Question

If $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius 2 units such that $AB=1,BD=2\sqrt{3}$ then $AD=$

• A. $\frac{3\sqrt{5}}{2}$
• B. $\frac{3\sqrt{5}-1}{2}$
• C. $\frac{\sqrt{5}+1}{2}$
• D. $\frac{3\sqrt{5}+1}{2}$

Answer 1

The correct option is D

$\frac{3\sqrt{5}+1}{2}$

Let $AD=x$ and $\angle BAD=\theta$

By Sine Rule, in$\Delta BAD,\frac{BD}{\sin \theta }=2R$

$\Rightarrow \sin \theta =\frac{2\sqrt{3}}{2\times 2}\Rightarrow \sin \theta =\frac{\sqrt{3}}{2}\therefore \theta ={60}^{\circ }$

By Cosine Rule, in $\Delta BAD,\cos {60}^{\circ }=\frac{x^2+1^2-(2\sqrt{3}{)}^2}{2\times x\times 1}$

$\therefore x=\frac{1\pm 3\sqrt{5}}{2}$

$\because x>0\therefore x=\frac{1+3\sqrt{5}}{2}$