Billiard balls

A layer of ivory billiard balls of radius 6.35 cm is in the form of a square. The balls are arranged so that each ball is tangent to every one adjacent to it. In the spaces between sets of 4 adjacent balls other balls rest, equal in size to the original. These balls form in turn a second layer on top of the first. Successive layers of this sort form a pyramidal pile with a single ball resting on top. If the bottom layer contains 16 balls, what is the height of the pile.

Result

h =  39.641 cm

Solution:

$r = 6.35 \ \\ (2x)^2 = (6r)^2+(6r)^2 \ \\ x^2 = 18r^2 \ \\ x = \sqrt{ 18 } \cdot \ r = \sqrt{ 18 } \cdot \ 6.35 \doteq 26.9408 \ \\ h_{ 1 }^2 +x^2 = (6r)^2 \ \\ h_{ 1 } = \sqrt{ (6 \cdot \ r)^2 -x^2 } = \sqrt{ (6 \cdot \ 6.35)^2 -26.9408^2 } \doteq 26.9408 \ \\ h = 2 \cdot \ r+h_{ 1 } = 2 \cdot \ 6.35+26.9408 \doteq 39.6408 = 39.641 \ \text{ cm }$

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