Suppose ABCDEF is a hexagon such that AB=BC=C | vedclass

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(B) Let $r$ be the radius of the circle. The center of the circle subtends angles at the center corresponding to the chords of lengths $1$ and $2$.Let

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$\sqrt{\frac{7}{3}}$

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(B) Let $r$ be the radius of the circle. The center of the circle subtends angles at the center corresponding to the chords of lengths $1$ and $2$.Let $2	heta$ be the angle subtended by a chord of length $1$ at the center,and $2\alpha$ be the angle subtended by a chord of length $2$ at the center.Then,$\sin 	heta = \frac{1/2}{r} = \frac{1}{2r}$ and $\sin \alpha = \frac{1}{r}$.The sum of the angles around the center is $3(2	heta) + 3(2\alpha) = 360^{\circ}$,which implies $	heta + \alpha = 60^{\circ}$.Taking cosine on both sides,$\cos(	heta + \alpha) = \cos(60^{\circ}) = \frac{1}{2}$.Using the formula $\cos(	heta + \alpha) = \cos 	heta \cos \alpha - \sin 	heta \sin \alpha = \frac{1}{2}$.Since $\sin 	heta = \frac{1}{2r}$,$\cos 	heta = \sqrt{1 - \frac{1}{4r^2}} = \frac{\sqrt{4r^2-1}}{2r}$.Since $\sin \alpha = \frac{1}{r}$,$\cos \alpha = \sqrt{1 - \frac{1}{r^2}} = \frac{\sqrt{r^2-1}}{r}$.Substituting these into the equation: $\frac{\sqrt{4r^2-1}}{2r} \cdot \frac{\sqrt{r^2-1}}{r} - \frac{1}{2r} \cdot \frac{1}{r} = \frac{1}{2}$.$\frac{\sqrt{(4r^2-1)(r^2-1)}}{2r^2} = \frac{1}{2} + \frac{1}{2r^2} = \frac{r^2+1}{2r^2}$.$\sqrt{(4r^2-1)(r^2-1)} = r^2+1$.Squaring both sides: $(4r^2-1)(r^2-1) = (r^2+1)^2$.$4r^4 - 5r^2 + 1 = r^4 + 2r^2 + 1$.$3r^4 = 7r^2 \Rightarrow r^2 = \frac{7}{3}$.Thus,$r = \sqrt{\frac{7}{3}}$.

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

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