The side of each of the equilateral triangles in the figure is twice the side of the central regular hexagon. what fraction of the total area of the six triangles is the area of the hexagon?

The fraction is 1/4

Step-by-step explanation:

we know that

The area of an equilateral triangle, using the law of sines is equal to

where

x is the length side of the triangle

In this problem

Let

b ----> the length side of the regular hexagon

2b ---> the length side of the equilateral triangle

step 1

Find the area of the six triangles

Multiply the area of one triangle by 6

we have

substitute

step 2

Find the area of the regular hexagon

Remember that, a regular hexagon can be divided into 6 equilateral triangles

so

The area of the regular hexagon is the same that the area of 6 equilateral triangles

we have

substitute

step 3

To find out what fraction of the total area of the six triangles is the area of the hexagon, divide the area of the hexagon by the total area of the six triangles

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet

a

step-by-step explanation:

the answer is a