[Diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) $DB$ of [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) $ABCD$ is divided into three segments of length $1$ by [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) lines $L$ and $L'$ that pass through $A$ and $C$ and are [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) to $DB$. The area of $ABCD$, rounded to the one decimal place, is 

$\mathrm{(A)\ } 4.1 \qquad \mathrm{(B) \ }4.2 \qquad \mathrm{(C) \  } 4.3 \qquad \mathrm{(D) \  } 4.4 \qquad \mathrm{(E) \  }4.5$
Let $E$ be the intersection of line $L$ and $\stackrel{\longleftrightarrow}{BD}$. Because $AE$ is the altitude to the hypotenuse of right triangle $ABD$, we have $(AE)^2=BE \cdot ED$. Thus, $AE^2=(1)(2)\implies AE=\sqrt{2}$. Now we use $A=\frac{1}{2}bh$ on $\triangle ABD$ to get $[ABD]=\frac{1}{2}(3)(\sqrt{2})=\frac{3\sqrt{2}}{2}$. Now we have to double it to get the area of the entire rectangle: $2\left(\frac{3\sqrt{2}}{2}\right)=3\sqrt{2}\approx4.2,$ Which leads to the answer $\boxed{\text{(B) 4.2}}$.