Regular octagon inscribed in a square Problem: The corners of a 2 meter square are cut off to form a regular octagon. What is the length of the sides of the resulting octagon?
From the picture below, the octagon would form a right isosceles, specifically a right isosceles triangle on the corners. The sides of the octagon were set to "x" and the legs of the triangle were set to $\frac{x}{\sqrt{2}}$. Then add the following cuts of a side of the square: $\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 m, which results to x = 0.828 m.
My inquiry is that, from what I know or learned, a right isosceles triangle has an angle ratio of $45-45-90$ and a side ratio of $1-1-\sqrt{2}$ or in algebra: $x-x-x{\sqrt{2}}$. In the problem he set the hypotenuse as $x$ instead and the legs of the triangle as $\frac{x}{\sqrt{2}}$, which I think is fine. But shouldn't setting the hypotenuse as $x\sqrt{2}$ and the sides as $x$ should equal the first equation?
$\frac{x}{\sqrt{2}}$ + x + $\frac{x}{\sqrt{2}}$ = 2 should also equal $x + x\sqrt{2} + x = 2$ where 2 is the length of a side of a square. I don't think multiplying or dividing both sides by $\sqrt{2}$ is the answer as that would not satisfy both equations.
This sounds like an easy problem, but it it's confusing me. Sorry. 

 A: Let $x$ be the length of your octagon (as in the left picture), and $c$ the length cut from one side of the square edge (which is the $x$ in the right picture).
Then you've correctly stated that $x = \sqrt{2}c$. Now you solve
$$
c + x + c = 2.
$$
This is rewritten as
$$
2c + x = 2c + \sqrt{2}c = (2 + \sqrt{2})c = 2.
$$
Thus
$$
c = \frac{2}{2+\sqrt{2}}
$$
so that
$$
x = \frac{2\sqrt{2}}{2 + \sqrt{2}}.
$$
This final fraction is the length of the sides of the octagon.
In the right picture, everything has been scaled up by $\sqrt{2}$ so that the length of the sides of the octagon will be
$$
\sqrt{2} \cdot \frac{2\sqrt{2}}{2 + \sqrt{2}} = \frac{4}{2 + \sqrt{2}}.
$$
This is why they are unequal.

If you were to solve it with $x$ as in the right picture, then you have
$$
2x + \sqrt{2}x = 2
$$
so that
$$
x = \frac{2}{2 + \sqrt{2}},
$$
and then the length of the octagon is
$$
\sqrt{2}x = \frac{2\sqrt{2}}{2 + \sqrt{2}}
$$
exactly as we had calculated above.
Your confusion stems from using $x$ as a label for two different lengths in either diagram.
A: The two equations are not equivalent, and they give different results. In the first case, $x$ is the side of the octagon; in the second one, the side of the octagon is $x\sqrt{2}$.
The first one has the following solution: $x=2(\sqrt{2}-1)$.
The second one has solution $x=2-\sqrt{2}$, and to obtain the side of the octagon, you have to multiply it by $\sqrt{2}$, obtaining the same result.
