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The question is below Any help is appreciated sorry if I did something wrong this is just my first time using this. enter image description here

I tried to do it and I got either $\sqrt{31}$ or $\sqrt{32}$

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  • $\begingroup$ Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. This should be added to the question rather than in the comments. $\endgroup$ – José Carlos Santos Sep 25 '18 at 10:26
  • $\begingroup$ I good either root 31 or root 32 but I am not sure which is correct or if it is completely wrong $\endgroup$ – epix Sep 25 '18 at 10:28
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    $\begingroup$ I've added the image for you, in future however, please type out your question in full and only use images for diagrams as this makes your question searchable in the future $\endgroup$ – lioness99a Sep 25 '18 at 10:30
  • $\begingroup$ How do you get $\sqrt{31}$? Recall that in a regular hexagon, the radius of the circumscribed circle and the side are equal. $\endgroup$ – egreg Sep 25 '18 at 10:32
  • $\begingroup$ Okay thanks I will try it again $\endgroup$ – epix Sep 25 '18 at 10:33
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Well, you need to set up a right triangle, obviously, since you're using Pythagoras theorem.

The segment $OC$ should also be $2$, since $OBC$ forms an equilateral triangle. So, we're seeking to find segment $OV$.

We can figure this out by setting up a right triangle that is $OCV$, where we know two lengths; the base length is $2$, and the hypotenuse is $6$ since that is given to us.

$6^2 = 2^2 + x^2$ can be solved for $x$: $x = \sqrt {32}$, which calculated to $3$ significant digits is $5.66$.

Cheers.

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  • $\begingroup$ Thanks for the help just wanted to check :) $\endgroup$ – epix Sep 25 '18 at 10:40

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