



 
Hi Colleen, In my diagram $C$ is the center of the hexagon and the square. Hence $CD = 16$ inches and the measure of the three interior angles of the triangle $AEC$ is $60^{o}.$ Hence triangle $AEC$ is an equilateral triangle and the side length of the hexagon is $AE = CA.$ By the symmetry in the diagram angle $BDA$ is a right angle and hence triangle $BDA$ is a right triangle and \[\frac{BD}{DA} = \tan(\left(60^o\right) = \sqrt{3}.\] Thus \[DA = \frac{BD}{\sqrt{3}} = \frac{16}{\sqrt{3}} = 9.24.\] Hence the side length of the hexagon is $CA = CD + DA = 16 + 9.24 = 25.24$ inches. I hope this helps, 



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