smc

Let $E$ be the point of intersection of $L$ and $\overline{BD}$. Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$, triangles $ADE$ and $BAE$ are similar, giving $$\frac{AE}{BE}=\frac{ED}{EA},$$ and so $$\begin{array}{rl}AE& =\sqrt{BE\cdot ED}\\ & =\sqrt{(1+1)(1)}\\ & =\sqrt{2}.\end{array}$$ Thus, taking $BD$ and $AE$ as the base and perpendicular height, respectively, of triangle $ABD$, we may compute its area as $\frac{1}{2}(3)\left(\sqrt{2}\right)=\frac{3\sqrt{2}}{2}$. By symmetry, the area of the entire rectangle $ABCD$ is $$2\left(\frac{3\sqrt{2}}{2}\right)=3\sqrt{2}\approx (3)(1.4)=\boxed{4.2}.$$