1. ## equilateral hexagon

Find an example of an equilateral hexagon whose sides are all $\sqrt{13}\$ units long. Give coordinates for all six points.

( $\sqrt{13}\$/2 , 3)
(2+ $\sqrt{13}\$/2 , 0 )
( $\sqrt{13}\$/2 , -3 )
(- $\sqrt{13}\$/2 , 3)
(- $\sqrt{13}\$/2, -3)
(-2- $\sqrt{13}\$/2 , 0 )

2. You tell us. Is the distance the same between each consecutive pair of points? Where is it centered? What was your plan design?

Mathematics is not for guessing. Prove it!

3. I didn't check your solution...(sorry)
But I have some approach to solve the problem...

I think the easiest way of solving this problem will be to solve this complex equation:

$z^6=\sqrt{13}$

correct me if I wrong...

4. That would be a great way, but specific to the hexagon that can be constructed with equilateral traingles.

More generally, any regular n-gon could be done with a slight variation. Just a little scaling would be required.

5. I see my answer is correct. The origin is (0.0) and the distance between the consecutive coordinates is root 13.

6. Hi Veronica,
You need to mark off 60,120,180, 240,300,360 angles on your rad 13 radius. Draw the defining right triangle for each angle. Review the properties of 30-60-90 triangles. Proceed to id each angle x y coordinates

bjh

7. Hello, Veronica1999!

Find an example of a regular hexagon whose sides are all $\sqrt{13}$ units long.
Give coordinates for all six vertices.

A regular hexagon is comprised of six equilateral triangles.
Code:
          C           B
* - - - - - *
/ \         / \
/   \       /   \
/     \     /     \
/       \   /       \
/         \ /         \
D * - - - - - * - - - - - * A
\         /O\         /
\       /   \       /
\     /     \     /
\   /       \   /
\ /         \ /
* - - - - - *
E           F

Each line segment has length $\sqrt{13}.$

Let the center of the hexagon $O$ be at the Origin.

Then $A$ and $D$ are at: . $\left(\pm\sqrt{13},\:0\right)$

The $x$-coordinate of $B$ is $\frac{1}{2}\sqrt{13}$

The $y$-coordinate of $B$ is the altitude of $\Delta ABO.$
. . which is: . $\left(\frac{1}{2}\sqrt{13}\right)\left(\sqrt{3}\ri ght) \:=\:\frac{1}{2}\sqrt{39}$

Then $B,C,E,F$ are at: . $\left(\pm\frac{1}{2}\sqrt{13},\:\pm\frac{1}{2}\sqr t{39}\right)$

8. That's the spirit!! There are other variations of solutions, as indicated by other posters, but find those that fit your style. Generally, you would want to stick with more general solution types unless you have a specific application that can be benefitted from a more specific solution.