AMC10 2006 10B, Problem 20

In rectangle $ABCD$, we have $A=(6,-22)$, $B=(2006,178)$, $D=(8,y)$, for some integer $y$. What is the area of rectangle $ABCD$?

Solution: This solution is the same as Solution 1 up to the point where we find that $y=-42$. We build right triangles so we can use the Pythagorean Theorem.  The triangle with hypotenuse $AB$ has legs $200$ and $2000$, while the triangle with hypotenuse $AD$ has legs $2$ and $20$.  Aha!  The two triangles are similar by SAS, with one triangle having side lengths $100$ times the other! Let $AD=x$.  Then from our reasoning above, we have $AB=100x$.  Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\fbox{40,400}$.

Answer: 40,400