Question

The area of a cyclic quadrilateral $ABCD$ is $\frac{3\sqrt{3}}{4}\text{sq. units}$. The radius of the circle circumscribing the quadrilateral $ABCD$ is $1\text{unit}$. If $AB=1\text{unit},BD=\sqrt{3}\text{units}$, then the value of $BC\times CD$ is

• A. $2$
• B. $3$
• C. $2\sqrt{3}$
• D. $3\sqrt{3}$

Answer 1

The correct option is A $2$

Clearly, $\angle BOD=2\angle BCD=2C$
$\therefore \cos 2C=\frac{1^2+1^2-(\sqrt{3}{)}^2}{2\times 1\times 1}\Rightarrow \cos 2C=-\frac{1}{2}\Rightarrow C={60}^{\circ }$

Now,
$\angle A={180}^{\circ }-\angle C={120}^{\circ }\Rightarrow \cos {120}^{\circ }=\frac{A{D}^2+1^2-(\sqrt{3}{)}^2}{2\times AD\times 1}\Rightarrow -\frac{1}{2}=\frac{A{D}^2-2}{2AD}\Rightarrow A{D}^2+AD-2=0\Rightarrow (AD-1)(AD+2)=0\Rightarrow AD=1$

Now,
$\text{Area}\left(ABCD\right)=\text{Area}\left(\Delta ADB\right)+\text{Area}\left(\Delta BCD\right)\Rightarrow \frac{3\sqrt{3}}{4}=\frac{1}{2}\times 1\times 1\times \sin {120}^{\circ }+\frac{1}{2}\times BC\times CD\times \sin {60}^{\circ }\therefore BC\times CD=2$