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.

Correct result:

h =  39.6408 cm

Solution:

$$\begin{gathered} r=6.35 \\ (2x{)}^2=(6r{)}^2+(6r{)}^2 \\ {x}^2=18r^2 \\ x=\sqrt{18}\cdot r=\sqrt{18}\cdot 6.35\doteq 26.9408 \\ {h}_1^2+x^2=(6r{)}^2 \\ {h}_1=\sqrt{(6\cdot r{)}^2-x^2}=\sqrt{(6\cdot 6.35{)}^2-26.9408^2}\doteq 26.9408 \\ h=2\cdot r+h_1=2\cdot 6.35+26.9408=39.6408\text{ cm} \end{gathered}$$

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