The distance between the two parallel lines is 1 unit. A point A is chosen to lie between the lines at a distance 'd' from one of them Triangle ABC is equilateral with B on one line and C on the other parallel line. The length of the side of the equilateral triangle is

Options

A. `(2)/(3) sqrt(d^(2)+d+1)`
B. `2sqrt((d^(2)-d+1)/(3))`
C. `2sqrt(d^(2)-d+1)`
D. `sqrt(d^(2)-d+1)`

Answer

B

Explanation

From figure ltbr. `x cos (theta+30^(@))=d`
and `x sin theta=1-d`
Dividing `sqrt(3) cot theta=(1+d)/(1-d)`
squaring Eq. ii and putting the value of `cos theta` , we get
`x^(2)=(1)/(3)(4d^(2)-4d+4)`