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)

Respuesta :

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:

[tex] A=\frac{3\sqrt{3}}{2} a^2 [/tex]

Where [tex] A [/tex] is the area of the regular hexagon

[tex] a [/tex] is the side length of the regular hexagon

Also, it's Perimeter is given by:

[tex] P=6a [/tex]

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:

[tex] d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} [/tex]

Where [tex] d [/tex] is the distance.

[tex] (x_1,y_1) [/tex] and [tex] (x_2,y_2) [/tex] are the coordinates of points D and E respectively. (please note that interchanging the values of the coordinates will not alter the distance [tex] d [/tex])

Applying the above formula we get:

[tex] d=\sqrt{(8-4)^2+(0-0)^2} =\sqrt{4^2}=4 [/tex]

[tex] \therefore d=4 [/tex]

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

[tex] \therefore d=a=4 [/tex]

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

[tex] a_{small}=4-\frac{40}{100}\times 4=2.4 [/tex]

Thus, the area of the smaller hexagon will be:

[tex] A_{small}=\frac{3\sqrt{3}}{2} a_{small}^2=\frac{3\sqrt{3}}{2} (2.4)^2\approx14.96 [/tex] unit squared

and the new smaller perimeter will be:

[tex] P_{small}=6a_{small}=6\times 2.4=14.4 [/tex] unit

Which are the required answers.

Ver imagen Vespertilio