Fall 2021 AMC 10 B

Because $\angle RBC$ is complementary to both $\angle RCB$ and $\angle PBA$, those two angles are congruent. Therefore $△BAP\cong △CBQ$ by ASA, so $CQ=BP=13$. Let $d=CR$; then $QR=13-d$, so the Altitude-to-Hypotenuse Theorem yields $6^2=d(13-d)$, which has solutions $d=4$ and $d=9$. Because $QR<RC$, in fact $d=9$. It follows that the area of the square is $$B{C}^2=B{R}^2+R{C}^2=6^2+9^2=117.$$

Let $c=BQ$ and $s=AB$. Because $△BRQ$ is similar to $△BAP$, it follows that $\frac{6}{c}=\frac{s}{13}$, so $cs=78$. As before, $CQ=BP=13$, so by the Pythagorean Theorem, $s^2+c^2=169$. Then $(s+c{)}^2=169+2\cdot 78=325$, so $s+c=5\sqrt{13}$. Similarly $(s-c{)}^2=169-2\cdot 78=13$, so $s-c=\sqrt{13}$. Solving this system of equations yields $s=3\sqrt{13}$, and the area of the square is $s^2=117$.