A geometry problem by Yahia El Haw

A regular octagon is composed of $8$ congruent isosceles triangles. From the diagram, the included angle between the two equal sides is $\dfrac{360}{8}=45^\circ$ . For the area of the triangle, we use the formula $A=\dfrac{1}{2}ab \sin C$ . So the area of the octagon is $8\left(\dfrac{1}{2}\right)(r^2)(\sin 45)=4r^2\left(\dfrac{\sqrt{2}}{2}\right)=2r^2\sqrt{2}$ .

$\large \mu \left( \widehat { AOB } \right) =\frac { \pi }{ 4 }$

$\large \sigma \left[ AOB \right] =\frac { { r }^{ 2 }\sin { \cfrac { \pi }{ 4 } } }{ 2 } =\frac { { r }^{ 2 }\sin { \cfrac { \sqrt { 2 } }{ 2 } } }{ 2 } =\frac { { r }^{ 2 }\sqrt { 2 } }{ 4 }$

$\large \sigma \left[ orthogon \right] =8\cdot \frac { { r }^{ 2 }\sqrt { 2 } }{ 4 } =2\sqrt { 2{ r }^{ 2 } }$