Difference between revisions of "2012 AMC 10B Problems/Problem 12"
Flamedragon (talk | contribs) (→Solution) |
|||
Line 15: | Line 15: | ||
<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>. | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{AMC10 box|year=2012|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:16, 8 February 2014
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 , and . Point D is 20 meters due north of point C. The distance AD is between which two integers?
Solution
If point B is due east of point A and point C is due north of point B, is a right angle. And if , is a 45-45-90 triangle. Thus, the lengths of sides , , and are in the ratio , and is .
is clearly a right triangle with on the side . is 20, so .
By the Pythagorean Theorem, .
, and . Thus, must be between and . The answer is .
See Also
2012 AMC 10B (Problems • Answer Key • Resources) | ||
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.