Question 1209347
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Here is another way to solve this problem, this time getting an exact answer.<br>
Add 45-45-90 right triangles to alternating sides of the regular octagon to form a square.<br>
The hypotenuses of those triangles are edges of the regular octagon, so they each have edge length 1;  and so their legs all have length {{{sqrt(2)/2}}}<br>
The side length of the square is then {{{1+2(sqrt(2)/2)=1+sqrt(2)}}}.<br>
Put the four added triangles together with their right angles at a common point to see that the combined area of the four triangles is the area of a square with side length 1.<br>
Then the area of the regular octagon is the area of a square with side length {{{1+sqrt(2)}}}, minus the are of a square with side length 1:<br>
{{{(1+sqrt(2))^2-1^2=(1+2sqrt(2)+2)-1=2+2sqrt(2)}}}<br>
ANSWER: {{{2+2sqrt(2)}}} <<== typo corrected thanks to note from tutor @ikleyn<br>
(which to several decimal places is equal to the answer obtained by the other tutor, 4.828417...)<br>
NOTE: This problem shows a formula that is familiar to many geometry students who participate in math contests: the area of a regular octagon with side length s is<br>
{{{A=s^2(2+2sqrt(2))}}} <<== typo corrected thanks to note from tutor @ikleyn<br>