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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)

1 Answer

4 votes

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:


A=(3√(3))/(2) a^2

Where
A is the area of the regular hexagon


a is the side length of the regular hexagon

Also, it's Perimeter is given by:


P=6a

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:


d=√((x_1-x_2)^2+(y_1-y_2)^2)

Where
d is the distance.


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

Applying the above formula we get:


d=√((8-4)^2+(0-0)^2) =√(4^2)=4


\therefore d=4

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


\therefore d=a=4

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


a_(small)=4-(40)/(100)* 4=2.4

Thus, the area of the smaller hexagon will be:


A_(small)=(3√(3))/(2) a_(small)^2=(3√(3))/(2) (2.4)^2\approx14.96 unit squared

and the new smaller perimeter will be:


P_(small)=6a_(small)=6* 2.4=14.4 unit

Which are the required answers.

Regular hexagon ABCDEF has vertices at A(4, 4!3), B(8, 4!3), C(10, 2!3), D(8, 0), E-example-1
User Alesanco
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