2 Answers

Samura · Answer 1 · 2021-03-01T10:29:53+0000

Answer:

Perimeter of given regular hexagon is 48.5 ft.

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,

$\triangle EOD, \triangle DOC, \triangle BOC, \triangle AOB, \triangle AOF , \triangle FOE$ are equilateral
$\triangle s$ .

Area of an equilateral
$\triangle$ is defined as :

(√(3))/(4) * a^(2)

Where a is the side of
$\triangle$ .

Area of hexagon =
6 * (√(3))/(4)* a^(2)

We are given that area of hexagon = 169.74
ft^(2)

Let s be the side of hexagon.

$\Rightarrow 6 * (√(3))/(4)s^(2) = 169.74 ft^(2)\\\Rightarrow s = 8.08 ft$

A regular Hexagon is made up of 6 equal sides, so

Perimeter of a regular hexagon =
6 * side

Perimeter =
6 * 8.08

$\Rightarrow 48.5 ft$

So, perimeter of given regular hexagon is
48.5 ft .

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