Question 1209920
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When the two squares overlap to form a regular octagon, the regions inside the squares and outside the octagon are eight isosceles right triangles.<br>
Let x be the side length of each of those triangles; then x*sqrt(2) is the length of the hypotenuse, which is the side length of the octagon.<br>
The side length of each square is then 2x+x*sqrt(2).  Since the side length of the square is 1+sqrt(2),<br>
{{{2x+x*sqrt(2)=1+sqrt(2)}}}<br>
{{{x(2+sqrt(2))=1+sqrt(2)}}}<br>
{{{x=(1+sqrt(2))/(2+sqrt(2))=((1+sqrt(2))(2-sqrt(2)))/((2+sqrt(2))(2-sqrt(2)))=sqrt(2)/2}}}<br>
The side length of the octagon, x*sqrt(2), is then<br>
{{{(sqrt(2)/2)(sqrt(2))=1}}}<br>
The area of a regular octagon with side length s is<br>
{{{A=2s^2(1+sqrt(2))}}}<br>
The side length of our octagon is 1, so the area is<br>
ANSWER: {{{2(1+sqrt(2))}}}<br>