Question

A cyclic quadrilateral $ABCD,(\angle A<\angle C)$ of area $\frac{3\sqrt{3}}{4}$ is inscribed in an unit circle.If one of its side $AB=1$ and the diagonal $BD=\sqrt{3}$ then which of the following is/are true?

• A. $\angle A={60}^{\circ }$
• B. Length of $AD\text{is}2$
• C. $BC=CD=1$
• D. $BC>CD$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\angle A={60}^{\circ }$
B Length of $AD\text{is}2$
C $BC=CD=1$


From the property of circumcircle of a triangle we have ,
$R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}$

So,$\frac{\sqrt{3}}{\sin A}=2R,R=1\Rightarrow \angle A={60}^{\circ }\text{as}\angle A<\angle C\text{and both are the opposite angle of a cyclic quadrilateral}\Rightarrow \angle C={120}^{\circ }$
From cosine rule we have
Let $AD=x$
So, $\cos {60}^{\circ }=\frac{x^2+1-(\sqrt{3}{)}^2}{2\times x\times 1}=\frac{1}{2}\Rightarrow x^2-x-2=0x=2,-1\text{(x cant be negative)}\Rightarrow x=2=AD$

Area of quadrilateral $ABCD=\frac{3\sqrt{3}}{4}=ar\left(△ABD\right)+ar\left(△BCD\right)\Rightarrow \frac{3\sqrt{3}}{4}=\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{4}\times c\times d\Rightarrow c\times d=\mathrm{1............}\left(i\right)$
Now by cosine rule in $△BCD$
$\cos {120}^{\circ }=\frac{c^2+d^2-3}{2cd}=\frac{-1}{2}.........\left(ii\right)$
By solving $\left(i\right)$ and $\left(ii\right)$ we have
$c=d=1\Rightarrow BC=CD=1$