regular octagon ABCDEFGH has a side length of 2 cm. Find the area of square ACEG.

Regular octagon $ABCDEFGH$ has a side length of 2 cm.

Find the area of square $ACEG$.

$\text{ Let the side length of the square $=s$ } \\ \text{ Let the area of the square $=s^2$ } \\ \text{ Let $\angle ABC =135^\circ \qquad 135^\circ = 180^\circ-\dfrac{360^\circ}{8}$ }$

cosine formula:

$\begin{array}{|rcll|} \hline s^2 &=& 2^2+2^2-2\cdot 2\cdot 2\cdot \cos(135^\circ) \\ s^2 &=& 8-8\cdot \cos(135^\circ) \\ && \boxed{ \cos(135^\circ) \\ =\cos(180^\circ-45^\circ)\\ = -\cos(45^\circ)\\=-\dfrac{\sqrt{2}}{2} } \\ s^2 &=& 8-8\cdot \left( -\dfrac{\sqrt{2}}{2}\right) \\ \mathbf{s^2} &=& \mathbf{8+4\cdot\sqrt{2}} \\ \mathbf{s^2} &=& \mathbf{13.6568542495\ cm^2} \\ \hline \end{array}$

The area of square $ACEG$ is $\mathbf{\approx 13.7\ cm^2}$

Using the Law of Sines....we have that

(1/2) side length of  the square  / sin (67.5°)   = 2 / sin (90°)          .....multiply both sides by 2.....

side length of the square =  4 sin (67.5°)  =  4 √ [  (1  - cos (135°)  ) / 2 ] =  4 √  [ (1 + √2 /2) / 2 ]  =  4 √  [ (2 + √2)/4 ] 

So....the area of the square is the square of this  =   16 (2 + √2)/4  =  4 (2 + √2) units^2