2012 AMC 10B Problems/Problem 14
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 is . It follows that the area of the rhombus is:
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.