$18 \cdot 59$ Rectangle $A B C D$ has its diagonal $D B$ divided into three segments of length 1 each by two parallel lines $l$ and $l^{\prime}$, where $l$ and $l^{\prime}$ pass through points $A$ and $C$ respectively and are perpendicular to $B D$. The area of $A B C D$ is (保留一位小数)
(A) 4.1 .
(B) 4.2 .
(C) 4.3 .
(D) 4.4 .
(E) 4.5 .
[Solution] As shown in the figure, let $A E=a, E F=b$.
In the right triangle $\triangle A D B$, by the projection theorem, we have
$$
\begin{aligned}
& a^{2}=D E \cdot E B=1 \cdot 2=2, \\
\therefore \quad & a=\sqrt{2},
\end{aligned}
$$

Similarly, in the right triangle $\triangle A D F$, we have $a b=1^{2}$,
$$
\begin{array}{rlrl}
\therefore & b & =\frac{1}{\sqrt{2}} . \\
\therefore & & S_{A B C \mathcal{1}} & =2 \cdot\left[\frac{1}{2} \cdot(a+b) \cdot 1\right]+(a+b) \cdot 1 \\
& & =2\left(\sqrt{2}+\frac{1}{\sqrt{2}}\right)=3 \sqrt{2} \approx 4.23 \approx 4.2 .
\end{array}
$$

Therefore, the correct choice is $(B)$.