2006 AMC 10B Problem 20

The slope of $\overline{AB}$ is $\tfrac{178-(-22)}{2006-6}=\tfrac{200}{2000}=\tfrac1{10}.$ Since $\overline{AD}\perp\overline{AB},$ its slope is $-10,$ so $\tfrac{y+22}{8-6}=-10$ gives $y=-42.$ Then $AB=\sqrt{2000^2+200^2}$ $=200\sqrt{101}$ and $AD=\sqrt{2^2+20^2}=2\sqrt{101}.$ The area is $200\sqrt{101}\cdot2\sqrt{101}$ $=400\cdot101$ $=40{,}400.$ Thus, the correct answer is E.