



 
Hi Rob, In my diagram $w$ is the width of the hexagon (in your case $w = 8$ feet), and $s$ is the side length. Triangle $ABC$ is an equilateral triangle so the measure of angle $DBC$ is $60^{o}.$ The tangent of the angle $DBC$ is $\large \frac{CD}{DB}$ and the tangent of $60^o$ is $\sqrt{3}.$ hence \[\tan(60^{o}) = \sqrt{3} = \frac{CD}{DB} = \frac{w/2}{s/2} = \frac{w}{s}.\] Hence $w = s \sqrt{3}.$ Thus in your situation \[s = \frac{w}{\sqrt{3}} = \frac{8}{\sqrt{3}} = 4.6 \mbox{ feet}\] Penny 



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