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Difference between revisions of "2012 AMC 10B Problems/Problem 12"
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<math>31^2 = 961</math>, and <math>32^2 = 1024</math>. Thus, <math>\sqrt {1000}</math> must be between <math>31</math> and <math>32</math>. The answer is <math>\boxed {B}</math>.
 
<math>31^2 = 961</math>, and <math>32^2 = 1024</math>. Thus, <math>\sqrt {1000}</math> must be between <math>31</math> and <math>32</math>. The answer is <math>\boxed {B}</math>.
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Revision as of 13:15, 4 July 2013

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

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 $\boxed {B}$. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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