Question

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).

Answer 1

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:

$A=\frac{3\sqrt{3}}{2} a^2$

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:

$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

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:

$d=\sqrt{(8-4)^2+(0-0)^2} =\sqrt{4^2}=4$

$\therefore d=4$

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:

$a_{small}=4-\frac{40}{100}\times 4=2.4$

Thus, the area of the smaller hexagon will be:

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

and the new smaller perimeter will be:

$P_{small}=6a_{small}=6\times 2.4=14.4$ unit

Which are the required answers.