2012 AMC 10B Problem 12

We know $AB$ and $BC$ are perpendicular, so $$AB^2 + BC^2 = (10\sqrt 2)^2 = 200.$$ Also, as $\angle BAC = 45^\circ,$ we know that $\triangle ABC$ is an isosceles right triangle, so $AB = BC ,$ making $2AB^2 = 200.$ Thus, $AB = BC = 10.$ As such, we know that $$BD= BC+CD = 30.$$ Thus, by the Pythagorean Theorem, we have that $$\begin{aligned}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{aligned}$$ Thus, since $$31^2 \lt AD^2 \lt 32^2,$$ we have $$31 \lt AD\lt 32$$ Thus, the correct answer is B .