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The distance between the two parallel lines is 1 unit. A point A is chosen to lie between the lines at a distance 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

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a

(2/3)d2+d+1

b

⇒x=2d2−d+13

c

2(d2−d+1)

d

(d2−d+1)

answer is B.

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Detailed Solution

we have xcos(θ+300)=d...........(1)xsinθ=1−d ……………(2)Dividing (1) by (2), we have3cotθ=1+d1−dSquaring equation (2) and putting the value of cotθ , we have x2=13(4d2−4d+4)⇒x=2d2−d+13

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