2019 AMC 10B Problem 10

The area condition fixes the distance from $C$ to line $AB.$ If that distance is $h,$ then $$\frac{10h}{2}=100,$$ so $h=20.$ Thus $C$ must lie on a line parallel to $AB$ at distance $20.$ Since the perpendicular distance from $C$ to line $AB$ is $20$, both $AC$ and $BC$ are at least $20$. They cannot both equal $20$, because that would require the perpendicular from $C$ to meet line $AB$ at both distinct points $A$ and $B$. Hence $AC+BC>40$, contradicting the perimeter requirement $AC+BC=40$. Therefore no point $C$ works. Thus, the answer is A .