Difference between revisions of "2012 AMC 10B Problems/Problem 14"
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| + | == Problem == | ||
| + | Two equilateral triangles are contained in square whose side length is <math>2\sqrt 3</math>. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus? | ||
| + | |||
| + | <math> | ||
| + | \text{(A) } \frac{3}{2} | ||
| + | \qquad | ||
| + | \text{(B) } \sqrt 3 | ||
| + | \qquad | ||
| + | \text{(C) } 2\sqrt 2 - 1 | ||
| + | \qquad | ||
| + | \text{(D) } 8\sqrt 3 - 12 | ||
| + | \qquad | ||
| + | \text{(E)} \frac{4\sqrt 3}{3} | ||
| + | </math> | ||
==Solution== | ==Solution== | ||
[[File:2012_AMC-10B-14.jpg]] | [[File:2012_AMC-10B-14.jpg]] | ||
Revision as of 20:20, 8 February 2014
Problem
Two equilateral triangles are contained in square whose side length is 
. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?
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 | ||
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