2020 AMC 10A Problem 20

Place $A=(0,0)$ and $C=(20,0)$. Since $\angle ACD=90^\circ$ and $CD=30$, take $D=(20,30)$. The point $E$ is $(5,0)$, so line $BD$ has equation $y=2(x-5)$. Because $\angle ABC=90^\circ$, point $B$ lies on the circle with diameter $AC$: $(x-10)^2+y^2=100$. Intersecting with $y=2(x-5)$ gives $x=2$ or $10$. The convex quadrilateral uses $B=(2,-6)$. Then $[ACD]=\dfrac12\cdot20\cdot30=300$, and $[ABC]=\dfrac12\cdot20\cdot6=60$. The total area is $360$. Thus, D is the correct answer.