2012 AMC 10B Problems/Problem 12

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Problem

Point B is due east of point A. Point C is due north of point B. The distance between points A and C is $10\sqrt 2$, and $\angle BAC= 45^\circ$. Point D is 20 meters due north of point C. The distance AD is between which two integers?


$\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35$

Solution

[asy] unitsize(4); pair A=(0,0); label ("A",(0,0),W); pair B=(10,0); label ("B",(10,0),E); pair C=(10,10); label ("C",(10,10),E); pair D=(10,30); label ("D",(10,30),E); dot(A); dot(B); dot(C); dot(D); draw(A--B); draw(A--C); draw(A--D); draw(C--D); draw(B--C); [/asy] If point B is due east of point A and point C is due north of point B, $\angle CBA$ is a right angle. And if $\angle BAC = 45^\circ$, $\triangle CBA$ is a 45-45-90 triangle. Thus, the lengths of sides $CB$, $BA$, and $AC$ are in the ratio $1:1:\sqrt 2$, and $CB$ is $10 \sqrt 2 \div \sqrt 2 = 10$.

$\triangle DBA$ is clearly a right triangle with $C$ on the side $DB$. $DC$ is 20, so $DB = DC + CB = 20 + 10 = 30$.

By the Pythagorean Theorem, $DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}$.

$31^2 = 961$, and $32^2 = 1024$. Thus, $\sqrt {1000}$ must be between $31$ and $32$. The answer is $\boxed {B}$.

See Also

2012 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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