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Solution:
\[Area = side^2 = AB^2\]
ABD is isosceles triangle;
:- AD = BD
By Pythagoras theorem:
\[AD = r\sqrt2=BD\]
\[\Rightarrow BC = r(2-\sqrt2)\]
In triangle ABO:
\[\Rightarrow BO = r(\sqrt2-1)\]
and
\[AO = r\]
Now,
\[AB^2 = AO^2+BO^2\]
\[\Rightarrow AB^2 = r^2+r^2(\sqrt2-1)^2\]
\[\Rightarrow AB^2 = r^2(4+2\sqrt2)\]
\[\Rightarrow Area = r^2(4+2\sqrt2)\]
