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.
• B.
• C.
• D.

Answer 1

The correct option is B


$\theta +\varphi ={120}^{\circ }So\angle A={120}^{\circ }\therefore ABCDEFareconcycliccosA=cos{120}^{\circ }=\frac{1^2+2^2-F{B}^2}{2\left(1\right)\left(2\right)}\frac{-1}{2}=\frac{5-F{B}^2}{2\left(2\right)}\Rightarrow F{B}^2=7FB=\sqrt{7}$
Again by cosine rule
$cos(\theta +\varphi )=\frac{r^2+r^2-7}{2r^2}-\frac{1}{2}=\frac{2r^2-7}{2r^2}3r^2=7r=\sqrt{\frac{7}{3}}$