Question 1209347
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We cut a regular octagon ABCDEFGH out of a piece of cardboard.  
If AB = 1, then what is the area of the octagon?
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            Here is another way to solve the problem.



<pre>
Let R be the radius of the circumscribed circle around our octagon.


Let's find its radius via the side length s.


The octagon consists of 8 congruent isosceles triangles, having 
the common vertex in the center.


Each such a triangle is an isosceles triangle with the lateral sides of the length R
and the angle at the vertex of 45°.  Write the cosine rule equation for such a triangle 

    {{{s^2}}} = {{{R^2}}} + {{{R^2}}} - {{{2*R*R*cos(45^o)}}},

    {{{s^2}}} = {{{2R^2*(1-cos(45^o))}}},

    {{{R^2}}} = {{{s^2/(2*(1-cos(45^o)))}}}.    (1)


In our case with s = 1, the last formula takes the form

    {{{R^2}}} = {{{1/(2*(1-cos(45^o)))}}} = {{{1/(2*(1-sqrt(2)/2)))}}} = {{{2/(2*(2-sqrt(2)))}}} = {{{1/(2-sqrt(2))}}} = {{{(2+sqrt(2))/(4-2)}}} = {{{(2+sqrt(2))/2}}}.    (2)


Now the area of one such a triangle is

    {{{area[triangle]}}} = {{{(1/2)*R*R*sin(45^o)}}} = {{{(1/2)*R^2*sin(45^o)}}} = {{{(1/2)*((2+sqrt(2))/2)*(sqrt(2)/2)}}} = {{{(2*sqrt(2)+2)/8}}} = {{{(sqrt(2)+1)/4}}}.    (3)


For the area of the entire octagon, we should take the quantity (3)  8 (eight) times to get

    {{{area[octagon]}}} = {{{2 + 2*sqrt(2)}}}  square units.    <U>ANSWER</U>
</pre>

Solved.