triangle C H F}=\frac{3}{2}, S_{\triangle D H G}=\frac{3}{2}$, then $S_{E F H G}=S_{\square A B C D}-\left(S_{\triangle A G E}+S_{\triangle B E F}+S_{\triangle C H F}+S_{\triangle D H G}\right)=5$. Hence, the answer is $(C)$.