The most efficient stacking of 10 circles will be the 4-3-2-1 pattern, which will produce a figure like an equilateral triangle with 4 circles along each side. The length of one edge will be composed of two end segments and the distance between the points of tangency of the first and fourth circles on each side.
The distance between the first and fourth points of tangency will be 2 diameters and two radii, or three diameters in length 3(1.5)in = 4.5in
Each end segment is the long leg of a 30:60:90 triangle, which has the length ratios of x:x√3:2x. In this case x = 0.75 and therefore the length of the end segment we are interested in is (0.75)√3.
There are two such end segments, which when added to the distance between the first and fourth points of tangency makes the length (2)(0.75)√3 + 4.5 = (1.5√3 + 4.5)in
That is the base of the triangle. The height can easily be calculated by taking half of that length, which is now the short leg of a 30:60:90 triangle, and multiplying by (√3) which the length of the longer leg of that 30:60:90 triangle (and the altitude of the original equilateral triangle).
(√3)(1/2)[(1.5)√3 + 4.5) = 2.25 + 2.25√3
The area of the triangle is
1/2(bh) = (0.5)(1.5√3 + 4.5)(2.25 + 2.25√3) = (0.75√3 + 2.25)(2.25 + 2.25√3) =
(1.6875)√3 + (1.6875)(3) + 5.0625 + (5.0625)√3 =
10.125 + (6.75)√3) ≈ 21.82 sqin
the area of the ten cicrles is
(10)π(0.75)^2 ≈ 17.68 sqin
the difference between the two is the area not covered by the circles.
21.82 sqin - 17.68 sqin = 4.14 sqin
