Regular octagon

[Guest]

The size of the regular octagon is 6 units given in a picture. What is the grey area in the picture?

[shyamjayakannan]

First let's find [tex]\angle AHG=\frac{\text{total internal angle}}{8}=\frac{{180}^\circ(8-2)}{8}={135}^\circ[/tex]. So, [tex]\angle AHC={135}^\circ-{90}^\circ={45}^\circ[/tex].
Now, each of the 4 small triangles is an isosceles triangle with each small angle = [tex]{45}^\circ[/tex] and so the height and base are each = [tex]6\sin{{45}^\circ}=\frac{6}{\sqrt2}=3\sqrt2[/tex]. Hence, the area of each small triangle = [tex]\frac{1}{2}\times3\sqrt2\times3\sqrt2=9[/tex] and the area of each rectangle = [tex]6\times3\sqrt2=18\sqrt2[/tex].
So, total area = [tex]4\times9+2\times18\sqrt2=\boxed{36\left(1+\sqrt2\right)}[/tex]