Difference between revisions of "2012 AMC 10B Problems/Problem 14"
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Revision as of 21:19, 8 February 2014
Solution
Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length 
and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length 
. The formula for the area of an equilateral triangle of length s is 
. It follows that the area of the rhombus is:
Thus, answer choice D is correct.
See Also
| 2012 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
