Answer:
Perimeter of given regular hexagon is 48.5 ft.
Step-by-step explanation:
Let ABCDEF be the regular hexagon as shown in the attached figure.
O be the intersection point of the diagonals EB, FC and AD.
As per the property of regular hexagon, all the 6 triangles formed are equilateral triangles.
In other words,
[tex]\triangle EOD, \triangle DOC, \triangle BOC, \triangle AOB, \triangle AOF , \triangle FOE[/tex] are equilateral [tex]\triangle s[/tex].
Area of an equilateral [tex]\triangle[/tex] is defined as :
[tex]\dfrac{\sqrt{3}}{4} \times a^{2}[/tex]
Where a is the side of [tex]\triangle[/tex].
Area of hexagon = [tex]6 \times \dfrac{\sqrt{3}}{4}\times a^{2}[/tex]
We are given that area of hexagon = 169.74 [tex]ft^{2}[/tex]
Let s be the side of hexagon.
[tex]\Rightarrow 6 \times \dfrac{\sqrt{3}}{4}s^{2} = 169.74 ft^{2}\\\Rightarrow s = 8.08 ft[/tex]
A regular Hexagon is made up of 6 equal sides, so
Perimeter of a regular hexagon = [tex]6 \times side[/tex]
Perimeter = [tex]6 \times 8.08[/tex]
[tex]\Rightarrow 48.5 ft[/tex]
So, perimeter of given regular hexagon is [tex]48.5 ft[/tex].
Answer:
C. 48.5 ft
Step-by-step explanation:
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