Question

A cyclic quadrilateral $ABCD$ of area $\frac{3\sqrt{3}}{4}$ is inscribed in a unit circle. If one of its sides $AB=1$ and the diagonal $BD=\sqrt{3}$, then $AD+BC+CD=$

Answer 1

Using sine rule in $△ABC$, we get

$\frac{\sqrt{3}}{\sin A}=2R=2\Rightarrow \sin A=\frac{\sqrt{3}}{2}\Rightarrow A={60}^0$

Given $AB=x=1$

Using cosine rule in $△ABD$

$\cos \left(\frac{\pi }{3}\right)=\frac{x^2+y^2-3}{2xy}=\frac{1}{2}\Rightarrow y^2-y-2=0\Rightarrow (y-2)(y+1)=0$

$\because y\ne -1\Rightarrow y=AD=2$

Since $\angle A={60}^0,\angle C={120}^0$

$\therefore$ In $△BCD$

$3=p^2+q^2-2pq\cos {120}^0\Rightarrow 3=p^2+q^2+pq$ ...(1)

Also, area of quadrilateral $ABCD=\frac{3\sqrt{3}}{4}$

$\frac{1}{2}.1.2\sin {60}^0+\frac{1}{2}.p.q\sin {120}^0=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}pq$

$\Rightarrow \frac{\sqrt{3}}{4}pq=\frac{\sqrt{3}}{4}\Rightarrow pq=1\therefore pq=1,p^2+q^2=2,p,q>0\Rightarrow p=q=1\therefore AB=1,AD=2,BC=CD=1$