Here is my approach to finding this result. I've just started using Geoalgebra so this was a nice problem to try things in it on, here's the diagram I made for this problem (though it would be identical to hand drawing it in the space the question provides, but for me at least the computer is a lot neater!)
Firstly in an attempt to make this problem simpler I just focused on one quadrant of the octagon since we can find the area of half the quadrant (triangle $QPI$ in the diagram) and multiply that by $8$ to get the area of the entire octagon (by symmetry).
So we can solve this problem by finding the area of the triangle $QPI$ and then multiplying that by 8 to get the area of the entire octagon.
To find angle $\angle RPI$ we can use the tangent function:
$$ \tan{(\angle RPI)}=\frac{HQ}{PQ}=\frac{2}{1} \;\; \therefore \;\; \angle RPI =\tan^{-1}(2)\approx 63.45^\circ$$
Since triangle $PRI$ is right angled we can work out that $\angle PIR$ is:
$$\angle PIR =90^\circ -\tan^{-1}(2)=\tan^{-1}(\frac{1}{2})\approx 90^\circ-63.43^\circ =26.57^\circ \therefore \angle PIR = \tan^{-1}{\left(\frac{1}{2}\right)}$$
Now to find the area of the triangle we need to know the lengths $PR$ and $IR$. A way we can do this is by noting that if we look at triangle $RGI$ that $\angle RGI \equiv \angle PIR$ therefore:
$$ \tan{(\angle RGI)}=\tan{\left(\tan^{-1}{\left(\frac{1}{2}\right)}\right)}=\frac{1}{2}=\frac{2x}{1+x} \implies x=\frac{\frac{1}{2}}{2-\frac{1}{2}}=\frac{1}{3}$$
where $x$ is the length $PR$, the numerator is $2x$ since we know that $IR$ must be twice whatever $PR$ is since in order to make $\tan{(\angle PIR)}$ equal one half (which we know it must since $\angle RGI \equiv \angle PIR$) $\frac{PR}{IR}$ must equal one half hence $IR=2PR$.
Thus with $x$ solved for it's relatively straightforward now since $RQ$ is $2x$ (because triangle $QRI$ is isosceles) we can find the area:
$$ \text{Area of $\frac{1}{8}$ the octagon}=\frac{1}{2}\times 1 \times 2x=x=\frac{1}{3}$$
Now all we need to do it to multiply this by 8 to get the answer for the entire octagon:
$$\text{Area of Octagon}=8\times \frac{1}{3}=\frac{8}{3}\approx 2.67$$
Here is my approach to finding this result. I've just started using Geoalgebra so this was a nice problem to try things in it on, here's the diagram I made for this problem (though it would be identical to hand drawing it in the space the question provides, but for me at least the computer is a lot neater!)
Firstly in an attempt to make this problem simpler I just focused on one quadrant of the octagon since we can find the area of half the quadrant (triangle $QPI$ in the diagram) and multiply that by $8$ to get the area of the entire octagon (by symmetry).
So we can solve this problem by finding the area of the triangle $QPI$ and then multiplying that by 8 to get the area of the entire octagon. To find angle $\angle RPI$ we can use the tangent function: $$ \tan{(\angle RPI)}=\frac{HQ}{PQ}=\frac{2}{1} \;\; \therefore \;\; \angle RPI =\tan^{-1}(2)\approx 63.45^\circ$$ Since triangle $PRI$ is right angled we can work out that $\angle PIR$ is: $$\angle PIR =90^\circ -\tan^{-1}(2)=\tan^{-1}(\frac{1}{2})\approx 90^\circ-63.43^\circ =26.57^\circ \therefore \angle PIR = \tan^{-1}{\left(\frac{1}{2}\right)}$$ Now to find the area of the triangle we need to know the lengths $PR$ and $IR$. A way we can do this is by noting that if we look at triangle $RGI$ that $\angle RGI \equiv \angle PIR$ therefore: $$ \tan{(\angle RGI)}=\tan{\left(\tan^{-1}{\left(\frac{1}{2}\right)}\right)}=\frac{1}{2}=\frac{2x}{1+x} \implies x=\frac{\frac{1}{2}}{2-\frac{1}{2}}=\frac{1}{3}$$ where $x$ is the length $PR$, the numerator is $2x$ since we know that $IR$ must be twice whatever $PR$ is since in order to make $\tan{(\angle PIR)}$ equal one half (which we know it must since $\angle RGI \equiv \angle PIR$) $\frac{PR}{IR}$ must equal one half hence $IR=2PR$.
Thus with $x$ solved for it's relatively straightforward now since $RQ$ is $2x$ (because triangle $QRI$ is isosceles) we can find the area: $$ \text{Area of $\frac{1}{8}$ the octagon}=\frac{1}{2}\times 1 \times 2x=x=\frac{1}{3}$$ Now all we need to do it to multiply this by 8 to get the answer for the entire octagon: $$\text{Area of Octagon}=8\times \frac{1}{3}=\frac{8}{3}\approx 2.67$$