Difference between revisions of "2005 AMC 10A Problems/Problem 12"
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<math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math> | <math>4\cdot\frac{60}{360}\cdot\pi\cdot1^2 = \frac{4}{6}\cdot\pi = \frac{2}{3}\pi \Rightarrow B </math> | ||
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Revision as of 13:34, 13 August 2019
Problem
The figure shown is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length 
?
Solution
The area of the trefoil is equal to the area of a small equilateral triangle plus the area of four 
sectors with a radius of 
minus the area of a small equilateral triangle.
This is equivalent to the area of four sectors with a radius of 
.
So the answer is:
See also
| 2005 AMC 10A (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. 
