Question

The area of the regular hexagon is 169.74 ft2 what is the perimeter, rounded to the nearest tenth

Answer 1

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,

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

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

$\dfrac{\sqrt{3}}{4} \times a^{2}$

Where a is the side of $\triangle$.

Area of hexagon = $6 \times \dfrac{\sqrt{3}}{4}\times a^{2}$

We are given that area of hexagon = 169.74 $ft^{2}$

Let s be the side of hexagon.

$\Rightarrow 6 \times \dfrac{\sqrt{3}}{4}s^{2} = 169.74 ft^{2}\\\Rightarrow s = 8.08 ft$

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

Perimeter of a regular hexagon = $6 \times side$

Perimeter = $6 \times 8.08$

$\Rightarrow 48.5 ft$

So, perimeter of given regular hexagon is $48.5 ft$.