Question

Suppose $ABCDEF$ is a hexagon such that $AB=BC=CD=1\text{and}DE=EF=FA=2.$ If the vertices $A,B,C,D,E,F$ are concyclic, the radius of the circle passing through them is

• A. $\sqrt{\frac{5}{2}}$
• B. $\sqrt{\frac{7}{3}}$
• C. $\sqrt{\frac{11}{5}}$
• D. $\sqrt{2}$

Answer 1

The correct option is B $\sqrt{\frac{7}{3}}$

$\varphi +\theta ={120}^{\circ }$
$\angle A={120}^{\circ }\because ABCDEF\text{is concyclic}$
Using cosine Rule:
$\cos \angle A=\cos {120}^{\circ }=\frac{1^2+2^2-F{B}^2}{2\left(1.2\right)}$
$\Rightarrow FB=\sqrt{7}$
$\cos (\varphi +\theta )=\frac{r^2+r^2-7}{2r^2}\Rightarrow -\frac{1}{2}=\frac{2r^2-7}{2r^2}\Rightarrow r=\sqrt{\frac{7}{3}}$