There is a square. 5 circles are kept on the diagonal of the square such that the two extreme circles touch the two sides of the square.each circle has equal radius.find the ratio of side of hte squae to that of the radius.

• Draw a square, and make 5 circles in the diagonal in such a way that two corner circles could touch the two sides of the square.
  Let the length of the side of the square and radius of the circle are respectively a and r. Then diagonal of the square = a(sqrt2)
  since this diagonal contains 5 circles and a bit extra length in both the corners so relation will be
  a(sqrt2) = r(sqrt2) + r + 2r*3 + r + r(sqrt2)
  or a(sqrt2) = 2r(sqrt2) + 8r
  or r/a = (sqrt2)/[2(sqrt2) + 8]
  or r/a = 1/[2 + 4(sqrt2)]
• 13 years agoHelpfull: Yes(11) No(2)
• let the length of the each side of square be a and radius of the circle be r
  since 5 circles have been placed around the diagonal so
  length of diagonal of the square=sum of diameters of all the circles
  a*sqrt(2)=10r
  so a/r=10/sqrt(2)
  a/r=5*sqrt(2)
• 13 years agoHelpfull: Yes(7) No(5)
• @amitabh, a/r is asked, did not see. Just reverse the equation , its [2 + 4(sqrt2)]:1
• 13 years agoHelpfull: Yes(3) No(0)
• 7.6:1
  8r+2sqrt2r=sqrta
• 13 years agoHelpfull: Yes(2) No(0)
• @suchandra ravi,ur solution is perfectly alright but tell me will it be a/r or r/a???
• 13 years agoHelpfull: Yes(0) No(0)
• On diagonal 5 circle..and two circle touch two side of square.their diameter will be r(sqrt2)+r.And diameter of circle which is on the diagonal is 2r.
  length of diagonal is a(sqrt2)
  a(sqrt)=2(r(sqrt(2)+r)+3*2r
  r/a=1/[2+4(sqrt(2)]
• 13 years agoHelpfull: Yes(0) No(0)