$18 \cdot 116$ A trapezoid is divided into four triangles by its two diagonals. If $A, B$ represent the areas of the triangles formed by the two bases and the diagonals, then the area of the trapezoid is
(A) $2(A+B)$.
(B) $(\sqrt{A}+\sqrt{B})^{2}$.
(C) $A+B+\sqrt{A B}$.
(D) $A+B+\frac{A B}{A+B}$.
(China Junior High School Mathematics Correspondence Competition, 1984)
[Solution] As shown in the figure, if we denote $S_{1}=S_{\triangle E O H}, S_{2}=$ $S_{\triangle F O G}$,  
it is clear that $S_{1}=S_{2}$. Also, $\triangle E O F$ and $\triangle F O G$ have the same height,  
so $\frac{A}{S_{2}}=\frac{E O}{O G}$,

Similarly, $\frac{S_{2}}{B}=\frac{F O}{O H}$.
Since $E F \parallel H G$, we have $\frac{E O}{O G}=\frac{F O}{O H}$,
$\therefore \frac{A}{S_{2}}=\frac{S_{2}}{B}$, which means $S_{2}^{2}=A \cdot B$,
$\therefore \quad S_{1}=S_{2}=\sqrt{A \cdot B}$,

Thus,
$$
S_{\text {EFCHG }}=A+B+2 \sqrt{A \cdot B}=(\sqrt{A}+\sqrt{B})^{2} \text {. }
$$

Therefore, the answer is $(B)$.