Question

Suppose $ABCDEF$ is a hexagon such that $AB=BC=CD=1$ 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 $\sqrt{\frac{7}{3}}$

From given figure, it is evident that $3\varphi +3\theta ={360}^{\circ }⟹\varphi +\theta ={120}^{\circ }$.

Now, as $△ABO$ and $△AFO$ are isosceles, $\angle A={120}^{\circ }$.

In $△ABF$,

$\cos A=\frac{1^2+2^2-F{B}^2}{2\left(1\right)\left(2\right)}$

$\therefore -\frac{1}{2}=\frac{5-F{B}^2}{4}$

$\therefore FB=\sqrt{7}$

In $△OBF$,

$\cos (\theta +\varphi )=\frac{r^2+r^2-F{B}^2}{2r^2}$

$\therefore -\frac{1}{2}=\frac{2r^2-7}{2r^2}$

$\therefore r=\sqrt{\frac{7}{3}}$

This is the required answer.