2021 Fall AMC 10B Problem 15

Since $BR=6$ and $PR=7,$ we have $BP=13.$ The right-angle and square-angle chasing gives $\triangle PAB\cong\triangle QBC,$ so $CQ=BP=13.$ Because $BP\perp CQ,$ point $R$ is the foot of the altitude from $B$ to the hypotenuse $CQ$ of right triangle $BQC.$ Thus $$QR\cdot RC=BR^2=36$$ and $$QR+RC=CQ=13.$$ So $QR$ and $RC$ are $4$ and $9.$ Because $P$ lies on $\overline{AD},$ we have $AP\le AB.$ Congruence gives $BQ=AP,$ while the right-triangle projection formulas give $BQ^2=CQ\cdot QR$ and $BC^2=CQ\cdot RC.$ Hence $QR\le RC,$ so $QR=4$ and $RC=9.$ Therefore, $$\begin{aligned} BC^2&=BR^2+RC^2\\ &=6^2+9^2\\ &=117. \end{aligned}$$ The area of the square is $117.$ Thus, the answer is D .