Suppose $A B C D E F$ is a hexagon such that $A B=B C=C D=1$ and $D E=E F=F A=$ 2. If the vertices $A, B, C, D, E, F$ are concyclic, the radius of the circle passing through them is.

• A

  $\sqrt{\frac{5}{2}}$
• Correct answer
  B

  $[\mathrm]\sqrt{\frac{7}{3}}$
• C

  $[\mathrm]\sqrt{\frac{11}{5}}$
• D

  $\sqrt{2}$

Solution

$$ \theta+\phi=120^{\circ} $$ So, $$ \begin{array}{l} \angle A=120^{\circ} \quad \because \mathrm{ABCDEF} \text { are concyclic } \\ \cos \mathrm{A}=\cos 120^{\circ}=\frac{1^{2}+2^{2}-F B^{2}}{2(1)(2)} \\ \frac{-1}{2}=\frac{5-F B^{2}}{2(2)} \Rightarrow F B^{2}=7 \\ F B=\sqrt{7} \end{array} $$ Again by cosine rule $$ \begin{array}{l} \cos (\theta+\phi)=\frac{r^{2}+r^{2}-7}{2 r^{2}} \\ -\frac{1}{2}=\frac{2 r^{2}-7}{2 r^{2}} \\ 3 r^{2}=7 \\ r=\sqrt{\frac{7}{3}} \end{array} $$

Hence, the correct answer is option: 'B'.