My Problem Number Eight

In imaginary plane, four vertices of the quadrilateral are $A\equiv (1,0), B\equiv (0,1), C\equiv (-\frac{1}{2},\frac{\sqrt{3}}{2})$ and $D\equiv (-\frac{1}{2},-\frac{\sqrt{3}}{2})$ respectively. Let the line $CD$ cut the $x-$ axis at the point $E$ .

The required area $=|ABCD|$

$=|ABO|+|BCEO|+|EDA|$

$=\frac{1}{2}*1*1+\frac{1}{2}*(1+\frac{\sqrt{3}}{2})*\frac{1}{2}+\frac{1}{2}*\frac{3}{2}*\frac{\sqrt{3}}{2}$

$=\frac{3}{4}+\frac{\sqrt{3}}{2}=\boxed{1.616}$ .

$A\equiv (1,0), B\equiv (0,1), C\equiv (-\dfrac{1}{2},\dfrac{\sqrt{3}}{2}), ~D\equiv (-\dfrac{1}{2},-\dfrac{\sqrt{3}}{2})~and~O\equiv (0,0)$ .

$The ~required~ area ~=|ABCD| \\ =~~~|ABO|~~~+~~~|BCO|~~~~~+~~~~~~~|CDO|~~~~~~~~~~~~~~+~~~~~~~~|DAO| \\ =\dfrac{1}{2}*1*1~~+~~\dfrac{1}{2}*\dfrac{1}{2}*1~~+~~\dfrac{1}{2}*2*\dfrac{\sqrt{3}}{2}*\dfrac{1}{2}~~+~~2*\dfrac{1}{2}*\dfrac{\sqrt{3}}{2}*\dfrac{1}{2}$

$=\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{\sqrt{3}}{4} +\dfrac{\sqrt{3}}{4}$

$=\dfrac{3}{4}+\dfrac{\sqrt{3}}{2}=\large \boxed{\color{#3D99F6}{~~~1.616~~~}}$

The shrortest way I found was that of using 1/2 a b*sinΘ