imc

Note that $△APB\cong △BQC$ by ASA. ($\angle PAB=\angle QBC=90^{\circ },AB=CB,$ and $\angle PBA=\angle QCB.$) Then, it follows that $\overline{PB}\cong \overline{QC}.$ Thus, $QC=PB=PR+RB=7+6=13.$ Define $x$ to be the length of side $CR,$ then $RQ=13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR\cdot RC=B{R}^2.$ Substituting the given lengths, we have $$(13-x)\cdot x=36.$$ Solving, gives $RQ=4$ and $RC=9.$ We eliminate the possibility of $x=4$ because $RC>QR.$ Thus, the side length of the square, by Pythagorean Theorem, is $$\sqrt{9^2+6^2}=\sqrt{81+36}=\sqrt{117}.$$ Thus, the area of the square is $(\sqrt{117}{)}^2=117,$ so the answer is $\boxed{117}.$ Note that there is another way to prove that $CR=4$ is impossible. If $CR=4,$ then the side length would be $\sqrt{4^2+6^2}=\sqrt{52},$ and the area would be $52,$ but that isn't in the answer choices. Thus, $CR$ must be $9.$ Extra Note: Another way to prove $4$ is impossible. The side length of the square, $S$, is equal to $\sqrt{4^2+6^2}=\sqrt{52}$. Because $x=4$, $RQ=9$. Because $QB=\sqrt{R{B}^2+R{Q}^2}=\sqrt{6^2+9^2}=\sqrt{117}$ and $QB<S$ but $\sqrt{117}>\sqrt{52}$, we have proof by contradiction. And so $x=9$.