Question 1008122
C=length of diagonal, A=B=lengths of sides:
{{{A^2+B^2=C^2}}} Pythagorean theorem 
{{{A^2+A^2=C^2}}}
{{{2A^2=C^2}}}
{{{2A^2=(12in)^2}}}
{{{2A^2=144in^2}}}
{{{A^2=72in^2}}}
{{{sqrt(A^2)=sqrt(72in^2)}}}
{{{A=8.485}}}
Since the two diagonals meet at the back center of 
the original square, the side of the square
is 2 times the length of the sides of the triangle:
Side of square=2A
Side of square=2(8.485)
Side of square=16.97 inches
The player's "Encyclopedia" is wrong:
The official rules call for a square 17 inches with
diagonals that meet in the back center starting 8.5 inches
from the front of the plate. The sides of the triangles 
would be 8.5 inches, making the diagonal:
{{{sqrt(2(8.5^2))=12.02in}}} 
or 12.02 inches, not 12 inches