2020 AMC 10B Problem 8

Place $P=(-4,0)$ and $Q=(4,0).$ Since the area is $12$ and $PQ=8,$ the distance from $R$ to line $PQ$ is $3,$ so $R=(x,\pm 3).$ If the right angle is at $P,$ then $R=(-4,\pm 3),$ giving $2$ points. If it is at $Q,$ then $R=(4,\pm 3),$ giving $2$ more points. If the right angle is at $R,$ then $PQ$ is the hypotenuse, so $$\begin{aligned} &64=PR^2+QR^2 \\ &\quad =(x+4)^2+9 \\ &\quad {}+(x-4)^2+9 \\ &\quad =2x^2+50. \end{aligned}$$ Thus $x^2=7,$ giving $R=(\pm\sqrt7,\pm 3),$ another $4$ points. The total is $2+2+4=8.$ Thus, the correct answer is D .