Question

A cyclic quadrilateral ABCD of area $3\sqrt{3}/4$ is inscribed in a unit circle. If one of its sides AB= 1 and the diagonal BD = $\sqrt{3}$, find the lengths of the other sides.

Answer 1

$AB=1,BD=sqrt3,OA=OB=OD=1$
The given circle of radius 1 is also circum-circle of $△ABD$.
$\therefore R=1$ for $△ADB\therefore \frac{a}{sinA}=2R$
$\therefore \frac{\sqrt{3}}{sinA}=2R=2\therefore sinA=\frac{\sqrt{3}}{2}\therefore A={60}^0$
and hence $C={120}^0$.
Also by cosine rule on $△ABD$
$(\sqrt{3}{)}^2=1^2+x^2-2xcos{60}^0$
or $x^2-x-2=0$
$\therefore (x-2)(x+1)=0\therefore x=2$
Now the equation reduced to part (a) above.
$△={△}_1+{△}_2$
$\Rightarrow \frac{3\sqrt{3}}{4}=\frac{1}{2}(1.2.sin{60}^0)+\frac{1}{2}(c.dsin{120}^0)$
or $\frac{3\sqrt{3}}{4}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}cd$
$\therefore cd=1or{c}^2{d}^2=1$
Also by cosine rule on $△BCD$ we, have
$(\sqrt{3}{)}^2=c^2+d^2-2cdcos{120}^0=c^2+d^2+cd$
$\therefore c^2+d^2=2ascd=1$
$\therefore c^{2}and{d}^2$ are the roots of $t^2-2t+1=0$
$\therefore c^2=d^2=1$
$\therefore BC=1CD$ and $AD=x=2$.