2010 AMC 10A Problem 7

Below is the professionally curated solution for Problem 7 of the 2010 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 10A solutions, or check the answer key.

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Concepts:Pythagorean Theoremvector

Difficulty rating: 1220

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?

11

2\sqrt{2}

3\sqrt{3}

22

222\sqrt{2}

Solution:

From the diagram, we see that the distance traveled is the hypotenuse of a right triangle.

One of the legs is just 11 from running due north. The other leg is 12+12=2. \sqrt{1^2 + 1^2} = \sqrt2.

The final distance is then 22+12=3. \sqrt{\sqrt2^2 + 1^2} = \sqrt3.

Thus, C is the correct answer.

Problem 7 in Other Years