2006 AMC 10B Problem 20

1. The trip from $A$ to $B$. Going from $A = (6, -22)$ to $B = (2006, 178)$ is $2000$ steps right and $200$ steps up. That is the same direction as $20$ right and $2$ up, just $100$ times longer.
2. The trip from $A$ to $D$. Side $AD$ makes a right angle with $AB$ at the corner $A$. A direction at a right angle to "$20$ right, $2$ up" is "$2$ right, $20$ down": swap the two numbers and flip one sign. (Check with slopes: $AB$ has slope $\frac{2}{20} = \frac{1}{10}$, and $AD$ has slope $\frac{-20}{2} = -10$. Their product is $-1$, so they are at a right angle.)
3. $D$ has $x$-coordinate $8$, which is $2$ right of $A$. So the trip from $A$ to $D$ is exactly $2$ right and $20$ down, and $y = -22 - 20 = -42$.
4. Side lengths. By the Pythagorean theorem,
$$AD = \sqrt{2^2 + 20^2} = \sqrt{404}.$$
Side $AB$ is the same shape of trip, $100$ times as long, so $AB = 100\sqrt{404}$.
5. Area.
$$AB \cdot AD = 100\sqrt{404} \cdot \sqrt{404} = 100 \cdot 404 = 40{,}400.$$

The answer is $\boxed{\textbf{(E)}\ 40,400}$.

On a grid, a right-angle turn of the trip "$a$ right, $b$ up" is "$b$ right, $a$ down" (or the reverse). Here that makes the two sides of the rectangle the same trip at two sizes, so the square roots multiply away. Before using the distance formula twice, check whether the two sides are scaled copies of each other.

With slopes only: $AB$ has slope $\frac{200}{2000} = \frac{1}{10}$, so $AD$ has slope $-10$. Then $\frac{y + 22}{8 - 6} = -10$ gives $y = -42$. The lengths are $AB = \sqrt{2000^2 + 200^2} = 200\sqrt{101}$ and $AD = \sqrt{2^2 + 20^2} = 2\sqrt{101}$, and the area is $400 \cdot 101 = 40{,}400$.