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 concylic, the radius of the circle passing throught them is-

Options

A. `sqrt((5)/(2))`
B. `sqrt((7)/(3))`
C. `sqrt((11)/(5))`
D. `sqrt(2)`

Answer

B

Explanation

From the figure :
`sintheta=(1)/(2r)&sinalpha=(1)/(r)`
`3xx(2theta)+(2alpha)xx3=360^(@)`
`theta+alpha=60^(@)`
Now, `cos(theta+alpha)=(1)/(2)`
`rArr costheta.cosalpha-sintheta.sinalpha=(1)/(2)`
`rArrsqrt(1-(1)/(4r^(2)))sqrt(1-(1)/(r^(2)))-(1)/(2r)*(1)/(r)=(1)/(2)`
`rArrsqrt(4r^(2)-1)sqrt(r^(2)-1)-1=r^(2)`
`rArr(4r^(2)-)(r^(2)-1)=(r^(2)+1)^(2)`
`rArr4r^(4)-5r^(2)+1=r^(4)//+2r^(2)+1`
`rArr3r^(4)=7r^(2)`
`rArr3r^(4)=7r^(2)`
`rArrr^(2)=(7)/(3)`
`rArrr=sqrt((7)/(3))`