Three circles,each of radius 1,touch one anot | vedclass

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Question

(A) Let the centers of the three circles be $O$,$A$,and $B$. Since each circle has a radius of $1$ and they touch each other externally,the distance b

Vedclass · Accepted Answer

$2+\sqrt{3}$

Vedclass · Answer

(A) Let the centers of the three circles be $O$,$A$,and $B$. Since each circle has a radius of $1$ and they touch each other externally,the distance between any two centers is $1+1=2$. Thus,$	riangle OAB$ is an equilateral triangle with side length $2$.The height of this triangle from vertex $O$ to the base $AB$ is $h = \sqrt{2^2 - 1^2} = \sqrt{3}$.The distance $d$ between the two parallel lines is the sum of the radius of the top circle,the height of the triangle $OAB$,and the radius of the bottom circles.$d = r + h + r = 1 + \sqrt{3} + 1 = 2 + \sqrt{3}$.

Three circles,each of radius $1$,touch one another externally and lie between two parallel lines. The minimum possible distance between the lines is:

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