check the picture below.
so, notice, if we use one of those triangles in the octagon, since a circle has 360°, therefore 360/8 will give us 45° for each of those triangles at the center, as shown in your picture.
using the 45-45-90 rule, we can get the altitude of the angle, whose base is "r" btw, and since we have the altitude and base for it, we can simply use those to get the area of one triangle,
[tex]\bf \textit{area of a triangle}\\\\
A=\cfrac{1}{2}bh\qquad
\begin{cases}
b=r\\
h=\frac{r\sqrt{2}}{2}
\end{cases}\implies A=\cfrac{1}{2}(r)\left( \frac{r\sqrt{2}}{2} \right)\implies A=\cfrac{r^2\sqrt{2}}{4}
\\\\\\
\textit{and now for all 8 triangles that'd be }8\left(\cfrac{r^2\sqrt{2}}{4} \right)\implies 2r^2\sqrt{2}[/tex]
bear in mind that, the length of the radius is just "r", thus the area will be in "r" terms.
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on the one for the BC segment, check the second picture below.
using the 30-60-90 rule, we get the segment DC, which is the same for both triangles, and then we simply use the pythagorean theorem to find the hypotenues of the second triangle, which happens to be the segment BC.
