ABCD is a rectangle such that $\angle CAB=60°$ and AC = $a$ units. The area of rectangle ABCD is  _______.

1. $\frac{\sqrt{2}}{3}{a}^2$ sq. units
2. $\frac{\sqrt{3}}{2}{a}^2$ sq. units
3. $\frac{\sqrt{3}}{4}{a}^2$sq. units
4. $\frac{\sqrt{3}}{8}{a}^2$ sq. units

The correct option is C $\frac{\sqrt{3}}{4}{a}^2$sq. units
 

In $△ABC$, right angled at B
$sin60°=\frac{BC}{AC}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{BC}{a}\Rightarrow BC=\frac{\sqrt{3}}{2}a$

$cos60°=\frac{AB}{AC}$
$\Rightarrow \frac{1}{2}=\frac{AB}{a}\Rightarrow AB=\frac{a}{2}$

So, area of the rectangle 
$=AB\times BC=\frac{a}{2}\times \frac{\sqrt{3}}{2}a=\frac{\sqrt{3}{a}^2}{4}$