Can anyone help me with this Pythagoras Question?

Question

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

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

Answer 1

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.