2006 AMC 10B Problem 20

The displacement from $A$ to $B$ is $(2006 - 6,\ 178 - (-22)) = (2000, 200) = 100 \cdot (20, 2)$.

Side $AD$ is perpendicular to $AB$ at $A$. A vector perpendicular to $(20, 2)$ is $(2, -20)$ (or its negative), obtained by swapping the coordinates and flipping one sign. Since $D$ has $x$-coordinate $8 = 6 + 2$, the displacement from $A$ to $D$ has first coordinate $2$, so it is exactly $(2, -20)$ and $y = -22 - 20 = -42$.

Now $|AB| = 100\sqrt{20^2 + 2^2} = 100\sqrt{404}$ and $|AD| = \sqrt{404}$. The rectangle's area is
$$|AB| \cdot |AD| = 100 \cdot 404 = 40{,}400.$$

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

In coordinates, "perpendicular" is cleanest as a vector statement: $(a, b) \perp (-b, a)$. Here $AD$ turns out to be a scaled copy of $AB$ rotated by $90^\circ$, so the two side lengths share the factor $\sqrt{404}$ and the square root disappears from the area. Look for such proportional right triangles before reaching for the distance formula twice.

Slopes: $AB$ has slope $\frac{200}{2000} = \frac{1}{10}$, so $AD$ has slope $-10$, and $\frac{y + 22}{2} = -10$ gives $y = -42$. Then $|AB| = 200\sqrt{101}$, $|AD| = 2\sqrt{101}$, and the product is $400 \cdot 101 = 40400$.

• Getting the sign of y wrong (y = 2 instead of -42) by using slope +10 for the perpendicular, or computing the diagonal instead of two sides.
• Dropping a factor of $10$ somewhere and answering $4040$, which is offered as a decoy.