2020 AMC 12A Problem 18

Place $A = (0,0)$ and $C = (20, 0).$ Since $\angle ACD = 90^\circ,$ $D = (20, 30),$ and $E = (5, 0)$ because $AE = 5.$ Since $\angle ABC = 90^\circ,$ $B$ lies on the circle of radius $10$ centered at $(10, 0).$ Line $DE$ is $(5 + t,\, 2t);$ substituting gives $t^2 - 2t - 15 = 0,$ so $t = 5$ or $t = -3.$ For $E$ to lie between $B$ and $D,$ take $t = -3,$ giving $B = (2, -6),$ a distance $6$ below line $AC.$ Then $[ACD] = \tfrac12 \cdot 20 \cdot 30 = 300$ and $[ABC] = \tfrac12 \cdot 20 \cdot 6 = 60,$ so the total area is $360.$ Thus, D is the correct answer.