Anoop managed to draw 6 circles of equal radii with their centres on the…

2025

Anoop managed to draw 6 circles of equal radii with their centres on the diagonal of a square such that the two extreme circles touch two sides of the square and each middle circle touches two circles on either side. Find the ratio of the side of the square to the radius of the circles. Assume root of 2 is 1.4.

  1. A.

    9:1

  2. B.

    6.2 :1

  3. C.

    10.4 :1

  4. D.

    7.6 :1

Attempted by 3 students.

Show answer & explanation

Correct answer: A

Concept

When several circles of equal radius r are placed in a row so that each is tangent to its neighbour, the distance between the centres of any two tangent circles is 2r (the sum of the two radii). For n such circles in a row there are exactly (n − 1) such gaps between consecutive centres, not n. Separately, when a circle of radius r is tangent to both arms of a right-angle corner (as at a vertex of a square), its centre lies on the angle bisector — here, the diagonal — at a distance r√2 from that vertex, since the perpendicular distance from the centre to each arm is r and the bisector makes a 45° angle with each arm.

Application to this question

  1. Let the side of the square be a and the radius of each circle be r. The diagonal of the square = a√2.

  2. Since the two extreme circles touch two sides of the square at each corner, each extreme circle's centre is at a perpendicular distance r√2 from the nearest corner, by the concept above.

  3. There are 6 circles in a row, each tangent to its neighbour, so there are (6 − 1) = 5 gaps between consecutive centres, each of length 2r. Total span between the first and the last centre = 5 × 2r = 10r.

  4. So the diagonal = (corner offset) + (span between extreme centres) + (corner offset) = r√2 + 10r + r√2 = 10r + 2r√2.

  5. Equating the two expressions for the diagonal: a√2 = 10r + 2r√2.

  6. Divide throughout by √2 (rationalising 1/√2 = √2/2 first, so the √2 ≈ 1.4 approximation is applied only to a surd term, keeping the arithmetic clean): a = 2r + 10r/√2 = 2r + 5r√2.

  7. Substitute √2 = 1.4: a = 2r + 5r(1.4) = 2r + 7r = 9r.

  8. So a : r = 9 : 1.

Cross-check

Each of the other offered ratios corresponds to a specific miscount of the span between the extreme circles' centres (using the sum of all six radii, using all six diameters, or one gap too few), and only the correct gap-count for 6 tangent circles reproduces a ratio that is actually among the options offered — confirming a : r = 9 : 1.

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