2012 AMC 10B Problem 12

1. Draw it. $B$ is due east of $A$ and $C$ is due north of $B$, so the corner at $B$ is a right angle.
2. Triangle $ABC$ has a right angle at $B$ and a $45^\circ$ angle at $A$, so its third angle is also $45^\circ$. It is a $45$-$45$-$90$ triangle, and its two legs are equal.
3. In a $45$-$45$-$90$ triangle, the long side (hypotenuse) is a leg times $\sqrt2$. Here the long side is $AC = 10\sqrt2$, so each leg is $10$: $AB = 10$ and $BC = 10$.
4. $D$ is $20$ more meters north of $C$, on the same north-south line through $B$. So $BD = 10 + 20 = 30$.
5. Triangle $ABD$ has a right angle at $B$, with legs $AB = 10$ and $BD = 30$. By the Pythagorean theorem,
$$AD = \sqrt{10^2 + 30^2} = \sqrt{100 + 900} = \sqrt{1000}.$$
6. To place $\sqrt{1000}$, compare with nearby squares: $31^2 = 961$ and $32^2 = 1024$. Since $961 < 1000 < 1024$, $AD$ is between $31$ and $32$.

The answer is $\boxed{\textbf{(B)}\ 31\ \text{and}\ 32}$.

"Due east" and "due north" always make right angles, so the picture is two right triangles sharing the leg $AB$. The $45^\circ$ angle turns the given hypotenuse $10\sqrt2$ into two legs of $10$ with no extra work. To find which whole numbers a square root sits between, square the nearby whole numbers instead of estimating decimals.

Since $B$ is east of $A$ and $C$ is north of $B$, the angle at $B$ is a right angle. With $\angle BAC = 45^\circ$, triangle $ABC$ is a $45$-$45$-$90$ triangle whose hypotenuse is $AC = 10\sqrt 2$, so both legs are $AB = BC = 10$.

$D$ lies $20$ farther north of $C$ along the same vertical line through $B$, so $BD = 10 + 20 = 30$. Triangle $ABD$ is right-angled at $B$ with legs $10$ and $30$:
$$AD = \sqrt{10^2 + 30^2} = \sqrt{1000}.$$
Since $31^2 = 961 < 1000 < 1024 = 32^2$, the distance $AD$ is between $31$ and $32$.

The answer is $\boxed{\textbf{(B)}\ 31\ \text{and}\ 32}$.