Question 1113612
<br>
Obviously you mean a regular dodecagon; otherwise the problem can't be solved....<br>
Let A, B, C, D, and E be consecutive vertices of the dodecagon; with sides AB and DE on adjacent sides of the square.<br>
Let F be the corner of the square determined by the extension of sides AB and DE of the dodecagon; that is, AF and FE are portions of adjacent sides of the square.<br>
Draw segments from vertex C of the dodecagon parallel to AF and FE, forming a small square in the corner of the big square.  The area outside the dodecagon and inside the square now consists of that small square and two small right triangles.<br>
The exterior angle of a regular dodecagon is 30 degrees; that means the small right triangles are 30-60-90 right triangles.  The hypotenuse of each of those triangles is a side of the dodecagon, with length 2.  Then the lengths of the legs of the triangles are 1 and sqrt(3).<br>
Then by symmetry the full side length of the large square is
{{{1+sqrt(3)+2+sqrt(3)+1 = 4+2*sqrt(3)}}}<br>
Then the area of the square is {{{(4+2*sqrt(3))^2 = 28+16*sqrt(3)}}}<br>
----------------------------------------------------------
<br>
Alternate solution method....<br>
Trigonometry makes this easy.<br>
Let AB be a side of the dodecagon which lies on a side of the square, and consider the triangle OAB where O is the center of the figure (center of both the square and the dodecagon).<br>
If M is the midpoint of AB, then OM is an apothem of the dodecagon; and the length of the apothem is half the side of the square.<br>
Angle AOB is 30 degrees, so angle AOM is 15 degrees.<br>
With AB being 2, AM is 1, so the length of the apothem is {{{cot(15)}}}.<br>
Then the side length of the square is {{{2*cot(15) = 2(2+sqrt(3)) = 4+2*sqrt(3)}}}, as before.