Question 1209185
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<pre>

An obvious trivial observation is that a regular octagon consists of 8 isosceles
congruent triangles with the common center. Each triangle has lateral sides of the length x,
equal to the radius of the circle, and the angle between two lateral sides is 

             360/8 = 45 degrees.


Therefore, the area of each separate triangle is  

    {{{(1/2)*x*x*sin(45^o)}}} = {{{(1/2)x^2*(sqrt(2)/2)}}} = {{{x^2*(sqrt(2)/4)}}}.


Hence, the area of the octagon is 8 times this,  or  {{{2x^2*sqrt(2)}}}.


Then the area of the circle outside the octagon is  

    {{{pi*x^2-2*sqrt(2)*x^2}}} = {{{(pi-2*sqrt(2))*x^2}}}.


It is the expression you want to get.
</pre>

Solved.