Question

Ten circles of equal radius are arranged as follows.Four circles are arranged in a row side by side touching one another.Three circles are arranged in the same way but on the four circles.Two circles are arranged on the three circles and the remaining one circle is arranged above the two circles.If r is the radius of each circle,then the height of the last circle i.e. top circle from the horizontal ground is. A:(2+2√3) B:(2+3√3) C:(1+3√3) D:2√3

Answer 1

see, when you arrange all the circles, the line joining centre will form a equilateral triangle.
so, side length =6r
half of side is 3r
now, applying pythagoras theorem, 
height= √((6r)²-(3r)²)
          =√(36r²-9r²) =√(27r²)
height = 2r√3