Five Hexagons

Then we can see $\displaystyle (2x) \times 6 + 4x = 6 \ units$

$\displaystyle \Rightarrow x = \frac {6}{16} \ units \ = 0.375 \ units$

Turn the large unit regular hexagon by $30^\circ$ counterclockwise. Let the center of large unit hexagon be the origin $O(0,0)$ .

Then the bottom vertex of the top small hexagon is at $A \left(\frac {\sqrt 3}2s, 1-\frac 52s\right)$ . And the line $CA$ is on $\dfrac {y-1+\frac 52s}{x-\frac {\sqrt 3}2s} = \dfrac 1{\sqrt 3} \implies y = \dfrac x{\sqrt 3} - 3s +1$ .

Similarly, the top vertex of the bottom small hexagon is at $B(0,2s-1)$ , and $CB$ is on $y = - \dfrac x{\sqrt 3} + 2s-1$ . Then the $x$ -coordinate of $C$ is $\dfrac {2x}{\sqrt 3} = 5s-2 \implies x = \dfrac {\sqrt 3(5s-2)}2$ . The line $CD$ has a length of $\dfrac {\sqrt 3}2-x$ and this fit in $\dfrac {3\sqrt 3}2s$ . Therefore,

$\begin{aligned} \frac {\sqrt 3}2 - x & = \frac {3\sqrt 3}2s \\ \frac {\sqrt 3}2 -\frac {\sqrt 3(5s-2)}2 & = \frac {3\sqrt 3}2s \\ 1 - 5s + 2 & = 3s \\ \implies s & = \frac 38 = \boxed{0.375} \end{aligned}$

Answer from Figure is 6/16=0.375 For formal proof, write two equations, one for conservation of length and another for area:

(1)3s-b=1, s is side of green hexagons.

(2) Large hexagon area= 5 (green hexagon area of size s)+8 (equilateral triangle of size s)+2* (trapezium smaller)+4*(trapezium larger)

Solve the two equations for s and b.

Answer is s=3/8