Regular hexagon abcdef has vertices at a(4, 4! 3), b(8, 4! 3), c(10, 2! 3), d(8, 0), e(4, 0) and f(2, 2! 3). suppose the sides of the hexagon are reduced by 40% to produce a similar regular hexagon. what are the perimeter and area of the smaller hexagon rounded to the nearest tenth? explain how you found your answer. (psa: " ! " is suppose to be square root sign)

Since the given hexagon is a regular hexagon all it's sides will be of equal length. Now, we know that the Area of any regular hexagon is given by:

Where is the area of the regular hexagon

is the side length of the regular hexagon

Also, it's Perimeter is given by:

Thus, all that we need to do is to find the side length of any one of the sides and to do that let us have a look at at the data of vertices points given and find out which points are definitely adjacent to each other and are also easy to calculate.

A quick search will yield that D(8, 0) and E(4, 0) are definitely adjacent to each other.

Please check the attached file here for a better understanding of the diagram of the original regular hexagon. Points D and E indeed are adjacent to each other.

Let us now find the distance between the points D and E using the distance formula which is as:

Where is the distance.

and are the coordinates of points D and E respectively. (please note that interchanging the values of the coordinates will not alter the distance )

Applying the above formula we get:

We know that this distance is the side length of the given regular hexagon.

Now, if the sides of the given regular polygon are reduced by 40%, then the new length of the sides will be:

Thus, the area of the smaller hexagon will be:

unit squared

and the new smaller perimeter will be:

unit

Which are the required answers.

126.66 m²

step-by-step explanation:

the pertinent area formula is a = πr², where r is the radius.

here, a = π(6.35 m)² = 40.32π, or approximately 126.66 m²