# 2010 AMC 10A Problems/Problem 7

## Problem 7

Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run? $\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ \sqrt{2} \qquad \mathrm{(C)}\ \sqrt{3} \qquad \mathrm{(D)}\ 2 \qquad \mathrm{(E)}\ 2\sqrt{2}$

## Solution

Crystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from traveling North for one mile, and her current destination is $\sqrt{2}$ miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to $\sqrt{((\sqrt{2})^2+1^2)}$, which is equal to $\sqrt{3}$. The answer is $\boxed{C}$ $[asy] import olympiad; draw((0,0)--(0,1)); draw((0,1)--(0,1.7071067811865476), dotted); draw((0,1)--(0.7071067811865476, 1.7071067811865476)); draw((0.7071067811865476, 1.7071067811865476)--(1.4142135623730951,1)); draw(anglemark((0.7071067811865476, 1.7071067811865476),(0,1),(0,1.7071067811865476))); label("45^{\circ}", (0,1.25), NE); draw((0, 1)--(1.4142135623730951,1), dotted); draw((1.4142135623730951,1)--(0,0), green); [/asy]$

~IceMatrix

## See Also

 2010 AMC 10A (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 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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