Difference between revisions of "2012 AMC 10B Problems/Problem 14"
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| − | Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length sqrt | + | Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length <math>\sqrt{3}</math> 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 <math>2\sqrt{3} - 2</math>. The formula for the area of an equilateral triangle of length s is <math>\frac{\sqrt{3}}{4}s^2</math>. It follows that the area of the rhombus is: |
| − | 2( | + | <math>2*\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = 8\sqrt{3} - 12</math> |
| + | |||
| + | Thus, answer choice D is correct. | ||
Revision as of 02:44, 8 February 2013
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.
