Question 1113612
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Go to Wikipedia and look in this its article https://en.wikipedia.org/wiki/Dodecagon 

talking about dodecagon (it is better one time to see than 100 times to hear).


Dodecagon has 12 sides. We consider a REGULAR dodecagon.


Look in the figure in this article.


In my solution, I consider the scheme when <U>4 vertices of the regular dodecagon coincide with the midpoints of square sides</U>.


I will work with 3 sides of the regular dodecagon that are located in QIII.


These three sides are sloped at 15°, 45° and 75° to the horizontal line.


The sides are of 2 units long each.


It means that half of the square side length is this sum


{{{a/2}}} = 2*cos(15°) + 2*cos(45°) + 2*cos(75°) = {{{2*cos(pi/12) + 2*cos(pi/4) + 2*cos(5pi/12)}}} = 3.866.


Hence, the square side length = 2*3.866 = 7.731.


Then the square area = {{{a^2}}} = {{{7.731^2}}} = 59.772.


<U>Answer</U>.  The square area = 59.772 square units (approximately).
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Notice that in his solution @greneestamps considers different scheme of the regular dodecagon inscribed to the square.