Hexagon ABCDEF

In the regular hexagon ABCDEF, the diagonal AE has a length 8cm. Calculate the circumference and the hexagon area.

Result

o =  27.713 cm
S =  55.426 cm2

Solution:

u=8 cm n=6 A=360/n=360/6=60 ∘  sin⁡A=u/2:a  a=u2/sin⁡(A∘→rad)=u2/sin⁡(A∘⋅ π180 )=82/sin⁡(60∘⋅ 3.1415926180 )=4.6188  o=n⋅ a=6⋅ 4.6188=16 3≐27.7128=27.713  cm u = 8 \ cm \ \\ n = 6 \ \\ A = 360/n = 360/6 = 60 \ ^\circ \ \\ \ \\ \sin A = u/2 : a \ \\ \ \\ a = \dfrac{ u }{ 2 } / \sin ( A ^\circ \rightarrow rad) = \dfrac{ u }{ 2 } / \sin ( A ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ ) = \dfrac{ 8 }{ 2 } / \sin ( 60 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ ) = 4.6188 \ \\ \ \\ o = n \cdot \ a = 6 \cdot \ 4.6188 = 16 \ \sqrt{ 3 } \doteq 27.7128 = 27.713 \ \text{ cm }
S1=a2⋅ 34=4.61882⋅ 34≐9.2376 cm2  S=n⋅ S1=6⋅ 9.2376=32 3≐55.4256=55.426 cm2S_{ 1 } = a^2 \cdot \ \dfrac{ \sqrt{ 3 } }{ 4 } = 4.6188^2 \cdot \ \dfrac{ \sqrt{ 3 } }{ 4 } \doteq 9.2376 \ cm^2 \ \\ \ \\ S = n \cdot \ S_{ 1 } = 6 \cdot \ 9.2376 = 32 \ \sqrt{ 3 } \doteq 55.4256 = 55.426 \ cm^2

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