ABC is a right triangle with right angle at B; AC $=$  2 units, BC $=$  1 unit and BD is perpendicular to AC. The area of the rectangle with BD as one of its diagonal is

Given that , $AC=2$ units , $BC=1$ unit

In $\Delta ABC,\angle B=90°$

$\therefore$ $A{C}^2=A{B}^2+B{C}^2\Rightarrow AB=\sqrt{3}$ units and

We know that,

$ar\left(\Delta ABC\right)=\frac{1}{2}\times AB\times BC$ $=\frac{1}{2}\times CA\times BD$

$\Rightarrow \frac{1}{2}\times \sqrt{3}\times 1=\frac{1}{2}\times 2\times BD$

$\Rightarrow BD=\frac{\sqrt{3}}{2}$ units.

In $\Delta BCD,C{D}^2=B{C}^2-B{D}^2\Rightarrow CD=\frac{1}{2}$

In $\Delta ABD,A{D}^2=A{B}^2-B{D}^2\Rightarrow AD=\frac{3}{2}$

$ar\left[\Delta BCD\right]=\frac{1}{2}\times BC\times FD=\frac{1}{2}\times CD\times BD$

$\Rightarrow 1\times FD=\frac{1}{2}\times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{4}$

perpendicular $ar\left[\Delta ABD\right]=\frac{1}{2}\times AB\times DE=\frac{1}{2}\times BD\times DA$

$\Rightarrow \sqrt{3}\times DE=\frac{\sqrt{3}}{2}\times \frac{3}{2}\Rightarrow DE=\frac{3}{4}$

$\therefore ar\left(DEBF\right)=DF\times DE=\frac{\sqrt{3}}{4}\times \frac{3}{4}=\frac{3\sqrt{3}}{16}$ sq. units