Hexagon Strip

A regular hexagon is composed of $6$ equilateral triangles. The area of one equilateral triangle of side $s$ is $\dfrac{s^2\sqrt{3}}{4}$ . We assume that the length of one side of hexagon is $2$ . Then the area of the hexagon is

$A_{hex}=(6)\dfrac{2^2\sqrt{3}}{4}=6\sqrt{3}$

Now, by cosine law

$x^2=2^2+2^2-2(2)(2)\cos 60$ $\implies$ $x=2\sqrt{3}$

The area of the shaded region is

$A_s=1(2\sqrt{3})=2\sqrt{3}$

Finally,

$\dfrac{A_s}{A_{hex}}=\dfrac{2\sqrt{3}}{6\sqrt{3}}=$ $\color{#D61F06}\boxed{\dfrac{1}{3}}$

If we draw a diagonal of the green parallegram, we will dividing it into two parts, each is $\frac{1}{6}$ of the the original hexagon, thus, the area of the green part is $\frac{1}{6}+\frac{1}{6}= \frac{1}{3}$ .

If we draw a line to split the hexagon in half between two vertices and label the outer edge B2 and the dividing line parallel to it as B1 forming two trapezoids with height h, we can find that the total area of the hexagon is h(B1+B2). By splitting the shaded area into two congruent triangles, we can find the total shaded area to be B2h. This gives us the ratio of shaded area:total area=[B2h]/[h(B1+B2)]=B2/[B1+B2]. Since this is a regular polygon, we know that B1=2*B2 giving the ratio of 1:3.

The parallogram shaded has the same area as a rectangle. Two rectangles form the midsection and the two outer triangles are each 1/2 of a triangle. Thats a total of 3 rectangles. Therefore 1/3

Sides of the hexagon = 2.
The perpendicular of the equilateral triangles= P.
Draw the line from the middle of the bottom line on the octagon M through the center O to the middle of the top line of the octagon T. The shaded area now is divided into two identical triangles with common properties 2p, 1, 60 , 30
Shaded area = 2( 2P * 1/2) = 2Psq
The area of the hexagon is six triangles with base 2, height P.
6( p* 2/2) = 6Psq.
I have not found how to write notation nor diagrams on the laptop I am using.
It is a simple construction

I held my finger up that best represented the colored in area and then covered the 2 un colored sides and when given the choices of coverage deduced 1/3 waa the closest to my visual estimate.....1/3 is also the amt of time it took to figure it out compared to the amt of time to type the answer.

2x(2apx(L/2)/2)= apL, the area of the hexagon is 6Lap/2= 3Lap...... apL/3apL= 1/3